elementary-number-theory

[wanton viper]

so the edges of the polygon are made by this equation? or am i being lost again

[torn escarp]

the reason the coefficients pop out is due to the p-adic absolute value's competitivity

the edges not necessarily

it might help to show me a concrete example you're looking at and confused with

[wanton viper]

its with puiseux-series, but it should the same idea, the roots of P(y) have the valuation of 1 or -1/3 and i dont understand how exactly these values show up as the edges

[torn escarp]

I can show you how to derive the slope if that's what you want to see too

yeah sure

it has 3 roots of valuation -1/3 and one root of valuation 1

I don't know what you mean about "edges" exactly

[wanton viper]

i got the hang of how to use the polygon, i want to understand it better

[torn escarp]

like where the slope changes or the slopes themselves

maybe I should just walk through how I derive the newton polygon, maybe that helps?

[wanton viper]

sure

[torn escarp]

I take a polynomial and imagine I'm in the algebraic closure so I have all the roots

so I have say, the root x, with f(x)=0 and I see that 0 = |f(x)| = |a_nx^n + ... +a_0|

|.| is your ultrametric absolute value here

|x| = e^{- ord(x)} to be clear

so the ultrametric absolute value has a specific property that if $|c_1+c_2+\cdots+c_n|=0$ then it must be that there's an $i \ne j$ that makes $|c_i|=|c_j|$, so we can prove that real quick

[sterile pumice]

Merosity

[torn escarp]

assume not, then pick the maximum $|c_i|>|c_j|$ for all $j \ne i$, then $$0=|c_1+c_2+\cdots+c_n|=|c_i| \ne 0$$

[sterile pumice]

Merosity

[torn escarp]

because the strongest win properties of the ultrametric absolute value forces for |x|>|y| that |x+y|=|x|

that gave us a contradiction, so that means some |c_i|=|c_j|, we can then just use that with the terms of the polynomial

$$|a_ix^i| = |a_jx^j|$$

[sterile pumice]

Merosity

[torn escarp]

rearranging this gives you the slope formula, maybe writing it in terms of the order would help

or maybe this doesn't feel deep enough to see it pop out of the algebra after this point, and that's just unsatisfying to you

I've said a lot of stuff so probably I lost you at some point too, so feel free to ask

[wanton viper]

i got this for once, i have the slope formula. maybe im just stupid and dont see it. im going to experiment a little more on paper, but you gave me a couple of ideas

thx :D

[torn escarp]

yeah, good luck, feel free to ping me if you have more questions

[wanton viper]

alright

[torn escarp]

there's more to say, you could look at Alain M. Robert's A Course in p-adic Analysis chapter 6 section 1.6 for some more info

[wanton viper]

im still a little lost.
where is the context between the slope (e.g. -3, for i=0 and j=1) and the geometric shape.
I draw the polygon as the lower half of the convex hull of the points (i, v(a_i)).
the values for i and j only work if they are "on" a line of the polygon

mainly its this part here, why must (i, v(a_i)) and (j, v(a_j)) belong to an edge of the newton polygon

@torn escarp

[torn escarp]

@wanton viper that was part of this argument here, which led to two terms being equal, you can think of the "smaller terms" are the ones that get thrown up above the newton polygon, and aren't part of the convex hull

another book to look at would be Gouvea's p-adic numbers book he goes in depth too, but

to see that the valuation lies on the newton polygon, that's all here to do with the ultrametric absolute value's properties, I think it's probably the main thing that's confusing, the fact that |a|>|b| implies |a+b|=|a| when you might suspect the weaker statements |x+y|<=|x|+|y| or |x+y|<=max(|x|,|y|), we actually get equality

[wanton viper]

i think i start to understand it now. any slope between point a and b would give me a result that equals $i \cdot v(y_0) + v(a_i) = j \cdot v(y_0) + v(a_j)$ but for the requirement that $i \cdot v(y_0) + a(a_i)$ is also the minimum for all i, the slope has to be on the edge of the polygon, where it is the steepest it could get

[sterile pumice]

[Hyp] Deutschland

[wanton viper]

@torn escarp

[native monolith]

surely the hint should contain mod 7 not mod 6??

then you get $10^4\equiv 1 \mod{7}$ and $10^100=10^{98}\times 2^2 \times 5^2$ so the power is a multiple of 4 so the answer is thursday

[sterile pumice]

Max..

[native monolith]

but i dont get why they used mod 6 in the hint

7 days means we need 7 least residues, mod 6 has only 6 least residues

[crimson forum]

hi i have a question how would one write x^2+y^2+x=c where c<0 as the sum of squares??

[torn escarp]

complete the square of x^2+x like the last problem

[crimson forum]

(x+1/2)^2+y^2-1/4=c but then the problem would be we move 1/4 to the other side and wed get (x+1/2)+y^2=c+1/4

my q is "observe that the polynomial can be represented as a sum of squares of polynomials. Conclude that the equation has no real solutions and therefore no integer solutions." so if i have -1/4 when completing the square i cant conclude that x+1/2)^2+y^2-1/4 is >0 if that makes sense

because of the -1/4

but i have no idea on any other way of making it sum of sqs?

[torn escarp]

well c+1/4 is a sum of squares so it must be >=0

so c+1/4 >=0 makes -1/4 <= c < 0, about as good as you can do

[crimson forum]

hmm but then im not sure i fully understand becuase arent we tryna make contradiction that c>=0 but we are told c<0 ==> no int sols

[torn escarp]

if c>=-1/4 it has real solutions, no amount of factoring can change that

[crimson forum]

so it a trick q?

[torn escarp]

well, is there any more information

if c is an integer that would be enough to solve it this way, but for all I know it's a rational number

since the interval [-1/4,0) contains no integers

actually I don't think that's enough, I'm distracted but

maybe there's a typo or some information you left out on accident maybe? I'm in a call right now sorry

[crimson forum]

hmm lemme show u the problem

theres no notes for this but then im thinking itd just make the 4th one also wrong

[crimson forum]

alright no worries if u in call

but yh thats the thing theres no proper notes at all on this eithr which is annoying

[crimson forum]

ok i think im just overthinking this a lot

im just gnna go with c being an integer and gna use what u sed about completing square then (x+1/2)^2+y^2>=1/4 so (x+1/2)^2+y^2-1/4 must also be >=0 if we rearrange as that works n makes sense

Thanks so much for the help and time too @torn escarp 🙂

[torn escarp]

yeah you're welcome, glad to hear you sorted it out

[minor nexus]

Given that n=p_1*...p_t, t is even.
Could someone help me with this? Seems that the answer is 0, but I can't strictly prove it. Tried saying that sqrt n > p_1...p_t. If I prove, that there is no dividers of n between sqrt n and p_1...*p_t, then the answer is easy to get, but Its not true

[torn escarp]

do the primes need to be distinct in n? for instance, n=4=2*2 gives you mu(1)=1

[minor nexus]

Yeah, sorry for the late answer. p_i != p_j, i != j

[minor nexus]

Okey, I solved it

[vernal void]

Hey guys!

Any tips on starting this guy?

[dusty dragon]

Note that aa’ = 1 (mod p), so 1 = (1/p) = (aa’/p) = (a/p)(a’/p)

Multiply both sides by (a’/p) and use the fact that (a’^2/p) = 1 to get the result

[runic token]

wait what does it mean with the parentheses

[dusty dragon]

It's the Legendre symbol

[ \left(\frac{a}{p}\right) = \begin{cases}0 & \text{if } p \mid a, \ 1 & \text{if } a \text{ is a non-zero quadratic residue mod }p, \ -1 & \text{if } a \text{ is a non-zero quadratic non-residue mod }p.\end{cases}]

[vernal void]

cvisnt this enough?

oh wait

[dusty dragon]

Indeed

Once you're there, you're effectively done

[ 1 = \left(\frac{a}{p}\right)\left(\frac{a'}{p}\right) \implies \left(\frac{a'}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{a'^2}{p}\right) = \left(\frac{a}{p}\right)]

[runic token]

oh, neat

[vernal void]

@dusty dragon thank you.

Any ideas for this one?

[torn escarp]

what kind of stuff have you been learning about in class, trying to not use crazy ideas to solve this if I can help it

[vernal void]

basically like up to this stuff

we're on pythagorean triples rn!

[torn escarp]

maybe try translating f(x-b/(2a))

idk, I think there's some easy trick to do this problem, I'm a bit busy right now and tomorrow but maybe I'll find time to think about it if no one else pops in

[vernal void]

@dusty dragon any idea ❤️

[ivory fable]

Yo guys

isn’t time always the input?

independent

just hear me out in this one

[fluid steppe]

that doesnt sound like number theory

[ivory fable]

well where can i find theory

or do i just hover to help

[torn escarp]

try #math-discussion I guess, but what you're saying sounds wrong

[fluid steppe]

maths is theory

[ivory fable]

I remember seeing a channel with math theories

I’ll just go to help

[fierce knoll]

no you don't even neet time on your axis's, for example y = 2x where x is apples and y is money, there isn't any time in this equasion

[verbal cedar]

say b and d are integers, b>0, and d is fixed such that 0≤d≤10. if b divides both d and 10-d, then what can I say about b?

[dusty dragon]

If b divides d and 10 - d, then b divides 10. To see this, note that 10 - d = bk for some integer k. So 10 = d + bk = bm + bk for some integer m. In other words, 10 = b(m + k) => b | 10

Since b > 0, this means that b \in {1, 2, 5, 10}

@verbal cedar in case you still needed it

[verbal cedar]

ah ofc. that's so obvious now that i see it lol

[dusty dragon]

it happens lol

I sometimes forget that gcd(a, b) = d implies that d | a and d | b

[verbal cedar]

is there a way to know which numbers will have a square root in a modular ring? say for Z/nZ. are there tricks that work for certain kinds of n? such as if n is prime?

[dusty dragon]

I guess another way to phrase your question is to determine which numbers are quadratic residues? If x \equiv y^2 (mod n), then x has a square root y in Z_n?

[dark flax]

what grade is this for?

[north pagoda]

as suggested by this being in the "early university" section, the main target audience of this section is university-level elementary number theory questions

but if youre a high school student and have a number theoretic question, you can ask it here

(if it doesn't fit, we'll redirect you)

[torn escarp]

factor n into powers of primes mod p^k, then you can reduce to the prime mod p and solve there, then hensel lift back up to p^k, then combine with the chinese remainder theorem.

if you're just checking if it has a square root, you just need to reduce to the prime and show that it satisfies the condition of hensel's lemma (which is a sufficient, but not necessary, there's a bit more general version of hensel's lemma that you can use)

[onyx rock]

hi i am new

k

[gray sequoia]

How should I study number theory for olympiad, if I can't solve most of the problems?

I'm reading modern olympiad number theory, but the book is getting harder and harder, almost can't keep up with the pace

[sacred junco]

What does this notation mean?

[dusty dragon]

It doesn't look like standard notation; card(...) is sometimes used to signify the cardinality of a set, but more context is probably needed to make out what the rest means

[north pagoda]

i would wager its meant to be the cardinality of the set of functions ℚ → ℝ?

usually thatd be denoted ℝ^ℚ, not ^ℚ ℝ

but same gist

[flint schooner]

help i need to solve and reduce A, B and C

[onyx granite]

Is the sieve of eratosthenes still relevant?

[north pagoda]

this is not number theory, try #prealg-and-algebra or a help channel

[north pagoda]

define "relevant"

[onyx granite]

We can use programming tools to find if a number is prime or not.

[north pagoda]

the sieve of eratosthenes is definitely not the fastest way to determine primality - its very bad at that

it also isnt a particularly fast way to determine the nth prime (the reason it exists), but it isnt that bad either

its merits are being simple and intuitive

[onyx granite]

I meant to ask if we still use it in today's world of mathematics

[north pagoda]

it is used as an example

it is not used, like, algorithmically

besides maybe in teaching intro programming students how to use loops

[onyx granite]

Hmm

[north pagoda]

that said, sieves ARE used in proofs

the most famous example here is probably euler's proof of the riemann zeta primes identity

https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function#Proof_of_the_Euler_product_formula

you wont find too many opportunities to prove things using the sieve of eratosthenes specifically

but sieving in general is a powerful technique in number theory

the modern "optimized" version of the sieve of eratosthenes is https://en.wikipedia.org/wiki/Sieve_of_Atkin

[onyx granite]

That gives a lot of insights, very helpful, ty!

[north pagoda]

i say "optimized" but this is only asymptotically

for non-astronomical inputs the sieve of eratosthenes is faster

[onyx granite]

What are some nice applications of number theory other than cryptography?

Is it more inclined to being 'pure mathematics'?

[north pagoda]

it's very pure, yes.

indeed, prior to the invention of cryptography, it was seen by many mathematicians as the "purest" field

turns out that cryptography is actually very useful

and a lot of developments in number theory have inspired developments in other stuff that turn out to be useful

(one can draw a direct lineage from early analytic number theory to most of modern mathematical physics, for example)

(because it inspired a lot of the development of complex analysis and differential geometry)

but number theory itself isn't all that applicable.

[onyx granite]

[inner anchor]

also in terms of theoretical interest, the sieve of eratosthenes-legendre, which is just a formalized version used to simply count the number of primes instead of determining primality, is a good starting point into more powerful sieve methods

[shy cedar]

Guys any book recommendation to start number theory my own?

Qualifications: almost a highschool passout

[sacred junco]

did you check #books-old

[stiff rivet]

i like silverman's friendly introduction to number theory

[shy cedar]

Oh I didn't, will

[shy cedar]

Thanks will check it out for sure

[shut scaffold]

If I am to perform a calculation with "3-digit chopping", 22/7 becomes 3.14 or 3.142? I think it's 3.14, but I'm sanity checking myself. 🙂

[brisk skiff]

wdym by 3 digit chopping

[shut scaffold]

It's out of my Numerical Analysis book. Truncation vs rounding.

[brisk skiff]

oh ok nvm

[runic token]

legal pdf 👍

wait

fuck

wrong book lmfao

[uneven granite]

How do this

[west sun]

Does this set notation describe all positive even numbers? If not, how is it wrong?

[lilac elk]

in order for x/2 to belong to N. x/2 must be an integer, let's say k

x/2=k, then x=2k. which is an even number as long as k is an N number

[weary patio]

Hi there, I have a question, but I'm not sure it fits in here directly (since I'm not professionally trained), please redirect me if it isn't appropriate (trying to avoid posting in adv)! My question concerns the argument for claiming that "there aren't enough names for the reals, since countable infinite text (names) with a finite alphabet can't be mapped onto the reals (via diagonal argument)." While I see that much ink has been spilled about that already, I would like to ask for help pointing me in the right direction to answer this question. In many discussion the argument quickly changes to definability or computability. Where can I look up how to properly answer this question (possibly that giving names necessarily leads to definability) and the connection between these topics (name, model & proof theory, definability). Thanks in advance!

[torn escarp]

I think this is more a #proofs-and-logic type of thing, you might get better reception there, or they might forward you somewhere better.

[weary patio]

@torn escarp thanks a lot!

[untold moat]

i have to prove that this will never give me a prime number , n is Natural number

how do i do it ?

[buoyant mason]

try simplifying it and factoring it

[untold moat]

n*11+55

and then what ?

[buoyant mason]

how could you factor it?

[untold moat]

idk

hhhhhhh

please

[buoyant mason]

you are overthinking it if you cannot figure it out

[untold moat]

i know it look simpel but like i have no ideas come to my head if how i should do it

[buoyant mason]

what are the factors of 55?

[untold moat]

5*11

[buoyant mason]

ok sure

n*11+ 5*11

any factorization now?

[untold moat]

11(n+5)

you mean

[buoyant mason]

yep

[ruby cloak]

someone help

[untold moat]

what should i do after that hhhhhhhhhhhhhhhhhhhhhh

[ruby cloak]

my hw

[buoyant mason]

the problem is done lol

11(n+5) is not prime for any n

[untold moat]

but how

this could be not it right

[buoyant mason]

11 times something greater than 1 is not prime

[untold moat]

ohhhhhhhhhhhhhhhhh

i see thank you

[buoyant mason]

yep ^-^

[karmic notch]

I also like this book, and A Book of Abstract Algebra because of the short chapters that introduce one idea at a time and exercises related to those ideas that were introduced. I'm wondering if there are other books like these two that I could learn from, but for other topics like combinatorics and such.

[torn escarp]

the question seems incomplete, you refer to 'x' but never use i

[full thicket]

nvm i have solved it

[dark zealot]

if a man has to work 5 days in 31 days, without working 2 consecutive days, how many arrangements are possible for his days of work?

[dusty dragon]

it might be easier to find the number of arrangements where the man has to work 2 consecutive days

[dark zealot]

i tried

too difficult

[dusty dragon]

Think of it similar to how you would solve the problem where two people have to sit next to each other

(i.e. start by fixing the 2 consecutive days and then figure out how many arrangements you have to fit the remaining 3 days)

[wind glacier]

How can we find all positive integer solutions of the equation $2(x^3-x)=y^3-y$

[sterile pumice]

Amorous aka Lucifer

[sacred junco]

is analysis number theory?

[midnight forge]

No

[midnight forge]

Try factoring both expressions and reducing modulo 6?

Idk if it’ll work but just an idea

[wind glacier]

Im not able to do ,much even after this

[serene socket]

It's an elliptic curve as verified by Sage

[runic token]

maybe do $2x(x^2-1)=y(y^2-1)$ and then break these up into smaller factors and use coprimality to set different factors equal to each other to get solutions?

[sterile pumice]

valley

[runic token]

not sure if it'll work, js a thought

[next abyss]

i have a set with two elements {a,b} and im looking for a set thats something like a hybrid between the powerset and all possible permutations of a set. id like to theoretically have all arrangements (permutations) of elements, such that:

{
  {a}, {b}, all "1-element-permutations"
  {a,a}, {a,b}, {b,a}, {b,b}, all 2-element-permutations
  {a,a,a}, {a,a,b}, {a,b,a}, {a,b,b}, {b, a, a}, ...,  "3 element permutations"
  {a,a,a,a}, {a,a,a,b}, ....,  "4 element permutations"
  ... up to k-element permutations
}

whats the mathematical operation im looking for here?

or structure?

[buoyant mason]

maybe something about permutations of multisets?

[next abyss]

thanks, i didnt find out but figured out how to implement it in python!

[buoyant mason]

ah great ^-^

[coarse carbon]

Idk if its correct in english but when they ask u to define a function explicitly what do they mean?

I have this problem

And they ask u to define f^-1 explicitly

And this is the solution and i didnt get why the teacher use a b c and why is he using the congruence system

Do we need like the direct image of the x’s for it to be defined explicitly is that what we should do?

[harsh solstice]

Define a function explicitly means to write down the output you get explicitly , in this case we know that there is a isomorphism f^- , we just didnt write it down explicitly as opposed to f which is clearly defined as f(x+165Z)=(x+3Z,x+5Z,x+11Z) <-- (we have an explicit formula for the output in terms of x)

notice f^- is from (Z3* ,Z5* , Z11* ) to (Z165*) meaning we take 3 elements from each set (in this case we call them a+3Z ,b+5Z and c+11Z)
and send them to (x+165Z)
or more concretely f^-(a+3Z,b+5Z,c+11Z)= x+165Z , now we insert f on both sides and we get f(f^-(a+3Z,b+5Z,c+11Z))=f(x+165Z)
since f is bijective we get (a+3Z,b+5Z,c+11Z)=f(x+165Z) =(x+3Z,x+5Z,x+11Z)

notice that a+3Z=x+3Z implies a and x are in the same congruency class modulu 3 , and thats where we get x cong a (mod3)

[coarse carbon]

Aw man

Thanks

That was a really good explanation

Btw i saw u like measure theory i have a class and im struggling in it can i ask u later if im stuck on something? Were just having basic stuff on it

[harsh solstice]

ye of course i can try to help if i can ,altho i have started learning it not so long go lol ,currently working on outer measures and caratheodory's criterion proof

[coarse carbon]

We're just having really basic stuff

So can I dm u when I need something with it?

[drifting imp]

Have I done this right?

i put triple no for injective,surjective and bijective since the relation R is not a function ?

actually mbe it is functional

and surjective?

[harsh solstice]

i Think its a better idea to post in #advanced-analysis , there is plenty people here more competent than me in measure theory , but i usually do lurk on the analysis channels so ile notice if you post something that i can help with.

[coarse carbon]

Alright thank you 💕

[marble tusk]

does there exist a quick way to compute

$$\sum_{d=1}^{d=n} n%d$$

[sterile pumice]

yoohoo

[marble tusk]

like a closed from, or an almost closed form

<@&286206848099549185>

[inner anchor]

by the division algorithm, we have $n \pmod{d} = n - d \left\lfloor \frac{n}{d}\right\rfloor$

[sterile pumice]

Pappa

[inner anchor]

http://oeis.org/search?q=1,1,4,3,8,8&language=english&go=Search

you can find some asymptotics here

[dense scroll]

Does anyone have a hint?

[dusty dragon]

If d | (3 * 10^499), then d | (3 * 2^499 * 5^499). You can now independently choose the number of 2's, 3's, and 5's to appear in your factorisation

[dense scroll]

Thanks!

[dense scroll]

Does anyone have a hint for doing this induction? I can't figure it out

[dusty dragon]

Note that $a_{n + 1} = 2^{2^{n + 1}} - 1 = (2^{2^n})^2 - 1$. This is just a difference of two squares, one of its factors being exactly $a_n$.

[ashen gull]

This may be a stupid question, but do the relative prime numbers of a number n generate the prime numbers as n tends to infinity?

[next abyss]

for reference, it was the "kleene star" i was looking for:

https://en.wikipedia.org/wiki/Kleene_star

[unkempt void]

Not really sure you mean by "generate the prime numbers"

You'll always omit all multiples of n for any n

And whenever you get an even number then only odd numbers can be coprime to it and so on

[buoyant mason]

ohh cool 🙂

[wind glacier]

Given that $\phi=\frac{1+\sqrt{5}}{2}$. Let $$n=\frac{1}{1}+\frac{1}{1+\phi}+\frac{1}{1+\phi+\phi^2}+\frac{1}{1+\phi+\phi^2+\phi^3}+\dots$$
The value of $\lfloor2n\rfloor+\lceil2n\rceil$ is $\dots$

[sterile pumice]

Amorous aka Lucifer

[wind glacier]

Please anyone help

[willow steppe]

guys i am in 7th grade and i dont understand any of this help pwease

[stoic bear]

Think about bounding it somehow

[still flare]

Welcome

[fiery zenith]

https://i.imgur.com/X8yKmPk.png

Hint 🙏, I've seen the answer before but forgot

iirc the general topic was determinants

if we set $f(x) = lhs - rhs$, we get $f(x) = (x-1)^2(x^4+2x^3+4x^2+2x+1)$

[fiery zenith]

but i dont think thats the way

ah found it, blumming am-gm

[fervent grove]

for the record, this could actually work

note that we must have x>0 for a solution

but then if x =/= 1, we see that (x-1)^2>0 and (x^4+2x^3+4x^2+2x+1) > 0, so f(x) > 0

so the only solution is x=1

[fiery zenith]

How do you get the 2nd inequality with the quartic

[fervent grove]

x > 0 and all the terms have positive coefficients

[fiery zenith]

🤔

[fervent grove]

did I misread something

[fiery zenith]

im thinking

right x > 0 is necessary (missed that earlier). nice one

(and yh, Q shouldve stated x in R)

[fair skiff]

so x>=0 obv has no solutions (except x=1), to show the same with x<0:
x^4+2x^3+4x^2+2x+1=(x+1)^4-2(x^3+x^2+x)=(x+1)^4-2x(x^2+x+1)

x^2+x+1>0 for Vx, but when x<0, -2x(x^2+x+1)>0, and we’re done

oh wait x>0 is necessary

Lol

[crisp finch]

Its necessary becsuse in the original equation the LHS is strictly positive.

[fair skiff]

yeah whoops

all that work for nothing lolz

[crisp finch]

Its fine

[fair skiff]

happens all the time 😩 sometimes i delve too deeply and miss a trivial insight

[crisp finch]

A nice tip before doing algebraic manipulations is trying to input 0 and 1

and then notice degrees

The degree on LHS is gigantic while RHS is half that

so if we know that they are equal at 1 then we can assume they dont intersect again after that.

the way i can justify this is sorta like this

we know the degree of LHS is 6 so we can say that the equation is approximately x^6+C=4x^3

[fair skiff]

it definitely simplifies the problem + provides groundwork to a potential solution

[crisp finch]

yeah how much time did you have for the problem?

[fair skiff]

i think i spent like 5 minutes on it, did it while eating breakfast

i just saw it randomly and decided to give my thoughts

but as you can see, i missed the big picture ;-;

[crisp finch]

no its no big picture thing

its fine rly

can I give u an equation?

[fair skiff]

okay maybe big picture wasn't the right phrase - i missed a simpler approach to prove that x > 0 is necessary

[fair skiff]

for what reason?

[crisp finch]

(x^2y^2+x-1)=3y-1

idk why not

what are the solutions

[fair skiff]

x^2y^2+x-1=3y-1?

[crisp finch]

yea lol

[fair skiff]

there are infinitely many solutions, do you want a general form or smth

[crisp finch]

yes

[fair skiff]

i think i got

$y\neq{0}, x=\frac{-1\pm\sqrt{12y^3+1}}{2y^2}$

[sterile pumice]

EzMoneyBois

[crisp finch]

yeah this is right

[tacit heart]

Why is this true $$\frac{2015^{2017} + 2^{2017} - 2017}{2017}\equiv -1 + \sum_{k = 0}^{2017}(-1)^k(-2)^{k}2^{2017 - k} \pmod{2017}$$

[sterile pumice]

kappa

[tacit heart]

or what formula is this

idrk where to put this so tell me if this is misplaced

[fair skiff]

consider $2015^{2017}=(2017-2)^{2017}$

[sterile pumice]

EzMoneyBois

[fair skiff]

@tacit heart

[stoic bear]

I think -1 from -2017/2017 and then consider the binomial formula expansion of something related to the remaining expression

[tacit heart]

Hey

[tacit heart]

Thanks

[tacit heart]

Nvm I don’t understand, could you elaborate

[fair skiff]

$2015^{2017}+2^{2017}-2017

=(2017-2)^{2017}+2^{2017}-2017

=\left(\sum_{i=0}^{2017}(2017)^i\cdot(-2)^{2017-i}\cdot\binom{2017}{i}\right)+2^{2017}-2017

=\left(\sum_{i=1}^{2017}(2017)^i\cdot(-2)^{2017-i}\cdot\binom{2017}{i}\right)+(-2)^{2017}+2^{2017}-2017$

[sterile pumice]

EzMoneyBois
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fair skiff]

can you finish from there?

[short lion]

math

[tacit heart]

Yea thanks

[leaden swan]

<@&268886789983436800>

[paper lion]

Thanks

[lean path]

what is the 100s digit of 44^44

[knotty anchor]

you posted in the other thread. modulo 100 and you can just use eluers totient theorem to simplify

[knotty anchor]

Is this from OTIS?

[tacit heart]

You were also curious about the summation so I asked. I am kn07

[opal dome]

I'm uncomfortable with this justification for the set of rationals being countable because it can be shown that a bijection exists between the natural numbers and the grid of p/q for all p, q in the integers

Namely, if there's redundancy in this construction, how is there a bijection?

lol like believe they're countable ... but this argument with the redundancies has me feeling 😕

[night quail]

you could say that the above shows Zx(Z/{0}) is countable, and you can construct Q to be a set of equivalence classes of Zx(Z/{0}) for the relation (n,m) ~ (p,q) <=> mq = np, and so should also be countable

well the above actually shows NxN is countable, but you could build the same argument for Zx(Z/{0}) im sure

[opal dome]

sorry for being slow to respond, im still relatively new to a lot of this jargon so brb as i figure out what equivalence classes are lol

[night quail]

it's essentially saying "get rid of all the redundancies" like 1/2 and 2/4

the idea is that Q will then be a "subset" of Zx(Z/{0}), and since we showed Zx(Z/{0}) is countable, Q must be as well

[opal dome]

ah thanks for the high level strategy

[night quail]

note that the construction of Q above is not actually a subset of Zx(Z/{0}), but it's... isomorphic (to a subset)?

[torn escarp]

I suppose one way to think of it is, since N is in Q you have |N| <= |Q|, but this picture shows |Q| <= |N| so putting those together |N|=|Q|

[buoyant mason]

@flint ivy i didn't know where to put this since the channel closed but for your problem here i was thinking of completing the square twice to rewrite it as 6(4x+1)^2 - (12y+2)^2 = 2, then this is almost an equation of form 6a^2 - b^2 = 2

and if you know pell equation stuff that can be solved

but the solutions will still be pretty complicated

and only describing the ones with a = 1 mod 4 and b = 2 mod 12 i think will make it even more complicated 🥹

[fervent grove]

I mean (a,b) = (1,2) is a solution
the rest of the solutions for (a,b) you can find by multiplying (sqrt(6)+2) by some powers of the fundamental unit in Z[sqrt(6)] as in normal Pell equation theory; I think the fundamental solution is (a,b) = (2,5)
if you actually write out what your solutions for (a,b) are, then you're going to get some recurrences to generate your solutions. to add congruence conditions on (a,b) to solve for (x,y), one should be able to "solve these recurrences" modulo an integer because they're periodic modulo any integer, so you can just pick out the indices you want

this is probably not the most efficient way to solve this, but it'll work

[west glade]

I mean you can just jump over any repeated numbers. f(1)=1/1, f(2)=2/1, f(3)=1/2, f(4)=1/3 and then f(5)=3/1. you'll obviously still hit all of them

[pine badge]

Hello, I am not formally trained in math. Is there a name for generating all of the factors of a given number based on that number's prime factors?

[silver oak]

well no?

there's no short equivalent for "generating all of the factors of a given number based on that number's prime factors"

[pine badge]

Thanks, not necessarily a shortcut, but just a name for the process.

[silver oak]

that's what i mean?

there's no other way to say it

well, you're looking for a powerset of a multiset, but it's barely a synonym

[languid spade]

Is there a formal name for the field created by taking the remainder of an irreducible polynomial in another field? More formally, the field $\mathbb{G}=\mathbb{F}[x]/p$ where $p(x)$ is an irreducible polynomial in $\mathbb{F}[x]$.

[sterile pumice]

TheUnTamed

[still flare]

Yes, the sometimes-used name for the specific construction where we quotient out by the ideal generated by an irreducible polynomial is called Kronecker's construction.

These days I don't think people would think of it as any more special than any other quotient of a ring. (except of course that it is a maximal ideal)

[languid spade]

Thanks!

[coarse carbon]

And this is my work

The result is weird

I tried to use fermat's little theorem

I factorised 5642 into something ×16

Which is 17-1

Idk if its the right way

But at the end I found that the remainder is 0

[torn escarp]

there's a trick for divisibility by 4 by checking the last two digits is divisible by 4, since 5642 = 56*100 + 42 we know 100 is divisible by 4, but 42 is not divisible by 4. That means you have made some mistake here at least.

[coarse carbon]

I found that 5642= 352×16+10

And 1035125 is congruent 12 mod 17

So I tried to use the fermat's little theorem

But I think it doesn't work here

[pine badge]

What are power sets and multisets ?

[odd quest]

Not sure this goes here but is anyone aware of why you couldn’t take a harmonic mean of any real numbers that aren’t zero as long as the sums dont cancel out?

[runic token]

my fav proof that q has the same cardinality as n uses the cailkin wilf tree (i spelled that wrong!) because it's v pretty

[runic token]

oh fuck i js looked it up i was rly close to the name

[torn escarp]

seems like the wrong channel, also should be no reason that wouldn't work. Might be easier to say if you describe what you're actually looking at though.

[odd quest]

Yea wasn’t sure if this actually had some more complex to do with it? Just looking at hippocampus volume rates over time

[torn escarp]

well, what's the specific problem you're running into

the only reason you'd be unable to do it is if you're dividing by 0

I think usually the values are positive in order to be meaningful, so I don't think that should be an issue in practice

[odd quest]

Ok just wanted to see if I was going crazy every “blog”/se question was like you can’t use negative numbers and I couldn’t see a good reason for it in an interpretation level

[torn escarp]

you should have just asked this from the start

[flint ivy]

hi uh what's pell's equation?

[buoyant mason]

it's just a type of diophantine equation

https://en.wikipedia.org/wiki/Pell's_equation

so what i was trying to say was your equation is almost of that form

after some manipulation

and also you probably can't get around doing something like this, the solutions to your equation look like pell equation solutions

[flint ivy]

ohh

got it

thx

[buoyant mason]

yep ^-^

[dense scroll]

Does anyone have a hint for c? I can't find a pattern/integers for which the statement should hold

[torn escarp]

odd+even is odd

[dense scroll]

Thanks!

Would you also happen to have a hint for this one?

[torn escarp]

show that it's 0 mod 2, mod 3, and mod 7 individually, since 42=2*3*7

[dense scroll]

ye so I wanted to do it using CRT but we're technically not 'at' that point yet, so I want to see how to solve this without CRT

[fervent crest]

This isn’t really crt

Think of it as 2, 3, 7 all divide the number, hence since these are distinct primes, x_n is a multiple of 42

If you don’t believe that try to show that if a | n and b | n where a, b are coprime then ab | n

[slate juniper]

what makes a equation diophantine?

[leaden swan]

Restrict solutions to Z and you have a diophantine equation

[slate juniper]

ah okay

[fervent crest]

Thus it's diophantine if you make it diophantine

[slate juniper]

huh I wonder if there's a diophantine analogue for differential equations, I guess you cant since you dont really have an analogue of Z for a frechet space

[slate juniper]

Wouldn't it be more accurate to call the solutions diophantine rather than the equation? because it's just any algebraic equation no?

diophantine solutions

[leaden swan]

You mean discrete analog?

[slate juniper]

no like diophantine differential equations but I guess you cant really have those (or rather that doesnt really make sense)

[elder rain]

Does anyone have their own set of rules when it comes to numerical notation?

[sleek spire]

how do I show that ${2^n \choose j}$ for $1 \leq j \leq 2^n$ is divisible by $2$?

[sterile pumice]

pure for president

[sleek spire]

do I just literally expand it

and show that there's a 2 in the top number

[fervent crest]

One thing you can do is consider $(x+1)^{2^n}$ modulo 2

[sterile pumice]

Music major

[fervent crest]

Now expand this using the fact that you’re in modulo 2, and expand it using binomial theorem

[prime sparrow]

Here is a combinatorial proof:
${2^{n} \choose j}$ counts the number of ways of choosing j subsets of an n element set. For each such choice there is a complementary choice of j subsets by choosing the complements of each set. This mapping shows that the number is divisible by two.

[sterile pumice]

chmonkeynumber1fan

[fervent crest]

What if you have A B where A B are disjoint and union to the n elements, then the mapping will not give you a different collection of subsets though

[fervent grove]

(you can find a few proofs of a stronger statement here: https://en.wikipedia.org/wiki/Lucas's_theorem. there is also a proof by bashing everything out with computations of nu_p(n!), if I remember correctly)

(in particular, there is a combinatorial proof there)

[sleek spire]

(this should've been j < 2^n btw)

[sleek spire]

this seems kinda ironic to me lol

cuz what I'm tryna do is show that $x^{2^n} + 1$ is irreducible in $\bZ[x]$

[sterile pumice]

pure for president

[sleek spire]

I used eisenstein's criterion and am almost done but just have to show that this holds

[sleek spire]

but yeah can't I just do this lol

,, {2^n \choose j} = \frac{(2^n)!}{j!(2^n - j)!} = \frac{2^n (2^n - 1)!}{j!(2^n - j)!} = 2 \frac{2^{n - 1} (2^n - 1)!}{j!(2^n - j)!}

[sterile pumice]

pure for president

[fervent crest]

Ok so what you do is, notice that $(x+y)^2\equiv x^2+y^2\pmod{2}$, so $(x+1)^{2^n}\equiv x^{2^n}+1\pmod{2}$. Then by binomial theorem, $(x+1)^{2^n}=\sum_{i=0}^{2^n}\binom{2^n}{i}x^i$, so $x^{2^n}+1\equiv \sum_{i=0}^{2^n}\binom{2^n}{i}x^i\pmod{2}$, by comparing coefficients you see that $\binom{2^n}{i}\equiv0\pmod{2}$ when $1\leq i\leq 2^n-1$

[sterile pumice]

Whoever

[sleek spire]

is this wrong tho?

[fervent grove]

you should explain why the number of powers of 2 in your numerator is at least in the denominator

namely, for 2^{n-1}(2^n-1)! / (j! * (2^n-j)!)

[sleek spire]

wut?

[fervent grove]

namely, why is this an integer

[unkempt void]

Nice method whoever

[fervent crest]

Thx potato

[torn escarp]

nice indeed

[torn escarp]

$$(1+x)^{z+p^n} = (1+x)^z (1+x^{p^n})$$
Compare coefficients on the binomial expansion of $x^i$ for $i<p^n$
$$\binom{z+p^n}{i} = \binom{z}{i} \mod p$$
It's periodic with period $p^n$, take $z=0$ for the special case.

[sterile pumice]

Merosity

[fervent crest]

Ohh ok

[torn escarp]

extra credit: show this forms a basis for p^n periodic maps

extra extra credit: prove Mahler's theorem

[fervent crest]

Ok buddy

[fervent crest]

Tbh this just feels like a corollary of lucas

[torn escarp]

Mahler series are kind of an analogy for fourier series in the p-adics

[torn escarp]

yeah you might be right, if I recall lucas's thm is the same except you decompose the number into base p so that all the digits come out, which is more of a hassle

like the same in terms of doing it by generating functions

[white lion]

the first is asking for the sum of the divisors whilst the second one is asking for the number of divisors.

$3718 = 2^1 x 11^1 x 13^12$

thats it's canonical form

so the second equation J(3718) it is [multiplication notation] (a_i+1)

from i=1 to r

so what I got was (a_1+1)(a_2+1)(a_3+1)(a_4+1)(a_5+1)(a_6+1) = (1+1)(0+1)(0+1)(0+1)(1+1)(2+1)= (4)(3)=12

am i correcr?

[white lion]

$3718 = (2^1)(11^1)(13^2)$

[sterile pumice]

jayz

[white lion]

This is its canonical form, and by the formula that tells us the number of divisors

[sterile pumice]

jayz

[white lion]

I got 12 is my anwser correct?

[night perch]

what is step 1

[low vapor]

may i please get some help with this?

[torn escarp]

Hint: the first and last number are of opposite parity

[low vapor]

so one of those expressions has to be 2?

[torn escarp]

Exactly😎

[mint coral]

How to find relative distance from 8

[upper gale]

Hi!
(it is in French, sry), question (1), I need to prove that U with multiplication of complex numbers is a finite group (hope it is the right term)

I did something but I'm stuck to prove that U is a finite set.

z^n = 1 can be written like this: exp( 2k*i*pi /n), for all 0 <= k <= n-1

[sacred junco]

do you know the fundamental theorem of algebra every polynomial of degree n with complex coefficient have exactly n roots

z^n - 1 = 0 has n solutions

$e^{\frac{2k\pi}{n}}, 0\leq k\leq n-1 $ the number of elements in that group is the number of k values ${0,1,\cdots,n-1}$

[sterile pumice]

Amer

[upper gale]

Yes, I saw it during my first year of university

So, do I need to say something like that?
| U | = n, because [insert fundamental theorem of algebra]
=> U is a finite set => (U, x) is a finite group

[sacred junco]

After saying it is a finite set show it is a group with multiplication

If you multiply any two elements in U it shoud stay there

And it has the identity element

[upper gale]

Got it!

[signal cedar]

How can I see that $\sum _{n=1}^\infty\frac{\mu (n) d(n)}{n^s }=\prod _p (1- \frac{2 }{p^s})$?

[sterile pumice]

bert

[unkempt void]

I'd not say it's Gauss lemma, more just a general algebraic thing

Not sure how much algebra you know, but the point is that the map Z -> Z/pZ gives us a map Z[x] -> (Z/pZ)[x]

(These are both ring homomorphisms - that is, they are compatible with addition and multiplication (and send 1 to 1 etc))

[torn escarp]

I try to think backwards from any given n, what terms contribute to it from the product? In particular if n is not squarefree, then it won't appear in the sum as we expect since mu(n)=0. So now suppose that n is squarefree. You get a -2 for each prime factor, the -1 is what you normally see in the mobius function and the 2 is the same as counting that 1 and p are prime factors.

[torn escarp]

I feel like there's a simpler statement, like: suppose f has no root mod n, but f has a root r in Z. Since f(r)=0, then f(r)=0 mod n, contradiction.

[sacred junco]

Logic

[unkempt void]

Okay, so another way of saying this is that evaluating a polynomial and then taking it mod p is equivalent to taking it mod p and evaluating

So if f in Z[x] has a root, so does its image in (Z/pZ)[x]

[torn escarp]

Your phi is not well defined at x=0

[unkempt void]

also like saying `x' there means like

this just is the polynomial 1

that isn't a polynomial with integer coefficients

also like you are using φ as like a function Z -> Z or something there in that last line, when it maps polynomials to polynomials

Like φ has a given description (reduction mod n)

Well what you've said is a bit vague

You seem to be using φ to mean two different things

Okay sure

So okay you seem to mean like

a polynomial p where p(0) = 0 and p(a) = 1 for all non-zero a, right?

and no such polynomial p exists

Where does that come from

But yes, that cannot be a homomorphism

[unkempt void]

Okay sure so there is more detail and it's from this

I'll make it more explicit lol

Okay so we have a map $\phi: \mathbb Z \to \mathbb Z/n\mathbb Z$ which is reduction mod $n$ and this induces a map $\bar \phi: \mathbb Z[x] \to (\mathbb Z/n\mathbb Z)[x]$. Then for any $a \in \mathbb Z$ and $f \in \Z[x]$, $\bar \phi(f)(\phi(a)) = \phi(f(a))$

[sterile pumice]

potato

[unkempt void]

okay oops i've used phi and bar interchangeably to mean reduction mod n

[sterile pumice]

potato
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[unkempt void]

but it becomes a pain writing it all out like this lol

but yes tl;dr evaluation and reduction mod n "commute" in the relevant sense

[torn escarp]

It isn't in textbooks?

That's not quite the same thing though

That's a stronger result, since a polynomial factoring doesn't necessary mean it has a root

Like a quartic factoring into two irreducible quadratics

[unkempt void]

Yeah

Nobody would write it like what I said though lol

but i wasn't sure if you were confused about like making it fully explicit

[errant dragon]

whhhhhhhao

whats this math

[thin cedar]

it means anything that doesn't require graduate level understanding about there but the line is fuzzy

[torn escarp]

It's like you went on a tangent but rejoined back on path

So you went on a secant

[upper gale]

Hi, I have a question: assuming G, a group and H, a subset of G
To prove that H is a subgroup of G.
Which way "is mostly better" than other one?
-> definition of subgroup
-> Subgroup theorem (Idk how to spell this theorem but you know what I meant)

[sacred junco]

When it is easy to show a-b is an element of the H I will do it but sometimes it is hard to verify that so i would show using the long way

[unkempt void]

Tbh usually easiest to just use the definition in my experience

[upper gale]

okay! ty both for your review

[rustic escarp]

is there a way to write the sum of the first n odd numbers in summation notation ?

[rustic escarp]

yea that's the general form of every odd number

but if I wanted to write it in summation notation this wouldn't mean that

[sacred junco]

$\sum_{k=1}^n (2k+1) = (1 + 3 + \cdots + [2n+1])$

[sterile pumice]

Amer

[rustic escarp]

https://tenor.com/view/kronii-blank-stare-nodding-clueless-buffering-loading-gif-22937863

I'm a retard

thanks btw

[rustic escarp]

ok so I'm trying to prove that the sum of the first n odd numbers adds up to n^2

when using the sum of [2k + 1] I get (n + 1)^2

but when using [2k - 1] I get n^2

actually nvm I think I get what's wrong

I guess both are the general form of odd numbers

but one gives the odd number a[k] and the other gives the odd number a[k + 1]

[sacred junco]

Hey guys can someone explain how in fermats little theorem, How is a^p ≡ a (modp), simplified to a^(p-1) ≡ 1 (modp)

[rustic escarp]

so getting (n + 1)^2 makes sense since we have an extra odd number in the sum

[sacred junco]

nvm i figured it out

[loud maple]

Could someone please help me out with the following question?
Let S={x>1: lim(n->∞) frac(x^n)=0} where frac(x) is the fractional part of x. Determine the cardinality of S.

Apologies for the poor formatting.

S ⊂ ℝ, if this was not clear from the question.

[uncut pumice]

The cardinality is infinite since all integers greater than 1 are in it

[loud maple]

It is quite trivial to observe that S is infinite. I was asking how to determine whether it is countable or uncountable.

[uncut pumice]

Here's my proof that the only rationals in 𝑆 are the integers. Let 𝑥 > 1, 𝑥 = 𝑎 + 𝑏 where 𝑎 ∈ ℤ, 𝑏 = frac(𝑥) ∈ (0, 1). Let's assume 𝑥 ∈ 𝑆,

Given an 𝜀>0, we can find 𝑁 such that ∀ n > 𝑁, frac(xⁿ)< 𝜀.
Let 𝑚 > 𝑁, we have frac(𝑥ᵐ) < 𝜀. Let xᵐ = 𝑐 + frac(𝑥ᵐ), where 𝑐 ∈ ℤ.
frac(𝑥ᵐ⁺¹) = frac( (𝑐 + frac(𝑥ᵐ)) (𝑎 + 𝑏) ) = frac( 𝑎𝑐 + (𝑎 + 𝑏) frac(xᵐ) + 𝑐𝑏 ) = frac((𝑎 + 𝑏) frac(𝑥ᵐ) + 𝑐𝑏) = frac((𝑎 + 𝑏) frac(𝑥ᵐ) + frac(𝑐𝑏))

where the last equality comes from the observation that frac(𝛼 + 𝛽) = frac(𝛼 + frac(𝛽)).

If 𝑏 ∈ ℚ with denominator 𝑞, then frac(𝑐𝑏) ∈ {1/𝑞, 2/𝑞, 3/𝑞... 𝑞-1 / 𝑞} ,

We can make 𝜀 smaller than 1/𝑞.
We can also make 𝜀 small such that (𝑎 + 𝑏) frac(𝑥ᵐ) is smaller than 1/𝑞, such that 1/𝑞 < (𝑎 + 𝑏) frac(𝑥ᵐ) + frac(𝑐𝑏) < 1, and thus frac((𝑎 + 𝑏) frac(𝑥ᵐ) + frac(𝑐𝑏)) > 1/𝑞.

But this is contradicts the assumption that frac(𝑥ᵐ⁺¹) < 𝜀 < 1/𝑞.

I tried

My guess is that the irrationals are also not in 𝑆, and thus 𝑆 is just the set of all positive integers.

[sacred junco]

if x is an irrational, and it was in S
lim(x^n) in Z
lim(x^{n-1})x in Z
but thats false as integer times irrational is irrational
this should work 🥴 nvm

[loud maple]

If x is in S, then the limit (n->∞) (x^n) does not exist

[uncut pumice]

are you an engineering or physics major...? /joking /sorry

[loud maple]

Thanks, I'll read through this.

[loud maple]

This follows easily from the fact that x>1 ∀x ∈ S

[loud maple]

Where does the statement, "Given an ϵ>0, we can find N such that..." come from?
Other than that, the proof seems to be valid.

[loud maple]

However, I am not so sure about the validity of this guess, as I do remember reading in some text about the existence of a class of irrational numbers satisfying the property mentioned in the problem.

[uncut pumice]

I made a typo on the first line, corrected: it's from the assumption 𝑥 ∈ 𝑆, i.e. lim frac(𝑥ⁿ) → 0.

[uncut pumice]

did you read this?
https://math.stackexchange.com/questions/453673/fractional-part-of-rational-power-arbitrary-small
and this
https://mathworld.wolfram.com/PowerFractionalParts.html

Two interesting irrationals seem to be the golden ratio 𝜙 and the irrational 1 + √2, which have accumulation points 0 and 1 and thus are not elements of 𝑆

[loud maple]

Oh ok, makes sense. Thanks.

[loud maple]

No, I haven't read these yet. Thanks

[uncut pumice]

@loud maple I just realized the stackexchange question asked exactly the same thing as what I tried to prove.
I just typeset everything in a neater way and posted my answer there, if you want for future reference
https://math.stackexchange.com/a/4596429/1129672

[unkempt void]

cool q

[sacred junco]

hellow can you help me for this please, How to show that if a prime number different from 2 and 3 then it is congruent to 1 or -1 modulo 6, thanks

[uncut pumice]

Let 𝑝 be a prime > 3, if 𝑝 = 5 we see it's congruent to 1. So consider 𝑝 > 6, 𝑝 = 6𝑛 + 𝑘 for some 𝑛 > 1, 𝑘 ∈ {0, 1, 2, 3, 4, 5}. 𝑘 ≠ 0 because 6𝑛 + 0 = 2(3𝑛) is not a prime. 𝑘 ≠ 2 because 6𝑛 + 2 = 2(3𝑛 + 1) is not a prime. 𝑘 ≠ 3 because 6𝑛 + 3 = 3(𝑛 + 1) is not a prime. 𝑘 ≠ 4 because 6𝑛+4 = 2(3𝑛 + 2) is not a prime. Thus 𝑘 = 1 or 𝑘 = 5. Thus 𝑝 is congruent to 1 or 5. Note that 5 is congruent to -1 mod 6, so we can also say that 𝑝 is congruent to 1 or -1.

Also note the converse is not true: 25 is congruent to 1 mod 6, but 25 is not a prime.

[sacred junco]

thank you for your help but I have to find a form pn +K or not

[uncut pumice]

No. to say 𝑝 is congruent to 𝑘 mod 6 means 𝑝 = 6 𝑛 + 𝑘 for some 𝑛 ∈ ℤ, 𝑘 ∈ {0,1,2,3,4,5}.
4 options of 𝑘 can be readily eliminated because for example 6𝑛 + 0 always has divisor 2, which makes it unable to be prime. 6𝑛 + 2 = 2(3𝑛 +1) always has divisor 2 too. 6𝑛 + 3 = 3(2𝑛 + 1) always has divisor 3...

[loud maple]

Nice, thanks.

[loud maple]

Yeah

[brittle hill]

ame

wait im so sry this is the wrong server 😭

[sterile pumice]

person2709505

[latent gate]

Sorry, that should be "knowing 2003 and 10 are relatively prime"

My first thought was that this looks like Euclid's lemma, but that definitely isn't right

Or rather, the book I'm reading seems to suggest it follows without knowing that 2003 is prime

[unkempt zephyr]

Test for divisibility by the factors of 10.

[fervent grove]

it is more generally true that given gcd(a,b) = 1, if a | kb, then a | k

this is an application of Bezout's lemma

[latent gate]

Oh oops I didn't know that, thanks!

[latent gate]

As far as I can see, this shows 10 doesn't divide 2003, which is implied by gcd(10, 2003)=1

[unkempt zephyr]

Oh, I think I misread the question altogether. If my response was a little off the mark or seemed non sequitur, that's why. Sorry.

[latent gate]

Oh got it, no worries

[crystal plume]

i want to double check that the following function is an isomorphism

j : Z×Z→Z×Z, j(a, b) = (−b, a)

[solemn scaffold]

What will happen if Carmichael numbers were accidentally used in one or both of the $p, q$ pair that are used to form $N = pq$?

[sterile pumice]

rngwn

[solemn scaffold]

Note: Carmichael numbers are composite numbers that passes the Fermat's test on all of its coprimes smaller than itself.

[dark zealot]

if a=bq+r , gcd (a,r)=gcd(a,b)

?

euclidean division

[mild siren]

This is not true. For example, a=20, b=9, then q=2 and r=2, but gcd(a,b)=1 while gcd(a,r)=2. It is true, however, that gcd(a,b) always divides r.

[fervent grove]

you might want gcd(a,b) = gcd(b,r), which is true

[sacred junco]

Hello!

I am unsure what $\mathbb{Z}_n$ means in the context of number theory.

[sterile pumice]

trololol !

[plucky jacinth]

It probably means the integers modulo n

[torn escarp]

apply it again, j^2(a,b) = (-a,-b) then it's obvious if you do it 2 more times you get j^4(a,b)=(a,b) so j^3 is the inverse of j, which proves it's an isomorphism.

[tranquil shoal]

Z/nZ

so essentially the set containing every number that has remainder 0 when you divide by n (the set [0]), the set containing every number that has remainder 1 when you divide by n (the set [1]), and so on until you reach the set containing every number that has remainder n-1 when you divide by n (the set [n-1]), all of those sets are contained with in Z_n or Z/nZ

[serene roost]

anyone know how youd go about proving this?

[torn escarp]

I might try writing x^p = 1 + mn

[serene roost]

so what exactly would going through with that contradict?

[torn escarp]

look at the gcd, it's larger than 1 and divides x and n and try to apply that to this equation x^p = 1 +mn

[sacred junco]

thanks

[sacred junco]

thanks

[loud maple]

Does ∑(1/ϕ(n))² converge or diverge, where ϕ is the Euler totient function, and the sum is taken over all integers n≥1?

[sage fern]

there are an infinite number of numbers n for which totient(n) = 1, so it trivially diverges

@loud maple

wait, no

i'm a fool

i got it the wrong way round lol

infinite number of numbers n for which totient(n) = n-1

ok

ok well we can still look at it like this

consider 2n, then this has like n numbers which are not relatively prime to it, approximately

hmmmmm

ok this is actually interesting

ok so for even numbers it's basically like sum over 1/n^2 which converges

and then for odd numbers it should be less than sum over 1/n^2 because we should expect odd numbers to have fewer factors than even numbers

so overall it should converge i think

like, odd numbers have less factors, so more numbers below them are coprime to them, so totient(n) is larger, so 1/totient(n) is smaller

[loud maple]

It is quite reasonable to guess that it converges, that was my guess too, but I don't know how to even start a proof of this claim.

[sage fern]

oh, i've got something

[loud maple]

We need to demonstrate some nontrivial lower bounds for ϕ(n) in order to be able to proceed, I guess. The only lower bound I know is ϕ(n)≥(0.5×√n)

[sage fern]

ok so let (1/totient(n)) be p(n), and then the series is sum from n = 1 to infinity over p(n)^2

then when n = 2k, p(n) < 1/k

i think

[loud maple]

1/k is divergent anyway

[sage fern]

fuck

o wait

the series is squaring it

[loud maple]

Being less than a divergent series doesn't help to prove anything

[sage fern]

so it's 1/k^2

there

or ok no i'll bake it into the p(n)

ok so

let f(n) = (1/totient(n))^2

then the series is just sum from n = 1 to infinity over f(n)

then when n = 2k, it will share a factor with at least all the even numbers below it, so totient(n) =< n

wait

[loud maple]

So we get 1/ϕ² ≥ 1/n²

doesn't help

[sage fern]

ok that's worthless

i got it flipped originally is my problem

i always get it flipped, pain

[loud maple]

I tried searching math stack exchange for this, couldn't find anything.

Is the convergence/divergence of this series a well-known result?

[sage fern]

idk

i don't know the well-known results

ok

wait

if n = 2^k, n shares a factor with at most 2^(k-1) numbers below it; the even numbers

so t(n) = 2^(k-1)

[loud maple]

n=2^k implies ϕ(n)=n/2

[sage fern]

sure

[loud maple]

For k≥1

Therefore?

[sage fern]

so f(n) = 1/2^k

so summing over k, we get just 1

if we exclude n = 1

like, all the powers of 2

[loud maple]

What does the evaluation of this series at a proper subset of the natural numbers tell us about the entire series?

What can we conclude from this?

[sage fern]

so now here's my trick

let's do the same for all the primes

all the prime powers

disjoint subsets of the naturals

and i pray it will diverge

maybe? maaaybe?

[loud maple]

hmm

[sage fern]

actually maybe it could work for composite powers as well

[loud maple]

This converges if we sum it over all primes

[sage fern]

ok so for general powers, n = a^k, a a constant:

n shares a factor with at least a^(k-1) factors below it, all the multiples of a

so t(n) < a^k - a^(k-1)

so f(n) > 1/(a^k - a^(k-1))^2 = a^-2(k-1) * 1/(a-1)^2

wow this is tiny

yeah

hm

well we expect it to converge anyway

i guess?

[loud maple]

What is the conclusion here?

[sage fern]

the conclusion is that this will not diverge

[loud maple]

By giving a lower bound how can one show that something doesn't diverge?

[sage fern]

well the sum over the lower bounds doesn't diverge, to be precise

so we can't show that the actual series will diverge

it's a bad lower bound, is what i mean

[loud maple]

Right

Would this question be more appropriate for the advanced number theory channel?

[sage fern]

man idk what goes on in there

[loud maple]

I thought what I was asking was some trivial, well-known result that everyone would know except me because I'm a dum dum

[sage fern]

it's not the most fundamental hypothesis to consider

that said maybe it is well-known and i just also don't know it by coincidence, hell if i know

[loud maple]

Farewell it is, then. I leave for the advanced channel, wish me luck for my journey.

[analog onyx]

you aren’t alone

[sage fern]

luck!

[analog onyx]

my turn to ask a trivial well known question

[loud maple]

Any question I look at I just assume that it is high-school level, around 9th grade. Irrespective of what it is.

[analog onyx]

how would i go about showing 2^n is not congruent to 98 mod 100

[loud maple]

for what n?

[sage fern]

write out all of them until they start repeating 🙂

[analog onyx]

for all n

[loud maple]

Wait a minute, 98 mod 100 implies 2 mod 4

There, you have your answer

[sage fern]

yup

[analog onyx]

how’d you get there

[sage fern]

just take mod 4 of both

i mean it's like

98 mod 4 is 2

[loud maple]

For n>1, 2^n is 0 mod 4, trivial to check. Leaving the case n=1, which is 2 mod 100 and not 98 mod 100

[analog onyx]

surely i’d go about proving the first part of the statement with induction right

[sage fern]

no

you can do it outright

[loud maple]

100k+98=4(25k+16)+2

[loud maple]

Hence numbers of this form are not multiples of 4, which means they cannot be powers of 2 that are ≥4. And then you check that 2 is not 98 mod 100

[analog onyx]

okay that part makes sense

[loud maple]

Number theory is taught thoroughly to high schoolers in the US right? I heard that it is a standard course offered in high schools over there that basically everyone deals with around grade 9 or 10.

[analog onyx]

LOL i’m in my senior year of college and havnt had a formal course in NT

[loud maple]

Are you a math major?

[analog onyx]

yes

[sage fern]

that sounds like a lie

[loud maple]

Ohk, so I'm assuming you didn't do it in high school either?

[analog onyx]

no gosh no

highschool was like

algebra 1 and 2, geometry, precal calc and stats

[loud maple]

My intention wasn't to lie, I'm sorry. Maybe the person who told me this was not reliable.

[sage fern]

yes

i know

didn't mean to imply blame

[loud maple]

Algebra 1 and 2? Is that linear algebra and abstract algebra?

[sage fern]

no

it's like

quadratic formula

[analog onyx]

LOL gosh no

[sage fern]

ok so

[analog onyx]

bro what did you learn in highschool

[sage fern]

high school is for teenagers

i think you're thinking of university

or college

i think?

[loud maple]

I know very well what high school and college are, lol

[analog onyx]

although algebra 1 does has system of equations

[sage fern]

honestly i'm not even american

[loud maple]

I just heard from someone that US students do all this in high school

[sage fern]

but yeah there is no abstract algebra going on in high school

[analog onyx]

bro what

makes me wonder where these high schools are bc they aren’t near me

[loud maple]

Nothing at all, unfortunately .

High school was just coordinate geometry (2D: conic sections and all that, 3D: basic stuff with lines and planes and stuff), basic computational (non proof-based) stuff involving vectors, matrices, determinants, then trigonometry and results involving triangles,calculus(very basic integral and differential calculus, not even convergence and multivariate stuff) and then some stuff involving permutations, combinations and probability

[sage fern]

yes that sounds exactly the same as what would be expected in us afaik

[analog onyx]

i agree

[loud maple]

Entrance exam prep sucked the soul out of me. Couldn't even do proper math

[analog onyx]

i didn’t see any matrices and determinants though

but all the other stuff i saw

does this make sense

[loud maple]

Yeah, proof is valid.

[sage fern]

looks good

[analog onyx]

i feel like my proof writing is terrible but the math is right thanks to you two

maybe i should show 2^n is congruent to 0 mod for for all n > 1

even though that’s obvious

[loud maple]

Also write why 98 mod 100 implies 2 mod 4

[analog onyx]

i mean don’t you just apply mod 4 to 98

[loud maple]

I assume that points would be deducted for an incomplete proof right?

[analog onyx]

yeah absolutely

[loud maple]

The question itself was trivial, but we can't just write 'trivial' can we, if we want credit

[analog onyx]

i left it as an exercise to the grader one time but she didn’t like that

[loud maple]

If only we could

[analog onyx]

was it not as simple as taking 98 mod 4 though

[loud maple]

No

[analog onyx]

i’m gonna be honest i didn’t know you could reduce linear congruences like that

[loud maple]

Because if it was 98 mod x, where x is not a multiple of 4, then we can't proceed simply by doing this

[analog onyx]

but if we want to reduce 98 mod x where x is a multiple of y, we can

[sage fern]

it's like

[loud maple]

yeah

[sage fern]

x = 98 mod 100 only if x = 2 mod 4

idk

i can't explain this stuff simply any more lol

[analog onyx]

so i should say something like

since 100 is a multiple of 4, we take the congruence module 4 leaving us with 2^n ect

[loud maple]

Just write 100k+98 = 4(25k+16)+2

[analog onyx]

oh yes that’s what you did

[loud maple]

Which implies that 98 mod 100 implies 2 mod 4

[analog onyx]

yes

it feels kinda weird just keeping this is not congruent sign throughout

[sage fern]

ok so because phi(ab) = phi(a)phi(b), we essentially get that the series is:

1 + 1/phi(2)^2 + 1/phi(3)^2 + 1/phi(2)^4 + 1/phi(2)^5 + 1/phi(2)^2 * phi(2)^3 + 1/phi(7)^2 + ...

which is like 1 + a + b + a^2 + c + ab + d + ...

[loud maple]

Only when a and b are relatively prime

[sage fern]

wait i'm a fool

ohhhkay

totally wrong then

[loud maple]

This can't be an open problem though, right? It would have been much more famous if it was, I guess.

The result is probably sitting somewhere in an analytic number theory text.

[sage fern]

yes

[analog onyx]

what function are you using as phi

[sage fern]

totient

euler's totient function

[loud maple]

Unfortunately I haven't even started with analytic number theory.

[sage fern]

ok so excluding all the numbers such that the prime factorisation contains a power greater than 1:

(so excluding 4, 8, 9, 12, etc.)

the series is like 1 + 1/phi(2)^2 + 1/phi(3)^2 + 1/phi(2)^5 + 1/phi(2)^2 * phi(2)^3 + 1/phi(7)^2 + 1/phi(2)^2 * phi(2)^5 + ...

which is like 1 + a + b + c + ab + d^2 + ac +...

rearranging to 1 + ((a + ab + ac + abc + ...) + (b + bc + bd + bcd...) +

= 1 + a(1 + b + c + bc ...) + b(1 + c + d + cd + ...) +...

i think this rearrangement is legitimate because if it's convergent it's clearly absolutely convergent

we want to show it diverges though since we're missing terms

[loud maple]

And? What does this rearrangement give us?

[sage fern]

no clue

just a line of attack

wait no it's not complete

ok

[loud maple]

https://math.stackexchange.com/questions/4204432/evaluating-sum-n-1-infty-1-phin2

The series converges

[sage fern]

ah, just some high-strength lower bounds then

fine

[loud maple]

Yup, thanks to Ramanujan.

[sleek onyx]

i have literally no idea how to even start doing this

like where do formal derivatives even fit in

if you take the derivative of $m(x)^2$ and of $f(x)$ is it true that $m'(x)^2 | f(x)$?

[sterile pumice]

failingphysics

[sleek onyx]

so here’s what I have so far

if m^2 | f then there exists an n: f = m^2n

my thought is that if I take the formal derivative of both sides

then because m is nonconstant the degree of f’ will not equal the degree of (m^2n)’

but I just did that and it doesn’t seem to hold up

<@&286206848099549185> sorry to ping but I gotta finish this hw tonight so I am a little panicked

[leaden swan]

If you take the derivative you get that m divides 1
@sleek onyx

[torn escarp]

$f(x)$ has a multiple factor iff $f'(x)$ has the same factor, for instance $f(x)=g(x)^n h(x)$ for $n>1$ you have $$f'(x)=ng(x)^{n-1}g'(x)+g(x)^nh'(x) = g(x)^{n-1}(ng(x)^{n-2}g'(x)+h'(x))$$

[sterile pumice]

mOwOsity

[sleek onyx]

but we can factor $m$ out of both terms of the formal derivative no?

[sterile pumice]

failingphysics

[sleek onyx]

like if $f = m^2n$ then $f' = 2mm'n + m^2n' = m(2m'n + mn')$

[sterile pumice]

failingphysics

[sleek onyx]

so how do we get that $m$ divides 1? or what is the contradiction here?

[sterile pumice]

failingphysics

[leaden swan]

f'(x) is also $p^n x^{p^n-1} -1=-1$

[sterile pumice]

Drake

[leaden swan]

Since we are in (Z/p)[x]

[sleek onyx]

ahh ok

but $2mm'n + m^2n' = m(2m'n + mn')$ is not constant because none of those terms is guaranteed to be a multiple of a prime number?

[sterile pumice]

failingphysics

[leaden swan]

Well you m(2m'n+mn')=-1

Which means m has to divide -1

[leaden swan]

This is the set of all polynomials where the coefficients are from Z/p

[lethal spindle]

im kibda dumb ngl

kinda

[wintry dagger]

What are partial sums of 1/n^2

[torn escarp]

$$\sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}-\sum_{k=n+1}^\infty \frac{1}{k^2}$$
also can do a little work with either abel summation or approximating the sum as an integral to get something in terms of big O

[sterile pumice]

mOwOsity

[wintry dagger]

rub

didnt ask

no closed form?

this is trash to know

[torn escarp]

lol

[wintry dagger]

sorry for my behavior

how can i be more poliet

[torn escarp]

take it up with the analytic number theorists, this is the best they could come up with, I'm just the messenger

[wintry dagger]

hey @torn escarp

u want to talk later tonight?

math voicechat?

[torn escarp]

what do you want to talk about

[wintry dagger]

@torn escarp anything

maybe math?

[torn escarp]

sure I'm free now

[wintry dagger]

bruh i really appreciate u

but maybe tomorrow

quiet hours still

[wheat tinsel]

Where

@torn escarp So I think I could prove this

I was only interested in the case p=5, but I think these exact statements also hold for p any prime and 4 replaced with p-1

Also, I mention k>=3 because when k=2 you kind of have to treat it separately (because then 5^(k-2) is simply one and k-2=0), so its kinda annoying, but it is also true for k=2 and I guess for k=1, but these are no problem

[wheat tinsel]

Not entirely sure, but my main interest was in the case p=5, as I said. Tho to extend it to any prime p, you would need to prove some results that generalize the fact that the numerator of 1+1/2+...+1/(p-1) is divisible by p^2 and other facts that do not seem too intimidating, tho I wont try

In particular, if it were true for any prime p, then for n>= p^3 the denominator of H_n would be divisible by p, so this would yield another proof of the infinitude of primes xD

[gilded breach]

Let (n+1) is composite. Then there are natural numbers p, q such that 1 < p, q < n+1 and n+1 = pq.
Now p, q<n
Here, how can we derive p,q<n? Is it because n can't divide (n+1) so automatically none of p, q can be equal to n?

[loud maple]

Let p=n
Then q=1+(1/n)
Since 1+(1/n) is a natural number, n=1.
This implies than n+1=2, which is not composite.

[brisk skiff]

How can you quickly compute 2^103 (mod 1000) ?

[lilac lark]

= (2^100 * 2^3) mod 1000
= ((2^10)^10 * 8) mod 1000
= (1024^10 * 8) mod 1000
= (24^10 * 8) mod 1000
= (3^10 * 2^10 * 2^10 * 2^10 * 8) mod 1000
= (3^10 * 1024 * 1024 * 1024 * 8) mod 1000
= (3^10 * 1024^3 * 8) mod 1000
= (3^10 * 24^3 * 8) mod 1000
= (59049 * 24^3 * 8) mod 1000
= (49 * 24^3 * 8) mod 1000
= (49 * 8^3 * 3^3 * 8) mod 1000
= (10584 * 8^3) mod 1000
= (584 * 8^3) mod 1000
= (584 * 2^9) mod 1000
= (584 * 512) mod 1000
= 299008 mod 1000 = 8

[brisk skiff]

thanks

sorry i was asking for this problem

and i just realized how it did it

thx tho

[lilac lark]

oh k

[gilded breach]

Nice proof!

Thank you!

[loud maple]

welcome

[somber cobalt]

hey guys gotta question for Y'all. what does the meaning of "number" refer to? does it refer to the index or the value? perhaps both?

[wheat tinsel]

I don't think this is appropiate for this channel, I don't think this question has a definitive answer either

This is kind of philosophy of mathematics

If you want to know more, you can read Hamkins or Shapiro maybe. These are the only texts on the philosophy of mathematics I know

[torn escarp]

Use the Chinese remainder theorem and Euler's theorem. Since phi(5^3)=100 it simplifies very nicely

[brisk skiff]

yeah ik but how do you use that

Like how do you know that because 2^100 congruent to 1 mod 125 and 0 mod 8 that its congruent to 1 mod 1000

[unkempt void]

That isn't how it works

Some x being 1 mod 125 and 0 mid 8 specifies it uniquely and then we know we can just solve for it e.g. write x = 125a +1 and then take it mod 8 to get 5a +1 = 0 mod 8 so a must be like 3 I guess

[torn escarp]

It's 8 mod 125 and 0 mod 8, so it's 8 mod 1000.

[brisk skiff]

alr thx 👍

[visual umbra]

Hello, I need help with this
Let x, y, z ∈ Z be three integers that are solutions of Fermat's equation x³+y³=z³. Show that one of the integers x, y or z is divisible by 3.

[torn escarp]

I think looking mod 3 then mod 9 gives an answer, just a guess I haven't actually done that myself

Ok,||suppose none are divisible by 3, then the equation is 1+1=2 or 2+2=1 mod 3. Wlog take the first case, since second is negative of the first, then it becomes 1+1=8 mod 9, contradiction.||

[visual umbra]

Got it thanks

[wooden glen]

I'd say #groups-rings-fields for that.

[inner anchor]

By linearity it suffices to find the derivatives of g(x)^n for a polynomial g(x)

Which probably follows from induction

[wooden glen]

There's a nice general approach here, based on the principle of permanence. It goes like this:
If we fix the degrees of f and g, and then ask whether (f' o g)·g' = (fg)', then that is an equation between polynomials, which boils down to corresponding coefficients on each side being equal. But each of those coefficients is some polynomial expression in the coefficients in f and g -- we don't need to figure out the exact expressions here, it is enough to convince oneself they are polynomials with integer coefficients.
So for each output coefficient we have some polynomial equation.
Now we know from calculus that these equations are always true when the coefficients of f and g are real numbers. In particular they are true if all the coefficients of f and g are distinct algebraically independent transcendental numbers. But (by definition of algebraic independence) the only way for that to happen is if each of the equations reduces purely as an integer polynomial to 0=0. And that reduction then has to apply no matter which ring we're planning to get numbers to plug in from.

[pale fjord]

the answer's 0 but how did they get there

[brittle root]

wait why does

x=y=z=0 not work here

[torn escarp]

chevalley warning theorem guarantees at least p solutions

(there is a multiple of p many solutions, and since we know (0,0,0) is one, there are at least p-1 more)

[fervent grove]

one should be able to show by hand that there is a solution to x^2 + y^2 + 1 = 0 (mod p) [by hand, i.e. with Chevalley--Warning]
namely, x^2 takes on (p+1)/2 distinct values, as does -y^2-1, so there must be some overlap

[brazen python]

Hello

https://www.youtube.com/watch?v=5O08mG_6OEo&list=PL22w63XsKjqyg3TEfDGsWoMQgWMUMjYhl&index=2

[brazen python]

Can someone explain the intermediate steps here?

how does he get 2x_1 = 1 (mod 3) from 20x_1 = 1 (mod 3)

i have a feeling it's something to do with the addition and multiplication properties of congruences but i don't see it

[fervent grove]

20 = 2 (mod 3), so 2x_1 = 1 (mod 3) is equivalent to 20x_2 = 1 (mod 3)

[brazen python]

but if you take 10 = 1 (mod 3) and 2x_1 = 1 (mod 3) then 20x_1 = 1 (mod 3)

im not sure i understand how they're equivalent

[dusty dragon]

20x_1 = 18x_1 + 2x_1 = 3(6x_1) + 2x_1 = 2x_1 (mod 3). So 20x_1 = 1 (mod 3) also implies that 2x_1 = 1 (mod 3) (and vice versa), so they are equivalent

[vast dove]

For what values of n does there exists two distinct odd positive integers b such that 2n <= b < 3n?

Sorry

This is the correct question:
For what values of n does there exists two distinct odd positive integers b such that 2b <= n < 3b?

[torn escarp]

I guess since we just care about n, we might as well try to look at consecutive intervals to find them, since that maximizes their overlap. The intersection between [2b,3b) and the next interval [2(b+2), 3(b+2)) is 3b-2(b+2) = b-4, so in order for this to be positive we need b to be 5 or greater.

I guess next is trying to determine what integers are actually in the interval [2b+4, 3b) now for odd b >=5. I dunno if there's some great way of listing these numbers out, you can imagine them as the y values on the line y=2x+4 and up to but excluding the line y=3x. This sorta triangular segment extends out infinitely far and gets larger, seems to be the first number is n=14.

https://www.desmos.com/calculator/qapcyw23am

idk, do the lines get far enough apart that eventually every n large enough appears? that would be convenient

then it'd be a matter of determining that cutoff value and working out the finitely many examples below that

[vast dove]

Yes

n>17 works

[torn escarp]

yeah, it's not hard to show, I did it earlier but was distracted but the end result is 14, 18, 19, 20, and n>=22

21 doesn't work

[vast dove]

Oh 21 doesn't work

Ok next question, for a given n, how many b's can we find that satisfy the constraints: b is odd positive integer and 2b<= n < 3b?

[torn escarp]

interesting question, pretty fun, I think I might try to see if I can generalize this question/result while I'm thinking about it. What is this for

[vast dove]

I was trying to prove something else and needed this condition on b

[torn escarp]

I suppose we might try to repeat the process but instead of looking at the intersection of 2 consecutive intervals do k consecutive intervals, then we basically would have worked out all integers that occur at least k times

maybe a bit roundabout way of doing it, but I think finding these sets might not be too difficult that we could sorta do this

so like call the sets S_k, we determined S_2 = {14, 18, 19, 20, and n>=22} which I was able to describe as finitely many inequalities, so S_k in general is probably of the same form

$S_2={n : 4t+14 \le n \le 6t+14 \text{ for } 0\le t \le 1, \text{ and} \ n \ge 22 }$

[sterile pumice]

mOwOsity

[torn escarp]

for any n we can always just look at what the inequalities are for each S_k, since they should be pretty systematic and only finitely many of them to ever check

[vast dove]

Can you give an expression in terms of n? Or a lower bound?

[torn escarp]

I probably could yeah, I just haven't worked it out yet

the thing is, S_k is just as easy to work out as S_2 since k consecutive intervals really only depend on the first interval overlapping with the last interval, the intermediate intervals overlap by default

it should be easy to work out the lowest number in S_k and the first number N such that n>=N will appear

[torn escarp]

what was the other thing you were proving I'm curious what this is for

I'm about to be busy for about an hour or so, so I want to think about it while I'm busy, I might not finish grinding out the details here

[vast dove]

A certain proof by contradiction. Theres a certain 2 player game on Fibbonaci decompositions of a positive integers n that I won't define and I'm trying to prove player 1 always has a winning strategy for n large enough.

[torn escarp]

back, ok:
All elements that can be written in between at least k different b are:
n = 12k+2, 12k+6, 12k+7, 12k+8, and n >= 12k+10

from here it shouldn't be too bad to invert this as a function of n I think, but this is pretty good already

[torn escarp]

Ok think this is the final answer now, the [] is the Iverson bracket, it's 0 if the statement is false, 1 if true $$k = \left\lfloor \frac{n}{12}\right\rfloor -1 + \left[\left\lfloor \frac{n}{12}\right\rfloor \equiv 2, 6, 7, 8, 10, or 11 \mod 12\right]$$

[sterile pumice]

mOwOsity

[torn escarp]

that k(n) = exact number of intervals n falls inside

[vast dove]

thanks mero!

[torn escarp]

Yeah hopefully I didn't make any mistakes haha, you're welcome