elementary-number-theory

[sage fern]

does this even exist?

let x = ay; then ay^2 + a = 185, ie. a(y^2 + 1) = 185 and it's brute force

you expanded it wrong

[torn escarp]

fermat's little theorem basically makes the exponent mod p-1 which is where the 127-1 = 126 is coming from

[sterile pumice]

Ryam

[green hinge]

Does anyone mind giving me a hint on what to do here? I've been staring at this for more than an hour without any good ideas on how to tackle this.

[torn escarp]

try writing a=b+cn and see if you can manipulate 2^a - 1 into 2^b - 1 mod 2^c-1

keep in mind 2^c = 1 mod 2^c - 1 is really handy

also assume when I write 'mod' I'm applying it to both sides of the equation not just the term on the right side, just looks cleaner to write that way

@green hinge

[green hinge]

Thanks. I will see if that will work. @torn escarp

[torn escarp]

let me know if you can't solve it in the next 20 min or so

[green hinge]

Don't you mean 2^c = 2 (mod 2^c - 1 )? @torn escarp

Like 32 = 2 (mod 15)?

[torn escarp]

no

$$2^c -1 = 0 \mod 2^c-1$$
add 1 to both sides
$$2^c =1 \mod 2^c-1$$

[sterile pumice]

Merosity

[green hinge]

But that doesn't make sense. For example, 2^6 mod (2^5) - 1 is 2. @torn escarp

[torn escarp]

you're using two different numbers

is c=5 or c=6?

[green hinge]

Oh wait.

[torn escarp]

you're adding an extra power of 2 in there, but there isn't in what I'm writing

[green hinge]

I see my error. Sorry about that.

[torn escarp]

cool lol

[green hinge]

Mm. I am honestly still not sure what to do.

Oh, wait. It finally clicked.

I'm good now.

@torn escarp Thanks for your help.

[torn escarp]

you're welcome

[torn escarp]

you make any progress on your problem

[winged surge]

hello, consider the integer tuple solutions (x_1, x_2, .. x_n) of the equation sum(a_i * x_i) = k where gcd(a_i) = 1 and k is an integer. i wanted to know how flexible the solutions are. more particularly, is there a solution satisfying x_i <= x_i+1 for all valid i ? is there some resource i can refer to for this?

[pliant citrus]

Trying to prove that if 7|n but 7^2 doesn’t divide n, then n can’t be written as sum of two squares

I know it’s true just by criteria of sum of two squares on Wikipedia but not sure how to show its true

[pliant citrus]

Any suggestions?

[urban peak]

how can i prove that for pythagorean triple (a,b,c), c isnt divisible by any prime 3mod4?

[sage fern]

consider a^2 + b^2 mod 4

[torn escarp]

@winged surge Yeah look up Bezout's identity, the key thing is between any two pairs of numbers is you can solve a_i x_i + a_j x_j = 0 and then you can simply add or subtract multiples of this without affecting the solution to get the order on the x_i that you want

[urban peak]

didnt even consider simple thing. think i got it, thanks

[torn escarp]

start by supposing x^2+y^2=0 mod 7, then try to solve for y in terms of x we can subtract x^2 from both sides to get y^2=-x^2 mod 7. If either x or y is 0 mod 7, then so must the other, but that gives us a contradiction and would make divisible by 7^2. So it must be that neither is divisible by 7. Let's divide by x^2 to get (y/x)^2=-1 mod 7. But if we try to solve z^2=-1 mod 7 but there's no element that squares to make -1 mod 7. So there are no solutions.

[pliant citrus]

Not sure if I could’ve got thst myself but does make sense looking at it. Appreciate it

[torn escarp]

yeah takes a bit of time playing around, like just assume it's true and see the consequences, you're welcome

[quartz needle]

Find a perfect number that is divisible by 6 and has exactly 6 factors.

6k = 1 * 2 * 3 * x * y * z
What now?

[torn escarp]

that's not what 6 factors looks like

for instance 4 has 3 factors, 1, 2 and 4 but we can't write 4 = 1 * 2 * 4

[winged surge]

ohhh nice idea! thanks a lot, ill think about it.

[torn escarp]

even perfect numbers are of a very particular form involving a mersenne prime $2^p -1 $ and are exactly $n=2^{p-1}(2^p-1)$

[sterile pumice]

Meroseous

[torn escarp]

cool 👍

fortunately the number of divisors function is multiplicative, so $$\tau(2^{p-1}(2^p-1)) = \tau(2^{p-1})\tau(2^p-1) = (p-1 + 1)* (1+1) = p*2$$

[sterile pumice]

Meroseous

[torn escarp]

because we know this is 6, it forces p*2=6 so p=3

hopefully that is not too unfamiliar @quartz needle

[winged surge]

for p = 3, its 4*7 = 28 which is not divisible by 6

[quartz needle]

what's that odd letter

[winged surge]

tau?

[quartz needle]

yeah

what does it signify

[winged surge]

here tau(x) represents number of distinct factors of x

[quartz needle]

how does this follow

[torn escarp]

true, there is no perfect number that satisfies this

6 is the only perfect number divisible by 6 but only has 4 divisors, 1, 2, 3, 6

f being multiplicative means f(ab)=f(a)f(b) when gcd(a,b)=1

the number of divisors of a prime power p^k is k+1

they're 1, p, p^2, ..., p^k

[quartz needle]

ho

i made a typo

[torn escarp]

just correct here I don't wanna scroll up to an edit

[quartz needle]

Find a perfect number that is divisible by 4 and has exactly 6 factors.

[torn escarp]

cool, so we're good then

[quartz needle]

brb

[sage fern]

well that's trivial

[sand yarrow]

hello

[zinc steeple]

can anyone prove this

[sage fern]

https://en.wikipedia.org/wiki/Prime_number_theorem#Proof_sketch

[zinc steeple]

thanks

[zinc steeple]

is there a video proof

[stiff rivet]

a detailed proof is quite involved

i.e. it takes a while

you will find it in any introduction to analytic NT book, for example apostol's

(chapter 13 deals with this, but you need many tools)

[zinc steeple]

vielen danke

thank you

[fresh hound]

if p congruent to q mod n, then is it true that a^p congruent to a^q mod n?

nvm

[high cloak]

What formulas guarentee prime numbers. Doesnt have to generate all, just that is never generates composites.

Besides mills formula

[sage fern]

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

[stiff rivet]

f(x) = 2

[torn escarp]

oh yeah check this out, 100% better:
f(n)=4+(-1)^n

[swift shard]

Nah fams, I have one and it is even bijective.
f(n) = the nth prime

[sterile pumice]

TC159

TC159

[torn escarp]

this looks like you're reading it wrong

for a power of a prime, $\tau(p^a) = a+1$

[sterile pumice]

Meroseous

[torn escarp]

so $\tau(p^aq^b)=\tau(p^a)\tau(q^b) = (a+1)(b+1)$

[sterile pumice]

Meroseous

TC159

[torn escarp]

that's wrong

$\tau(p)=2$

[sterile pumice]

Meroseous

[torn escarp]

it has only divisors 1 and p

you put p+1

that's false

p^n has divisors 1, p, p^2, ..., p^n so it has n+1 divisors

you can make a table and fill it in if that helps to see

put powers of p and powers of q along both sides

and then make all the combinations inside

huh

ok pick n=1

your formula says t(p)=p+1

that's totally wrong

p can't have p+1 divisors

2 has 3 divisors?

no

lol

whatever

definitely

lmfao

I'm not gonna steal 5 euros from a child

[sacred junco]

[thorn wedge]

Y'all are funny

[warped summit]

im trying to for all int m>=3, (m^2 - 1) is not prime. i know if n = rs is not prime then 1 < r < n and 1 < s < n. using that template i have (m^2 - 1) = (m-1)(m+1) and since m>=3 i know (m-1)>=2 and (m+1)>=4 so i have both of those things being >1 but how do i prove (m-1) < m^2 - 1 and (m+1) < m^2 -1 ?

[wooden crater]

if p is a positive prime and p = rs, then either (r = p and s = 1) or (r = 1 and s = p)

if p = m^2 - 1 = (m - 1)(m + 1) is prime, apply the above to solve your problem

[warped summit]

in this case i want to show m^2 - 1 is not prime

and i have (m-1) > 1 and (m+1) > 1

i want to show (m-1) < m^2 - 1

and (m+1) < m^2 - 1

question is: how to do that

[wooden crater]

this result fully characterizes when m^2 - 1 is prime, which will be if and only if m = 2

[wooden crater]

showing this does nothing for you tho. and this is just true in general. for positive integers n and m, n <= nm and m <= nm

[warped summit]

ok i see

[wooden crater]

i don’t see how it helps you prove that m^2 - 1 is not prime

[warped summit]

so since m^2 - 1 a product of two integers in the form of 1 < m-1 and 1 < m+1 then m^2 - 1 is composite?

since m-1 < m^2 -1

and m+1 < m^2 - 1

?

[wooden crater]

yea that works. you don’t need the last two inequalities

[warped summit]

thanks

well i’m still a little confused

i want to say

m^2 - 1 = (m-1)(m+1)

and since

m>=3

we know

1 < m-1
1 < m+1

now i need to finish it off saying

m-1 < m^2 -1
m+1 < m^2 - 1

somehow

how do i justify that

[wooden crater]

why are you so concerned with the last two inequalities?

[warped summit]

because n = rs is compsoite if 1 < r < n and 1 < s < n

so i want to justify that in this proof that r < n and s < n where n = m^2 - 1

and r=m-1 and s=m+1

[wooden crater]

it’s from here. equality holds iff m = 1 or n = 1

[warped summit]

ah i see, so we have 2 <= m-1 and 4 <= m+1 so neither m-1 or m+1 is equal to 1 therefore m-1 nor m+1 can be equal to m^2-1

but how do i know they are strictly less than m^2 - 1

[wooden crater]

yes. if you want it more directly,

1 < m - 1 so multiply by m + 1 on both sides

same with 1 < m + 1. multiply by m - 1 on both sides

[warped summit]

remind me, what axiom are we using to say the sign of the inequality doesn't flip if i do these multiplications?

[wooden crater]

it’s a direct consequence of the way you define the usual order relation on the positive (or non-negative) natural numbers

[warped summit]

1 < a and 1 < b, then b * 1 < b * a

so i get my m-1 < m^2 - 1

and m+1 < m^2-1

via your method

thus getting what i wanted!

[wooden crater]

cool

[warped summit]

ty for your help

[simple gyro]

For gcd(x,y)=1 does x or y have to be prime

[sage fern]

no

consider 9 and 16

[buoyant pecan]

Is there some family of groups that have order $\binom{n}{k}$? I'm working on a problem for finding which primes divide a binomial coeffecient, and I think it might be useful to look for a group of size $\binom{n}{k}$ and look at it's subgroups.

[sterile pumice]

deadpan2297

[sage fern]

yes

let c be nCk

then Zc has such an order

more seriously, i don't think so

[rain crown]

uhh idk where intro to group theory goes but could someone help me with this problem?

[wooden crater]

which part?

[rain crown]

all tbh but i’m sure if you help me with one the rest will be easier

[wooden crater]

this is linear algebra right?

[wooden crater]

anyway, for 1. try showing that for two functions f and g in F and for any scalar c in R, that phi(cf + g) = c*phi(f) + phi(g) (this should be really easy, just apply rules of derivatives). once you do that, then you've shown that its a linear map. just some leads, this map will be surjective but not injective. can you think of why?

[rain crown]

it’s for intro to group theory class

not injective bc there’s not one unique y for each x? since like the derivatives of constants are all 0

[wooden crater]

right, the right idea

wording is poor

it should be there is not a unique x for each y. if there were not a unique y for each x then phi wouldnt even be a function (thats like saying phi(f) = g and phi(f) = -g)

[rain crown]

how to prove that S and T are isomorphic with only the cayley tables?

[stiff rivet]

once you figure out what the identity is, there are only 2 possible bijections to check

[sacred junco]

anyone looked into reappearances of subset fibonacci numbers in subsequent parts of the fibonacci series?

[sacred junco]

i don't know why but this feels like it fits perfectly at the front of subsequent sets in the series.

1x10 for each successive row down the line

maybe im just forcing something here but it feels like something is there

[stuck forge]

i don't understand

1 isn't a congruent number (that is to say an area of a right rectangle)

but a right rectangle which sides are 1 and 2 has an area of (1*2)/2, so ONE !!!

Erklären Sie mir bitte

[stiff rivet]

the third side of that triangle has length sqrt(5)

[stuck forge]

ah ok thx

[astral fiber]

i have an issue, what does the inf mean here

just the minimum value of f(x)-g(x)?

<@&286206848099549185>

[stiff rivet]

no, it's the infimum

a minimum might not exist

[stiff trench]

$\is (n^2+3n+1)^2-1$\ is divisible by 24

Could anyone help me prove this. I know that it is divisible by 24 if it is divisible by 3 and 8 but I am not sure what to do next?

Factor out?

apply the formula?

[sterile pumice]

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

[queen robin]

Just factor it

As a difference of squares

[stiff trench]

Yeah but I cannot get anything meaningful

$\to n^4+3n^3+11n^2+6n$\ It turns out to

[queen robin]

Use difference of squares, instead

You can factor it easily

[sterile pumice]

Epifor

[stiff trench]

a^2-b^2

right?

[queen robin]

Yes

[stiff trench]

I did though

$\to (n^2+3n+2)*(n^2+3n)$\

[sterile pumice]

Epifor

[queen robin]

Then?...

[stiff trench]

I guess multiply it out?

[queen robin]

No

Continue factoring it out

[stiff trench]

hm

like

$\to (n*(n+3)+2) * (n*(n+3))$\

[sterile pumice]

Epifor

[stiff trench]

oh wait

[queen robin]

Not really

[stiff trench]

n*m = 2
n-m = 3?

you mean like that?

or n + m in this case

[queen robin]

Yes

[stiff trench]

aight moment

I get

$\to ((n+1)(n+2)) * (n^2+3n)$\

[sterile pumice]

Epifor

[stiff trench]

one of them has to be divisible by 3

n, n+1 or n+2

but we cannot prove for 8

[queen robin]

We can

Express it as $n(n+1)(n+2)(n+3)$

[sterile pumice]

Campanha

[queen robin]

It is pretty straightforward

[stiff trench]

but if a number is divisible by 4 it doesnt neccessarily have to be divisible by 8

[queen robin]

You can break into cases, say n = 2k, n = 2k + 1, then either k is even or odd

For each case

The last part isn't necessary at all

Just break in n even, n odd

[stiff trench]

Sorry but I am not getting it. We express n as either an odd or even number, and what do I do with that?

maybe like

you mean like 4k^2

and if 4 is a factor of k^2 and multiplies all of them

nvm

[queen robin]

Suppose n is even, say n = 2k, for an integer k. Then n(n+1)(n+2)(n+3) = 4k(k+1)(2k+1)(2k+3), but k and (k+1) are consecutive integers, so 2 divides k(k+1). So, 8 divides n(n+1)(n+2)(n+3)

The case for an odd n is analogous

[stiff trench]

ah so since k and k+1 are consecutive, one of them is divisible by 2 and since we have 4k, it must be divisible by 4*2 = 8?

[queen robin]

Yes

[stiff trench]

what about the case for n = 2k+1

[queen robin]

Do it on your own

It is analogous

[stiff trench]

What does analogous mean?

[queen robin]

Means it goes by the same way

It doesn't change because there are 4 consecutive integers

If n is even, then 2 are even and 2 are odd. If n is odd, then 2 are even and 2 are odd

The parity is invariant

You don't even need to split into two cases (even and odd), because of this fact

[stiff trench]

yes but for odd, it doesn't always have to mean that it will be divisible by 8 or 4

$(4k+1)(k+2)(k+3)(k+4)$

[sterile pumice]

Epifor

[queen robin]

Your substitutions are wrong

$n = 2k+1$, not $n = k + 1$

[sterile pumice]

Campanha

[stiff trench]

oh yeah

yeah we see the same case

tysm

[queen robin]

yw

[karmic salmon]

\subsection*{(a) If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$}
Assume $\gcd(a^n,b^n) \neq 1$, then $\gcd(a^n,b^n) > 1$. If $\gcd(a^n,b^n) > 1$ then there exists some prime $p>1$ such that $p\vert a^n$ and $p\vert b^n$. If this is the case, then it must also be true that $p\vert a$ and $p\vert b$. However, this contradicts the fact that $\gcd(a,b)=1$. Therefore, $\gcd(a^n,b^n)=1$.

[sterile pumice]

-vertex

[karmic salmon]

\subsection*{(b) The relation $a^n\vert b^n$ implies that $a\vert b$}
Let $d=\gcd(a,b)$, then $d\vert a$ and $d\vert b$. Therefore, there exists some integers $r,s$ such that $a=dr$, $b=ds$. Furthermore, there exists some integers $x,y$ such that $d=drx+dsy \implies rx+sy=1 \implies \gcd(r,s)=1$. Hence, by the previous result $\gcd(a^n,b^n)=1$.

[sterile pumice]

-vertex

[karmic salmon]

do these look okay?

$a^n\vert b^n \implies d^nr^n\vert d^ns^n \implies r^n\vert s^n$. Since $r^n\vert s^n$, and $\gcd(r^n,s^n)=1$, $r^n=1$, $r=1$.\
\
Therefore, $a=d$, so $a=\gcd(a,b) \implies a\vert b$.

[sterile pumice]

-vertex

[karmic salmon]

oops, missed end of second proof when pasting

last sentence of the first part of (b) should say $\gcd(r^n,s^n)=1$

[sterile pumice]

-vertex

[sacred junco]

How do i prove that 3^a + 63 is divisible by 72

for which a is always even

[sacred junco]

any hints?

[stiff rivet]

what have you tried?

[sacred junco]

I have tried putting a as in like 2k and square rooting it but nothing

and setting 3^a + 63 = 72k

i also observed that 63 can be 64 - 1 which can be difference of squares

dats about it

[stiff rivet]

look at the equation modulo 72

notice that 63 = -9 mod 72

then try to work with a=2k

[sacred junco]

like write it as 3^2k + 72-9 = 72t?

which i got to
(3^k-3)(3^k+3) = 72*(t-1)

[stiff rivet]

huh

do you know how mod works?

$$3^{2k} + 63 \equiv 3^{2k} - 9 \pmod{72}$$

[sterile pumice]

Lochverstärker

[stiff rivet]

then do the "obvious" thing from there

alternatively ||induction on a probably works in the original statement||

[sacred junco]

I know what mod is just like isnt it trivial, as in you can represent it in other ways

[sacred junco]

how though

it doesnt make sense that 3^2k + 63 / 72 = 3^2k-9

[stiff rivet]

that is not how mod works

63 is congruent to -9 mod 72 as their different is divisible by 72

[sacred junco]

isnt like the first number the X second the residue and third Y for X/Y

[stiff rivet]

$a\equiv b \pmod{n} \iff n\mid (a-b)$

[sterile pumice]

Lochverstärker

[sacred junco]

ah yeah a-b

[stiff rivet]

and well, then a number is divisible by n iff it is congruent to 0 mod n

[sacred junco]

I need to brush up on mods

so like if 63 is congruent to -9 and if we add any number to both sides

its the same

oke

[stiff rivet]

sure but like

3^(2k) = 9^k

and 9 - 9 = 0

so you want 9^k to be congruent to 9 mod 72 for every k

[sacred junco]

I see

do you have any tasks on resources to further improve my mod understanding

[stiff rivet]

most of the time you just want to write down a congruence and change numbers by adding or subtracting the modulus (in this case 72)

e.g. 63 - 72 = -9, so they are congruent

and -9 is kinda simpler, and its easy to notice that 3^(2k) = 9^k, so that is what i did here

anyways, review the mod relation and what "substitutions" are legal

[sacred junco]

I see, thanks

[sacred junco]

Hello. I hope this is not an advanced topic. Can someone brief me up about Riemann's hypothesis?

Or recommend me a YouTube video or article?

Please ping. Thank you.

[stiff rivet]

it's about locating zeros (roots) of a complex valued function (or the analytic continuation thereof), the riemann zeta function
without a decent background in complex analysis, it's not really possible to explain what even the riemann zeta function is

if you want a good introduction for laypeople, i can recommend this video: https://www.youtube.com/watch?v=sZhl6PyTflw
it's german, but there are english subtitles @sacred junco
alternatively there is https://www.youtube.com/watch?v=sD0NjbwqlYw for pretty pictures

[sacred junco]

how does this happen
n+4 | 3n-2
n+4 | 3*(n+4)

[latent hare]

Are you supposed to find which numbers satisfy n+4 | 3n-2?
The first and second statement aren't related, the second one is always true since n+4 must divide 3*(n+4) and can be used to help you find the solutions to the first one. But without much context I can't know

[stiff trench]

$2^2 + 4^4 + 6^6....50^{50}$
Prove that this is not a perfect square

[sterile pumice]

Epifor

[stiff trench]

I have tried turning them into same factors or like grouping

but no relevance

Maybe mod 3, but not sure how to get it to that

[queen robin]

Mod 3 is the answer

All squares are either congruent to 1 or congruent to 0 modulo 3. Try reducing this expression modulo 3, you'll see something...

[stuck forge]

is it not modulo 4 ???

[queen robin]

The square of an integer is congruent to either 0 or 1 both modulo 4 and modulo 3

This question can be solved reducing the expression modulo 3

[sacred junco]

I'm not sure if this is the right channel

but

I've heard that in some fields of maths people sometimes define 0^0 to be 1

where and why do they do that?

[quartz needle]

Show that if p is a prime number greater than 3, p^2 -1 is divisible by 24.

So far I have only determined that (p-1)(p+1) is a product of two even numbers, and thus is divisible by 2

It also has to be divisible by 3, because p/3 must give the remainder of 1 or 2, so (p^2 - 1)/3 doesn't have a remainder

@scarlet geode what do you think?

[scarlet geode]

Look at the prime factorization of 24

2^3 * 3

[quartz needle]

this is enough?

[scarlet geode]

No

Just some idea

I am getting ready for work

In the shower rn

[quartz needle]

yeah, but i'd have to prove that it's divisible by 2^2 too

so far I have only proven 2, 3, and 6

[scarlet geode]

You need to prove it's divisible by 8 actually

[quartz needle]

don't i have to prove it for all prime factors?

[scarlet geode]

sqrt(8k + 1) = p

You proved it for 3

The other prime factor power is 8

If you show it's divisible by 3 and 8

Then it's divisible by 24

[quartz needle]

ik

dunno how to prove it

oh

it's a product of two even numbers

so it must be divisible by 8

thanks

[haughty lichen]

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

[quartz needle]

Find a perfect number divisible by 4 that has 6 divisors
x = 1 + 2 + 4 + a + b + c
What now?

[sage fern]

why do you believe this exists

[quartz needle]

it's a problem

so it clearly does

Find all the natural numbers that satisfy the equation 2y^2 + xy - x^2 = 35

[sage fern]

oh, wait

no i thought this was the same question you'd posted before

just check all the perfect numbers

[quartz needle]

how am I supposed to do this

[sage fern]

start from the smallest one and work upwards

[quartz needle]

are you suggesting that i check if the divisors of each number add up to it?

[sage fern]

no

just look at the list of perfect numbers

[quartz needle]

that defeats the point of the exercise, no?

[sage fern]

ehhhh

it's not a brilliant exercise

the other one is better

[quartz needle]

it's in a preparatory material for the highschool graduation exam, and it is stated that it has been used as an entrance exam to some school

[sage fern]

i think it just wants you to know what a perfect number is, ig?

[quartz needle]

it's a third exercise regarding perfect numbers

[sage fern]

well then it seems pointless

[stiff rivet]

0^0 = 1 is correct when you want to study objects via functions on them
we write A^B for the set of functions from B to A (think of ordering B some way and assigning an element of A to every element of B thus generating a |B|-tuple in A)
when using ZFC to model mathematics, we define the natural number n as a certain set of cardinality n, it follows that there are |n^m| many functions m -> n
now there is only one possible choice to model 0 as a set this way, namely as the empty set since this is the only set of cardinality 0 and since there is precisely one funtion from the empty set to itself (namely the empty function) it follows that 0^0 = 1

turns out this pov is "correct" in anything not analysis and often lets you extend theorems in e.g. combinatorics to weird edge cases

[quartz needle]

@sage fern you've said that the other one is better, but I still don't know how to solve it

[sage fern]

just look at the list of perfect numbers

[quartz needle]

I mean

Find all the natural numbers that satisfy the equation 2y^2 + xy - x^2 = 35

[sage fern]

oh, the other one

lemme see

what have you tried

[quartz needle]

nothing

[sage fern]

well

try something

[stiff rivet]

there is no list of all perfect numbers 🤔

[quartz needle]

well, i suppose i have to facotrize it in some way

[sage fern]

list of the first few, whatever

[quartz needle]

i have no idea how to facotrize it, though

[sage fern]

ok

so reading the question we see that it says 'natural numbers'

and the LHS does look hard to factorise

so this suggests to me that factorisation might not be sufficient, at least by itself

for whole-number equations, looking at mods is usually a good idea

[sage fern]

prove it

[quartz needle]

could you elaborate

[sage fern]

modular arithmetic? no?

[quartz needle]

yeah i know it, but i'm not sure how would i use it to solve this exercise

[sage fern]

try something

i mean i don't know whether this will work either, i haven't solved it

but it seems like a likely general direction

[quartz needle]

i'm not sure how would i go about doing so

btw here's another problem

show that for all natural numbers n, n^5 - n is divisible by 30
i got to
n(n-1)(n+1)(n^2+1)
so it has to be divisible by 3 and 2 (since it's divisible by 3 consecutive numbers)
but I still have to prove it for 5, and I'm not sure how would I go about doing so

[sage fern]

bruh

you'll never solve anything difficult if you hop from problem to problem like this

[quartz needle]

what's the point of staying on one problem for too long?

if i get stuck i move on

[sage fern]

to practice solving difficult things

[stiff rivet]

looking at mod will not work (at least not without super heavy machinery)
you can mostly use it to show no solutions exist as solutions in Z give you solution in Z/n for free
but this equation has solutions in Z/n for all n, so you cannot rule out solutions in Z easily

[quartz needle]

i might get an idea after finishing another one

[sage fern]

that's a good idea when in an exam, but not when practicing

it's worth trying a little longer

[quartz needle]

20min isn't enough?

[sage fern]

ehhh

ngl i wouldn't say you've actually been trying for 20 mins

i mean you've been on here

[quartz needle]

well, you've been on here, too, and are better than me at math, so that shows my choice was the right one

[sage fern]

i haven't really tried it yet either

it's your problem

[quartz needle]

well, i don't know how would i go about solving it, because I have never seen a problem like this

I mean

I would have to factorize it to something like (a-2)(b+3) = 5

that I can solve

but since idk how to factorize it, and don't have any other ideas, I'm stuck

[sage fern]

oh, wait a sec

lemme try something

oh, it does factorise

i'm a fool

[quartz needle]

?

[sage fern]

ok so

it'll factorise into something of the form

(ax+by)(cx+dy)

or really just

(x+ay)(bx+cy) WLOG

[quartz needle]

is there some trick to factorization

or do you just stare at a problem for long enough

[sage fern]

well you can try and find brackets of a form that will give something like what you have

[quartz needle]

yeah ik

but you came up with this form

[sage fern]

for example this

yeah

well i just thought

'what will you need to multiply together to get x^2, xy and y^2'

x by x, x by y and y by y

[sage fern]

so it'll be like this

[quartz needle]

ok, i got that
a = b = c = -1
so
(x+ay)(bx+cy) = (x-y)(-x-y) || *-1
(x-y)(x+y) = x^2 - y^2
x^2 - y^2 = 35

x^2 - y^2 + 35 = 0

and idk what to do

[sage fern]

that's not how you factorise it

[quartz needle]

ay * bx = xy, so a = b

[sage fern]

no

[quartz needle]

x * bx = -x, so b = -1

[sage fern]

ay * bx = ab(xy)

[quartz needle]

well

i mean that ay * bx in your form

is equivalent to xy

[sage fern]

what

[quartz needle]

well, i looked at your form and looked at the original equation

and tried to figure out the coefficients this way

since i didn't know what else to do

[sage fern]

after you edited it, that's right

so ab(xy) = xy

therefore ab = 1

not a = b

[quartz needle]

it

is the same

ab = 1 only if a = b

[sage fern]

why do you assume they are whole numbers

they might be, but they don't have to be

[quartz needle]

well

i think you've mentioned mod before

and that only works for whole numbers

[sage fern]

just because x^2, xy and y^2 are whole numbers doesn't mean the quadratic has to factorise with whole numbers

consider x^2 - 2 = 0

(x - sqrt2)(x + sqrt2) = 0

[quartz needle]

i was just going off what you have said before

[sage fern]

you didn't go off it rigorously

[quartz needle]

well

[sage fern]

keep going

[quartz needle]

if we don't assume that they are whole numbers

then idk what to do

[sage fern]

ab(xy) = xy

no

you worked out that ab(xy) = xy, so ab = 1

keep going

do the same for x^2 and y^2

their coefficients

[quartz needle]

becaus ethen you have x * bx = -x^2, so b = -1

[sage fern]

ok

[quartz needle]

but since you don't know that a = c, you can't figure out ay * cy

[sage fern]

ok so

you have:
b(x^2) = -x^2, ie. b = -1
ab(xy) = xy, ie. ab = 1
ac(y^2) = 2y^2, ie. ac = 2

that's sufficient

[quartz needle]

well

idk how to use it

[sage fern]

??

b = -1
ab = 1
ac = 2

solve it

[quartz needle]

oh

well

b = -1
and from
ab = 1
ac = 2
we know that b = 1/2 c
so c = -2

(x+ay)(bx+cy)
(x-y)(-x-2y) = 35
(x-y)(x+2y) = -35

[sage fern]

no

[quartz needle]

?

[sage fern]

what is a

wait hang on

[quartz needle]

-1

[sage fern]

yeah so

why is there no - in your brackets

where a should be

yeah

(x-y)(x+2y) = -35
(y-x)(x+2y) = 35

just to make things easier

and now it's just some brute force

[quartz needle]

how

[sage fern]

factors!

[quartz needle]

oh

that

35 = 7 * 5
(y-x)(x+2y) = 35
(y-x) = 7
(y-x) = -7
(x+2y) = 5
(x+2y) = -5
(y-x) = -5
(y-x) = 5
(x+2y) = -7
(x+2y) = 7

[sage fern]

almost

firstly you can quickly rule out some of these options since x and y are natural numbers, so they are non-negative

also 35 = 1 * 35

[quartz needle]

wdym

[sage fern]

(x+2y) = -7
this is impossible

[quartz needle]

35 * 1 = 7 * 5
(y-x)(x+2y) = 35

(y-x) = 7
(x+2y) = 5
3y = 12
y = 4
x= -3
FALSE

(y-x) = 5
(x+2y) = 7
re

(y-x) = 35
x+2y = 1
3y = 36
y = 12
x = -23
FALSE

(y-x) = 1
(x+2y) = 35
re
???

@sage fern

[sage fern]

yes

[quartz needle]

isn't it incorrect

[sage fern]

why

[quartz needle]

there exist number for which this is correct

[sage fern]

yeah? good

that's what you want

[quartz needle]

according to my solution there don't exist any

[sage fern]

what does re mean

[quartz needle]

the same as before

[sage fern]

(y-x) = 1
(x+2y) = 35
what's wrong with this

[quartz needle]

add them up

and you get 3y = 36

[sage fern]

ok

[quartz needle]

and as shown above, it doens't work

[sage fern]

why not

[quartz needle]

cause it would mean that x is negative

[sage fern]

no

if:
(y-x) = 35
x+2y = 1

x is negative

_ _
if:
(y-x) = 1
x+2y = 35

this is a different set of equations entirely

with different solutions

[quartz needle]

(y-x) = 35
x+2y = 1
(y-x) + (x+2y) = 35 + 1
3y = 36
y = 12
12 - x = 35
x = -23

(y-x) = 1
x+2y = 35
(y-x) + (x+2y) = 1 + 35
3y = 36
y = 12
12 - x = 1
x = 11

so one solution is (11, 12)

[sage fern]

yes

[quartz needle]

well

this isn't a solution

[sage fern]

yes it is

[quartz needle]

according to the book from which this question comes from, the solutions are
(x,y): (1,4), (3,4), (23, 12)

since we have had 4 systems of equations to solve, and one is incorrect, all are

[sage fern]

wait

hm

2y^2 + xy - x^2 = 35

(2y - x)(x+y) = 35

2y - x = 1, x + y = 35

3y = 35, y = 12

x + y = 35

x = 23

right so you screwed it up

ah you factorised wrong

[quartz needle]

2y^2 + xy - x^2 != (2y - x)(x+y) = 2xy + 2y^2 -x^2 - xy

[sage fern]

didn't notice bc it looked similar

2y^2 + xy - x^2 = (2y - x)(x+y) = 2xy + 2y^2 -x^2 - xy

[quartz needle]

thanks for help, anyways

i'm tired of this problem, though

show that for all natural numbers n, n^5 - n is divisible by 30
i got to
n(n-1)(n+1)(n^2+1)
so it has to be divisible by 3 and 2 (since it's divisible by 3 consecutive numbers)
but I still have to prove it for 5, and I'm not sure how would I go about doing so

[quartz needle]

<@&286206848099549185>

[quartz needle]

prove that if (a^2 + b^2)(c^2 + d^2) = (ac+bd)^2, then ad = bc
a^2 * c^2 + a^2 * d ^2 + b^2 * c^2 + b^2 * d ^2 = a^2 * c^2 + 2abcd + b^2 * d^2
so
a^2 * d^2 + b^2 * c^2 = 2abcd
no idea what to do now

show that if x^2 * y^2 + z^2 = 2xyz then z = xy
no idea what to do here

[sacred junco]

isn't the first line direct statement of cauchy schwartz inequality?

[quartz needle]

don't know it

[sacred junco]

and this is direct application of AM GM inequality

[quartz needle]

don't know it either

neither of these is required, anyways

the knowledge of them

[sacred junco]

i mean yeah this becomes a perfect square so

[quartz needle]

so either you are supposed to figure it out by yourself without having any knowledge of cauchy schwartz or am-gm, or there's some other method

[sacred junco]

see for this problem

we can re write it as (xy)^2 + z^2 - 2xyz=0

which is same as (xy-z)^2=0

[quartz needle]

oh

in that sense

got it

should have looked a bit closer, sry

[sacred junco]

number theory is all about approaches so don't worry just keep this in your catalogue

[sacred junco]

this is also similar to the above problem

by rearranging you can write (ad-bc)^2=0 and so

[quartz needle]

right

thanks

[sacred junco]

BTW you can read about AM GM ineq and Cauchy swartz ineq, they are simple and very powerful

[sacred junco]

happy to help mate👍

[quartz needle]

as for

show that if a + b = 1 and a^2 + b^2 = 7, then a^4 + b^4 = 31
i just solve this by calculating b^2, substituting it in, solving the quadratic, then substitutitng it again to solve for b, and then squaring both, right?

or is there any other method?

[quartz needle]

yeah, i'm aware of their usefulness, but haven't had the chance to learn them yet

a friend of mine recommended me this book called cauchy schwartz master class but apparently it uses calculus

so rip

[sacred junco]

i think this would require Binomial theorem

are you aware of it?

[quartz needle]

i recall learning it, but don't remember it anymore

i don't think combinatorics is required here, though

like, wouldn't my method work?

[sacred junco]

no not combinatorics i just need binomial theorem for expansion of (a+b)^4

[sacred junco]

yeah it definitely is correct

[quartz needle]

what's the point of yours?

faster?

[sacred junco]

it is just that sometimes extraneous roots creep in when we form quadratic equations so we have to be careful of that

[quartz needle]

what do you mean by extraneous roots?

[sacred junco]

ummm

let me think of an example to explain it just a moment

say we want to solve x=sqrt(x+2)

so we would probably square, rearrange and write x^2-x-2=0

and we get x=2,-1

but x should be positive because square root is always positive (since x= sqrt(x+2) so x must be greater than zero)

so x=-1 is called an extraneous root

[quartz needle]

got it

[sacred junco]

happy to help 😀

[quartz needle]

thanks

[sacred junco]

wdym? there are many people who have sqrt to be + or -
In such a case -1 would be a solution because:
x=sqrt(x+2)
x substitute for -1 (the supposed "extranious" root)
-1 = sqrt(-1+2)
-1 = sqrt(1)

and since the sqrt of 1 is plus or minus 1, the equation is true, because it is telling us it is equal to -1

[sacred junco]

square root, by convention is always non negative that is zero or positive

square root of 1 is +1 ONLY

but x^2=1 means x can be +1 or -1

[sacred junco]

square root is the inverse of squaring, and since there are always 2 numbers (positive and negative) that square to a positive, the square root must be plus or minus

It does not matter what we do by convention, it is plus or minus anyways.
using "convention" is like prioritizing an answer, that is bias in maths which we can't have

[stiff rivet]

this "prioritizing" is done frequently to turn sqrt into a function

one then distinguishes between the square root and solutions to some quadratic equation

if you dont do that notation like $\sqrt{x}$ doesnt really work, since it wouldnt be well defined

[sterile pumice]

Lochverstärker

[sacred junco]

why would anybody want to force sqrt to be a function

actually, that wouldnt even be sqrt anymore

sqrt is a multifunction, and there is no reason for one to turn it to a function, besides hiding answers to quadratics or higher

as a matter of fact, it is not only sqrt that is a multifunction, all roots have multiple outputs for 1 input

As for a good definition

a definition for sqrt is "find a number that multiplies by itself to give you the number under the root"

and since there are always + and - varients, it will always have 2 outputs

it is simply a consequence of the definition, that is not quite like other previous operations but it is simply something new that we dont like to experience

[stiff rivet]

[sacred junco]

square root is not multi functioned

for example plot y=sqrt(x) and y^2=x on desmos

square root is always positive and that is a convetion and this convention is utilised in framing some really good problems at competitive maths at school level

though yeah maths is not about problem solving only but i've been under this arena for quite some time this is what is going on so.........

[sacred junco]

Bruh, desmos arbituarily defines it to be positive

anyways, if it was a function, it would no longer be in inverse to squaring

btw, why the hell are we talking about square roots only

this applies to any root besides 0 and 1

and -1

even some super roots are multifunctions

[quartz needle]

show that for all natural numbers n, n^5 - n is divisible by 30
i got to
n(n-1)(n+1)(n^2+1)
so it has to be divisible by 3 and 2 (since it's divisible by 3 consecutive numbers)
but I still have to prove it for 5, and I'm not sure how would I go about doing so

[stiff rivet]

induction should work

there are nicer ways but induction should do

[quartz needle]

nicer ways?

i'd rather not use induction

[sage fern]

just hit it with modular arithmetic

it's trivial for n = 5k-1, 5k, 5k+1

so consider what happens for 5k+2, 5k-2

not even modular arithmetic

just... case-checking

and factorisation

[quartz needle]

k ty

[sacred junco]

(n-1)n(n+1) is always divisible by 6 you should just see if the expression is always divisible by 5

[quartz needle]

didn't write it because it's irrelevant, only 3, 5, and 2 matter

[stiff rivet]

yeah you can also just look at it mod 5 and check all cases, but its also not very nice
overall this is a finite problem and you can just check all cases

the "nicer" way is observing that n^5 - n = (n−2)(n−1)n(n+1)(n+2)+5n(n^2−1)

so its the sum of two numbers that are both divisible by 5

[sacred junco]

i think it really can't get shorter than that

nice job on that one👍👍

[sage fern]

meh, that's hard to see imo

[near carbon]

to prove that k in N is composite => 2^k - 1 is composite, can't you just say that 2^k - 1 = 1+2+...+2^(k-1) <=> 2^k = 2*( 1+2+...+2^(k-2)). but this expression is only non-composite if (1+2+...+2^(k-2)) is 1, which is true iff k = 2. but since k is composite, we have that k is not equal to 2, hence 2^k -1 must be composite?

[sage fern]

that works

[near carbon]

ok thanks

[quartz needle]

calculate all natural numbers a and b for which the following equality holds a^2 - b^2 = 24
a^2 - b^2 - 24 = 0
and i'm stuck

[sacred junco]

(a-b)(a+b)=24

now factor 24 into two parts

like 2x12, 3x8, 4x6 etc

and compare

like for example set a-b=2 and a+b=12 so on and just remember a and b are natural numbers

[quartz needle]

thanks

[sage fern]

ngl you shoulda been able to get that from when we did that other factorisation one

[quartz needle]

i was half asleep then, sry

on painkillers too

[sage fern]

mmmm

quality > quantity

[quartz needle]

yeah ik but i was being lazy with solving these questions ofr a long time

i was supposed to have been done with all that i wanted to do by the end of sept

[sage fern]

again, quality > quantity

you can't rush it

[quartz needle]

i know sry

[thorny trail]

Hi everyone. I have a homework question which I don't know how to start. It states: "Find a nonzero rational polynomial P(T) that has sqrt(5) + cbrt(7) cube root of 7 as a root". Any advice would help as I have no idea where I should start

I have discovered that Wolfram Alpha gives this polynomial, but I have no idea how it got this answer. Advice always helpful. Thank you!

[sacred junco]

i think the procedure is like this:

rewrite it as x-sqrt(5)=croot(7)

then cube on both the sides to elliminate the cube root from seven

now on RHS still has some terms with sqrt(5) so bring them all to one side

and then square so now the whole equation becomes radicle free

and now just rearrange and you would get this polynomial

[thorny trail]

oh interesting idea

I'll try that

Thank you for the tip

[sacred junco]

welcome👍

[thorny trail]

It worked. I really appreciate it @sacred junco . Thanks again for the help. The mistake that I made was in thinking that if I were to cube sqrt(5), I wouldn't get something simplifiable

[sacred junco]

Oo yeah that's a genuine one

anyways elementary number theory is all about approaches so you're all set now

[thorny trail]

I agree. Thanks again! Enjoy the rest of your morning/afternoon/evening

[sacred junco]

Thanks man( it's morning here). Same to you!

[quartz needle]

Show that for each k the number k(k+1)(k+9)(k² +1) is divisible by k.
By inspection it's clear that it's true for k 1,2,3,3,4,5,6,7,8,9,10, but I'm not sure how to prove it generally

[sacred junco]

Don’t overthink it

Why is it clear by inspection for those values of k?

[quartz needle]

Because when you substitute k for any of the digits, either of the terms becomes diviisble by5

[sacred junco]

What?

Ok never mind that

What does it mean for a number to be divisible by k?

I’m gonna go now but this problem is very quick if you know what it means for a number to be divisible by k and apply that definition to the scary looking number in the question

[quartz needle]

x | n => n = kx

[quartz needle]

Oh in that sense

I tried to do it this diff way because I recall there being some problem of this type in which.it wasn't possible

Sorry for asking this question

Should have looked at it for a bit

[sterile pumice]

marc

[vernal ibex]

Does the set S only contain 1 integer because a,b and n are constants, so b - na should be 1 integer only?

Or is n changing since it is every number which when multiplied by a > b?

[stiff rivet]

n is not a constant

in the construction of S it takes on every positive integer

[vernal ibex]

Yess.

Thanks.

[vernal ibex]

Is not m the smallest n here? Because the set consists of only b -na for different positive values of n, and m is the smallest of them?

[stiff rivet]

what

b - na gets smaller with larger n

[vernal ibex]

Yes it does.

Wait let me see.

So is m related to n in any way?

[stiff rivet]

yes

[vernal ibex]

Can you tell me how?

[stiff rivet]

m is the n that minimizes b - na

[vernal ibex]

Interesting stuff.

[stiff rivet]

but like

such an m does not really exist

[vernal ibex]

Can you please elaborate?

So.

[stiff rivet]

you look at the number b - na here

[vernal ibex]

m is the value which makes b - na the smallest positive integer in the set?

[stiff rivet]

so the numbers b-a, b-2a, b-3a, ....

[vernal ibex]

Yes.

[stiff rivet]

and if you assume this is always positive, there must be a minimal element

[vernal ibex]

Is this true?

[vernal ibex]

Yes.

[stiff rivet]

smallest positive integer, but yes

[vernal ibex]

Cool!

[vernal ibex]

So this is how m is related to n?

[stiff rivet]

sure

[vernal ibex]

Oooo.

Any other relation between them?

[stiff rivet]

no

[vernal ibex]

Cool!

Thanks!

[gilded breach]

Proof verification:

1. Prove that a set S of n elements has precisely 2^n subsets.
   Proof: Since the subsets are every combinations of the n elements, we get two choices while considering each element. We can either pick it or leave it. That's why there are 2^n possible subsets.
   QED.

2. Use question 1 to give a combinatorial proof of the formula:
   ${ 2^n = 1 + {n \choose 1} + {n \choose 2} + {n \choose 3} + ..... + {n \choose n} }$

[sterile pumice]

Autumn

[gilded breach]

Proof: In the right side, 1 is the number of subsets with n elements, ${n \choose 1}$ is the number of subsets with 1 elements, ${n \choose 2}$ is the number of subsets with 2 elements.. in this way, we get the total number of subsets of a set S with n elements. From exercise 1, we know the total number of subsets is $2^n$ for S. Thus, the statement in the question is proved.
QED.

[sterile pumice]

Autumn

[gilded breach]

Can anyone verify if the proofs are correct?

[stiff rivet]

you double count the number of subsets with n elements this way

[gilded breach]

I can't figure out how

[stiff rivet]

you said that $1$ is the number of subsets with $n$ elements, but the last summand is $\binom{n}{n}$ which is the number of subsets with $n$ elements as well

[sterile pumice]

Lochverstärker

[stiff rivet]

so you count those twice?

[gilded breach]

I saw that one too. The last one is the empty set

[stiff rivet]

what

[gilded breach]

The empty set can be a subset of set S, and it's only 1

Has my language created confusion?

[stiff rivet]

yes that is true, but its not what you said in the original post

if that is fixed, both proofs work

[gilded breach]

Okay thanks for seeing my proofs

[gilded breach]

I thought the proof would work even if I don't clarify that $n \choose n$ is the number of empty sets. I didn't specify the cases for $n \choose 3$, $n \choose 4$ etc., used dots to show they come next

[sterile pumice]

Autumn

[gilded breach]

Or do I have to specify for the empty set too?

[stiff rivet]

$\binom{n}{n}$ is the number of n element subsets by definition, so its weird to do it this way anyhow

[sterile pumice]

Lochverstärker

[stiff rivet]

actually you can just notice that $1 = \binom{n}{0}$...

[sterile pumice]

Lochverstärker

[stiff rivet]

so the sum is actually $\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}$

[gilded breach]

So (n 0) means the empty set?

[sterile pumice]

Lochverstärker

[stiff rivet]

it's the number of 0 element subset of a set with n elements

there is only one set with 0 elements, namely the empty set and it is a subset of every set

so \binom{n}{0} = 1

[gilded breach]

Alright, makes sense. It actually came to my mind before but I thought using n element set would be the same. I'm glad for being able to discuss, learned something new

[topaz ridge]

does it make sense to talk about $a(\mod n)$ as $n \to \infty$?

[sterile pumice]

gmod

[stiff rivet]

probably not in the way you are thinking

[topaz ridge]

then in what way?

[stiff rivet]

as a sequence (of numbers) it will be pretty boring

you can instead consider the sequence $(a + 1\bZ, a + 2\bZ, \dots)$ which is an element of the profinite integers $\varprojlim_{n}\bZ/n\bZ$ and this is then an interesting object

[sterile pumice]

Lochverstärker

[topaz ridge]

oh so the sequence ${a (\mod n)}_{n \in \mathbb{N}$?

[sterile pumice]

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

[topaz ridge]

I see, but what does the arrow under the limit represent?

tending to the front of the sequence?

0?

[stiff rivet]

its an inverse limit, its a category theory meme

[topaz ridge]

oh

[stiff rivet]

its some subring of $\bZ/1\bZ \times \bZ/2\bZ \times \dots$

[sterile pumice]

Lochverstärker

[topaz ridge]

currently learning about rings lol

didn't know subrings existed

makes sense tbh

[stiff rivet]

a subset that is also a ring

[topaz ridge]

yeah

[empty falcon]

hey for somthing like this:

2a^2 = 4a^4 (mod m)

is it ok to divide by 2 on both sides as long as m is never 2?

[abstract flax]

What do you mean m is never 2?

You can divide by 2 if 2 and m are coprime which means 2 has a multiplicative inverse

But think of it as multiply on both sides by the multiplicative inverse of 2, rather than dividing

[empty falcon]

thats fair

yeah i meant if m is always odd my bad

i have the smallest exposure to finite rings so please remind me why them being coprime makes 2 have a multiplicative inverse?

[empty falcon]

and also maybe explain exactly how thats relevant?

[abstract flax]

@empty falcon the gcd of two integers a,b is the smallest positive d that you can write as d = ax + by for integers x and y

[empty falcon]

right

[abstract flax]

So a has an inverse x mod b iff ax = 1 mod b for some x iff ax - 1 = by for some x, y iff 1 = ax + by for some x,y iff a and b have a gcd of 1

[empty falcon]

ok that i dont as well understand

[empty falcon]

particularly the ax - 1 = by part

[abstract flax]

That's the definition of congruence

[empty falcon]

oh i see it now

ok thanks good to know

[abstract flax]

👍

[empty falcon]

so then how would you use this to find that very inverse

[abstract flax]

Do you know euclidean algorithm?

[empty falcon]

ive barely been exposed to it

[abstract flax]

You use that

It allows you to find the gcd of any two numbers using division

Then you can reverse the equations

To write the gcd as a linear combination of your numbers

[empty falcon]

ok

[abstract flax]

So when you have 1 = ax + by then x is the inverse of a mod b

[empty falcon]

could you give an example for the inverse of 2 mod 3

[abstract flax]

That's a bit easy lol

[empty falcon]

yeah thats why i chose it

but using the algorithm

[abstract flax]

3 = 2(1) + 1

So 1 = 2(-1) + 3(1), then the inverse of 2 is -1 (which is the same as 2 mod 3), so 2 is its own inverse

[empty falcon]

wait how did you get 1 = 2(-1) + 3(1)

[abstract flax]

I subtracted 2(1) from both sides of the equation 3 = 2(1) + 1

Let's do inverse of 3 mod 35.

35 = 3(11) + 2
3 = 2(1) + 1

1 = 3 - 2 = 3 - (35 - 3(11)) = 3(12) + 35(1)

Then 12 is the inverse of 3 mod 35

[empty falcon]

ok

and this only works because you get gcd = 1 right ok awesome

[abstract flax]

Yes, if you don't get gcd = 1, there is no inverse

[empty falcon]

why might adding 1 to b and deviding by a not work tho

[abstract flax]

Huh?

[empty falcon]

so like (35 + 1)/3

12

3+1)/2

=2

[abstract flax]

It's a coincidence this works

Try finding the inverse of 9 mod 35

Oh nvm

Lol that works too 1 sec

Okay, for example

Inverse of 23 mod 35, won't work

This only works when you do 1 = ax + by, and y happens to be 1

[empty falcon]

how about adding up to what you need to

ummm

so like 23*2 = 46

thats like adding 11 then deviding

so it should be 2?

or no not at all

i see

[abstract flax]

What's the justification for doing this now? If you happen to add the right number then sure it will work lol

[empty falcon]

me seeing a pattern and seeing if it works lmfao

[abstract flax]

Euc Alg is best, or just trying all the options until you find the inverse lol

[empty falcon]

lol

ok

[sage fern]

23 * 32 = 736 = 1 mod 35

[empty falcon]

can something have more than one inverse

[abstract flax]

No

I mean, there are infinitely many integers that work

But they are all congruent

[empty falcon]

right

ok cool

[sage fern]

surely the key factor is that they're not coprime

[abstract flax]

I didn't mean it doesn't have an inverse, but that his way of finding it won't work there

[sage fern]

oh right

[empty falcon]

Ok another question
How would you go about proving that 1 is never congruent to 2a^2 mod 8k + 3

[empty falcon]

After some research I've found that this is the same question as "when is 2 a quadratic residue mod n"

It's a good thing I only care about n being prime

Because then I can use the result that 2 is only a quadratic residue mod p when p is +/-1 mod 8

So therefore my 8k + 3 is no problem if 8k + 3 is prime

[latent hare]

no

[sacred junco]

c=b , d=a is a counterexample

[inner crest]

set b=1

[gilded breach]

How to prove "The product of any n consecutive positive integers is divisible by the product of the first n positive integers" by using induction and the relationship $\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$ ?

[sterile pumice]

Autumn

[gilded breach]

I saw some inductive proofs, and I don't understand them.

If anyone knows how to prove it, can you share your proof?

[rigid thorn]

I need some help with a variation of stars and bars:

I want to find how to generate those tuples.

or preferably, count how many there are

in other words, I'm struggling to find the number of k-tuples which sum to n where order does not matter, which I believe is just the number of tuples there in that example. Stars and bars takes the number of permutations of those tuples, if I understand correctly.

$$
x_1 + x_2 + \dots + x_k = n ;; where ;; 0 < x_1 \leq x_2 \leq \dots \leq x_k
$$

[sterile pumice]

blorgon

[latent hare]

It's not that easy, see: https://en.wikipedia.org/wiki/Partition_(number_theory)

[wooden crater]

see if this helps

#proofs-and-logic message

[sterile pumice]

marc

[sterile pumice]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[sterile pumice]

gmod

[wooden crater]

what’s after tetration?

[topaz ridge]

I forget what its called

lemme find out

pentation

wait shit it exists

[wooden crater]

sweet

[topaz ridge]

but what about non-N inputs

do they even make sense

[wooden crater]

i was trying to figure out what like, H_5(3, 2) would be

[topaz ridge]

oh lol

[wooden crater]

cus i have just never pentrated? pentated?

lmao

💀

[topaz ridge]

but yeah idk I feel like negative n values would probably just be inverses of their respective operation

[wooden crater]

that sounds reasonable

[topaz ridge]

but I have no clue about non-Z values of n

[quick furnace]

penetrated

[wooden crater]

bro ann💀

[simple gyro]

What is number theory

[elfin sandal]

google

[sage fern]

the theory that numbers exist

[elfin sandal]

Dedekind cuts are elementary number theory confirmed \s

[gilded breach]

The study of properties of numbers

[gilded breach]

@wooden crater In your opinion, what is the difficulty level for this problem?

I'd like to hear others opinion too, if anyone is interested

[inner anchor]

pretty simple

you just need the fact that binomial coefficients are integers

[gilded breach]

I read a proof which uses the binomial coefficients are integers, but this one states using induction. I find this problem hard, perhaps I need to study more combinatorics

[wooden crater]

if math problems were hot sauce, mild

[cosmic sigil]

https://www.youtube.com/watch?v=O0wb12c1DN8

666 is a special number

happy halloween

[gilded breach]

lol

[noble pumice]

Idk if this is the right channel but here goes
You know how we use a base 10 number system, could base 1/2 be a thing?

[sage fern]

yes

[noble pumice]

What would it look like?

[sage fern]

just take base 2 and send the last digit before the decimal point to the last digit before the decimal point

swap the second last digit before the decimal point with the first digit after the decimal point

swap the third last digit with the second digit after the point

and so on

1 in base 2 is 1 in base 0.5

10 in base 2 is 0.1 in base 0.5

etc.

[noble pumice]

So like 100 in base 2 is 0.01 in base 1/2

Makes sense

[sage fern]

yes

[noble pumice]

Thanks

[sage fern]

np

the real fun is with repeating decimals in base 2

[noble pumice]

What about irrational numbers or do they make no sense?

[sage fern]

irrational numbers you would do the same thing, but it gets interesting

it's not even just irrational numbers

so say you have 1.010101...

this is 1 + 1/4 + 1/16 + ... = 4/3

but when you flip that into base 1/2, you get like.
...1010101

[noble pumice]

Yeah

[sage fern]

so it's fun

something something p-adic numbers? not sure

[noble pumice]

so could you have base pi for example

[sage fern]

i mean sure

it would be awful, but sure

have you heard of the golden ratio, phi

where phi = (1+sqrt5)/2

[noble pumice]

How would that work? Like use the approximation of pi or something else

Yeah I am aware of the golden ratio

[sage fern]

ok so a lot of what ppl say about it popping irl is a bit rubbish

but if you have base phi it has quite nice properties

idk off the top of my head what they are i just remember it's nice

[sage fern]

you would use the exact value of pi, but then you'd have to approximate all the other stuff

[noble pumice]

Did you read about it somewhere? If so I would like to look at that too

[sage fern]

wikipedia i think lol

[noble pumice]

Which wiki page was it?

[sage fern]

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