elementary-number-theory

[sacred junco]

oh the sequence is, what's the value of the counter at the place I'm at I see

[bronze breach]

Yep, you've got it

[sacred junco]

I think I might know some nice solution to this

[bronze breach]

I was happy since I came up with all of this independently, but I just now checked OEIS and of course they've already got it hahaha

[sacred junco]

hahaha nice

that's fun

[bronze breach]

What are you thinking?

[sacred junco]

I was thinking use some sine waves to bounce between counters or something but

it'll just boil down to the same thing ultimately

have you considered trying to generalize it to N counters?

[bronze breach]

Yes! That's where I want to go next

[sacred junco]

nice

I'll give it a try see what I can come up with

[bronze breach]

Sounds good, it'd be fun to compare the final functions

[sacred junco]

https://www.desmos.com/calculator/5hbxtri557

@bronze breach got it

[bronze breach]

I've almost got it

[sacred junco]

nice

[bronze breach]

@sacred junco Alright now I think I did it

[sacred junco]

cool

I'm curious to see how you did it but I guess you're trying to type up whatever monstrosity you got haha

mine's not nice that's for sure but it got the job done

[bronze breach]

To find the nth step of a sequence involving k containers, you can use:

There we go

And since it's the topic of the night, here we are as a singular expession:

That one's the monstrosity you predicted haha

[sacred junco]

I did some similar stuff

I used the ceiling function in the same way I think

but past that then I separated 'em apart using a fourier series which is something desmos could understand since it would probably spit errors at me if I tried a 0^0 thing

[bronze breach]

Actually, desmos handles that one perfectly well

https://www.desmos.com/calculator/2fzr1x3arj

[sacred junco]

fancy

learn something new every day I guess

[bronze breach]

What does your solution look like?

[sacred junco]

I linked it earlier

right when I tagged you

https://www.desmos.com/calculator/5hbxtri557

[bronze breach]

Ah I thought that was just a picture

[sacred junco]

oh no you gotta click the link not the picture or it does that

[bronze breach]

Your technique is really interesting. Where do you learn how to work with this?

[sacred junco]

with fourier series?

I just like PDEs

I'm basically like screening out the parts I don't want and leaving the parts in I do want

it might help to explain the l(x) I wrote in there

=tex \ell(x) = \frac{1}{A} \sum_{k=0}^{A-1} e^{\frac{i 2\pi}{A} k}

[sacred junco]

so basically what I did was make this simple series which is

=tex \ell(0) = 1

[sacred junco]

it's actually going to be A if you don't divide by A, so that's how I normalized it

every other integer value that's not a multiple of A will have the effect of basically putting them around a circle or regular polygon in an equal way so that they cancel out

think like, vectors in equilibrium

you could just as well have written this with modular arithmetic with a generator

at any rate I just take this thing which is now an indicator function at x=0 mod A

and then I multiply it by the stuff I want while having it 0 where I don't

my h function is just the opposite I'm really just doing 1 - l(x) so it just inverts itself

since most of my values are represented by this h function

hopefully that's like, mildly clear

oh I should have explained, since l(x) has only outputs of real values, 0 and 1, it's actually just the same as if I replace it with the real part of that

which is how I wrote cosine since I think desmos doesn't handle imaginaries

I didn't check

[bronze breach]

I wonder what the analogy to this process would be, given imaginary inputs.

"How many marbles are in the third cup after 5i steps"

Yeesh

[sacred junco]

wot lol

what's "i steps" look like

[bronze breach]

Hence the "yeesh" haha

I'm afraid I have zero experience with fourier series. I only know they're commonly used to approximate other functions.

[sacred junco]

yeah

it's hard to understand them without studying linear algebra if you haven't yet, definitely get on that

[bronze breach]

It's on my "to do" list 😅

[sacred junco]

I think the two most important things in math to study are calculus and linear algebra

and after a while they join together to form one subject lol

number theory is fun though too

either way you slice it, no matter what you do I think you can always find something from one field to use in the other though

basically each of these e^ikx terms are basis vectors for your space

and you're using dot products to find the projection of your function onto this space to get your coefficients

it's the same as a least squares problem in that you're approximating

but in math like, same thing goes for power series actually, your approximation can become exact

it's just a different basis in that case, {1,x,x^2,...}

[bronze breach]

I definitely prefer number-theory-style thinking over calculus-style thinking, if that makes sense

Although that may be because calculus feels like there's a higher barrier to entry before you can start working on proving things, whereas grade schoolers can consider problems such as "is any even number plus another even number also even?"

The sort of problem we've been working on now is the kind of thing I enjoy a lot

[sacred junco]

I know what you mean, I think each area of math sorta has its "ways of thinking" and it's hard to get into those separate mindsets

yeah I like this sort of problem style like you're saying too

I think I like to stay in the realm of math techniques that are like at around highschool level or so just cause the problems are more down to earth

I don't really care about counting rational points on elliptic curves or something like that personally

but I can be coerced into other stuff

yeah if you have some other problems you wanna work on I wanna try to do some

here's a problem I have enjoyed cause it's easy to say

find all prime triples that satisfy p^2+q=r^2

[bronze breach]

I think "highschool level" can be so appealing because, since you already have so much knowledge surrounding the question, you can really delve into it and appreciate its depth

The difficulty comes from the creativity required, not from needing to learn a ton of new material

find all prime triples that satisfy p^2+q=r^2

Uniqueness is interesting. I haven't quite gotten my hands dirty as far as rigorously proving that a set of solutions is the complete set of solutions for any problems yet

[sacred junco]

this problem only seems harder than it really is

[bronze breach]

Hm

[sacred junco]

you have any problems or whatever for me to work on while you're thinking about that one

or one you want to work on together while I'm here haha

[bronze breach]

I have a quick one I did a little while ago:

Let's say you're trying to find the square root of 12.

12 is closer to 4^2 than 3^2, so you assume that rounding sqrt(12) will give you 4, not 3.

Indeed this is correct. However, it doesn't apply to all real numbers. Which ones?

[sacred junco]

oh what I'm a bit lost on the reasoning going on

oh oh for some reason I thought you were using the factors of 12 = 4*3 and looking at sqrt(12) it being closer to 4^2 or 3^2 because of this lmao

just too much number theory wew

ok this is a fun problem nice

[bronze breach]

I have a handful of fun problems I've come up with/come across.

Some of them were really involved, at least for me haha. You know a problem is good when the answer doesn't hit you until months later

[sacred junco]

yeah definitely

I don't think a problem has to be hard or anything to be good that's for sure

hmm I came to some kind of contradiction but I will keep working on this problem you gave

this problem is not like supposed to be tricky is it

like since you say "all real numbers" when of course it doesn't work for negatives

[bronze breach]

cough
I hadn't thought of that, I immediately defaulted to positive real numbers when working on it

[sacred junco]

ok cool I did too but then I just came to a result that it's never true but, I probably just messed up my work but

just thought I'd double check haha

I guess while I'm on the subject, I know 0 and 1 are sorta weird

so if it's something like that I suppose I'll consider that

but I'm thinking there is a more broad range of possibilities

like I think many fractions are closer to 0 than 1, but if you square root them they get closer to 1

haha I'll just work on the problem I guess don't give it away I need something to do tomorrow while I'm bored at work

[bronze breach]

Sure, I also dislike spoilers

Speaking of, there's one unsolved problem I have. So far it's my longest-running one. I'm not sure if you were the one who saw it last time, but I'll bring it up in case you weren't.

This time I'd like to request no spoilers from you haha

Oh wait there's a formatting error

Actually nevermind that

Given any n, there exists some k where s >= 3 results in a fractional output from f(n, s)

So, for any n, all values of s before that k will result in an integer output of f(n, s). Everything after is a fraction irreducible to an integer

n and s are natural numbers

[sacred junco]

oi I don't know

[bronze breach]

Now, I am 99% sure that statement is true.

If it was false for some n, then that n would disprove the collatz conjecture haha

[sacred junco]

aha

[bronze breach]

Since this is a description I came up with to model the behavior of a number which tends towards infinity instead of 1 when placed through the rules of the CC

[sacred junco]

oooh interesting

well I don't think I could spoil it even if I tried

well, unless I found some way that kinda like invalidates the method or something maybe

=tex c(n) = \frac{n}{2} \frac{1+(-1)^n}{2}+(3n+1) \frac{1-(-1)^n}{2}

[stoic basin]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[sacred junco]

unrelated exactly but just a way to write the collatz function without piecewise since we were talking about it earlier

[bronze breach]

I can do you one better:

Collatz function defined for reals and complex numbers

[sacred junco]

haha that works too, but it's kinda like what I did to make my previous solution work too

really what you've written is kind of like the version before I simplified it down

[bronze breach]

I wish I wrote that

It's by someone who created a fractal based on the collatz function:http://yozh.org/2012/01/12/the_collatz_fractal/

[sacred junco]

http://prntscr.com/kpnzz6

ahh yeah he's doing how I was working the last problem too, it's the exact same method, maybe he has more stuff about solving things with this method since you were looking for it

do you know anything about p-adic numbers?

[bronze breach]

Those are the ones like ...9999999, correct?

[sacred junco]

yeah that's -1 in the 10-adics yep

I think you might appreciate one way of thinking about them that might be sort of how you think currently

[bronze breach]

Also sidenote but my sister painted me the fractal and I thought that was awesome lol

[sacred junco]

these kinds of things come up sort of naturally when trying to solve a congruence at higher powers of the modulus

oh hey that's really cool

[bronze breach]

:D

And by all means, please share

[sacred junco]

alright so maybe I'll try to show that one with ...99999

but it's kind of a boring example

Reminds me of

t!wiki bers slice

so let's say you want to write -1 mod 1000 as some natural number in the range [0,999]

well, it's kind of obvious that -1=999 so it's not exciting haha

but we could actually keep going to higher digits and we have -1 = ...9999 mod 10^k for all k

so the 10-adic number is just like saying, let's just make that the number itself

and now we can always "round it off" to get back any regular old congruence

more interesting examples exist though that you couldn't get normally

like for instance let's work on finding solutions to say

=tex x^2 = -1 \mod 5

[sacred junco]

so we can find some solution to this, and this corresponds to getting the first digit of a number in base 5

but this definitely won't correspond to a real number

since x^2=-1 is what we're looking at

[bronze breach]

(Sorry to interrupt... is the answer to your problem all pythagorean triples where a, b, and c are prime?)

[sacred junco]

nope

Oh I was done I think actually I guess I realized what I was going to explain some work I had done at one point

was I extended the collatz function to all 2-adic integers, then integrated the kth iteration of it over the entire set of 2-adic integers

and then took the limit as k-> infinity

I actually forget what the result was past this, since I wasn't actually interested in the collatz conjecture I just was using it to try to practice since it seemed like something I could do while learning how to integrate p-adic functions lol

but I think it would be kind of long to explain haha

[bronze breach]

"Integrating p-adic functions" sounds like a really interesting idea

I have to say, I hadn't realized the p was a variable. This is all new now

I just assumed it was a name for numbers that were "infinitely to the left"

So I'm not sure what you mean when you say p-adic functions.

And with something like x^2 = -1 being i, how is mod defined on imaginary numbers? What is the remainder of 5/i?

[sacred junco]

oh yeah there's a p-adic field for every prime

but if p isn't prime, you get a commutative ring with zero divisors

well the thing is, when you look at p-adic numbers you have to go back a step

and think of them as being something of a fork in the road

you are basically at the rational numbers, and then you can either make real and then make complex numbers

OR you can pick a prime and make p-adic numbers

so the solutions to x^2=-1 you get in the 5-adics is not the same as the solutions you get to x^2=-1 in the complex numbers

you could call it i though, as long as you understand you're in an entirely different field

the p-adics are not ordered on a number line, they're more like in a fractal structure, they can be associated to being the leaves of an infinite number tree

I guess it would help if I explained how you describe distance between p-adic numbers but

but maybe that'd be silly at the moment, I can show you how to compute digits of "i" if you're calling i to be one of the solutions to x^2=-1 in the 5-adics

so you do it one digit at a time, this is something called Hensel's lemma btw I can show you how to prove that later but the algorithm is what's useful to you

first off

x=2 will work

since 2^2 = -1 mod 5

now we add on another digit in base 5 and solve it again,

=tex (2+5a)^2 = -1 \mod 25

[sacred junco]

(I'm working in base 10 but you can see how a will be a digit in base 5 I hope)

so you square this and rearrange a bit

=tex 4 + 20a =-1 \mod 25 \ 5(1+4 a) = 0 \mod 25 \ 1+4a = 0 \mod 5 \ a= 1 \mod 5

[sacred junco]

and so here we have another digit

and so on, the lemma says you can always repeat this indefinitely, it's actually an induction argument that is proved in complete generality, supplied you can show some base case

anywho, it stands on its own as far as a method for using the chinese remainder theorem with it to boil down diophantine equations but I really rarely do that but Iw ouldn't mind trying to do this more

I guess you can then try to get the next digit in base 5, you'd have to do this btw

=tex (2+15 + b25)^2 \equiv 1 \mod 125

[sacred junco]

and then work through this to solve for b

you can see if you evaluate this mod 25 or mod 5 it collapses back down to the previous solutions

so you're never messing up your established digits

that's one thing that makes doing math in p-adics so clean, you carry to the left, while real numbers have this terrible issue where digits to the right affect those to the left

if your math only includes rational numbers, since the p-adics contain all rational numbers, you can just as well work in them instead

so back to your question if you wanted to evaluate 5/i you'd actually be able to do 5*i^{-1} since that would just be solving this same thing but starting with 3 as the first digit

actually nice thing is

well w/e I'm rambling lol

[bronze breach]

Reading that 5 times was a trip

But then it clicked

So as we continue to go up through powers of 5, we can find more and more digits for what our 5-adic i for x^2 = -1 is?

Does the number have infinite digits? Would any/all have infinite digits?

It seems like the definition is kind of forcing infinitely-sized numbers here, unless I'm misunderstanding

But I don't really see a case where you'll hit, say, 1231 and be done. Best case is something like ...00000001231, maybe?

[sacred junco]

yeah it does, and generically all p-adic numbers have "infinite digits to the left"

but they could be 0 too nothing wrong with that

yeah it does get you infinitely large numbers if you use the absolute value to measure how big your numbers are

so that's what I sorta skipped out on earlier when I decided I'd give you that concrete example just now

that fork in the road between going to real numbers means taking rational numbers and defining the absolute value you know and love

but the p-adic route means defining the absolute value in a different way

a way that's consistent with this sort of thing because it's useful to us

there is a sense in which x^2=-1 mod 5^n has a coherent solution for arbitrarily large natural number n

but this number just isn't a real number

so ok, here's how we define the absolute value for rational numbers

first write your rational number with the largest power of p factored out and relatively prime numbers a,b

=tex x = p^k \frac{a}{b}

[sacred junco]

generically speaking, k is just an integer

just meaning to say that cause it could very well be negative no problem

=tex |x|_p = \left| p^k \frac{a}{b} \right|_p = \frac{1}{p^k}

[sacred junco]

the subscript p is just to clarify, but if you're working with them normally you might just omit it cause it's just one extra thing to write and it's not that ambiguous at times

this seems kinda silly, like when I first saw this it seemed cheap, like ok you just throw away the whole number basically

but it's actually surprisingly full of structure that it makes a fractal when you use it to define distance in this very simple way

=tex d(x,y) = |x-y|_p

[sacred junco]

in fact you get a metric space outta it and then you can go on to do stuff like find the completion of this to make the p-adic numbers just like you can do it with the regular absolute value to get the real numbers

anyways it's quite strange but you have a lot of interesting properties like a stronger triangle inequality called the ultrametric inequality

it'd be good if I drew a picture of what this metric actually looked like so you can get a sense of distance

[bronze breach]

Ahh I have a lot of questions but I really need to head to bed

Could we continue another time?

[sacred junco]

sure

[bronze breach]

Also I think I got your earlier problem:

There are no triplets satisfying this. If there were, p^2 + q would have to be a prime number, but also a perfect square. This is impossible

[sacred junco]

oh why would this have to be a prime

there is at least one solution I'll say that much

let me make sure I said the problem right, I did say p^2+q=r^2 right

[bronze breach]

p^2 + q = r^2

r = sqrt(p^2 + q)

r is prime, thus sqrt(p^2 + q) is prime

And here's where I realize I was wrong and p^2 + q doesn't necessarily need to be prime. Gosh darn it```

[sacred junco]

haha]

I'll think about your problem with sqrts that's a good one

[bronze breach]

Lol alright. Feel free to ping/dm me when you get it

Until then, have a good night. Thanks for the chat :)

[sacred junco]

yeah later

[digital kettle]

hey

(this is probably more of algebra than nt)

oh wait

oops nvm

[sharp gulch]

#prealg-and-algebra

[digital kettle]

(didn't read the question properly)

Anyway, the question was

Let p(n) equal the product of the nonzero digits of n. S = p(1) + p(2) + ... + p(999). Find the largest prime factor of S

Have fun! (even though it is an aime problem, it's PRETTY easy)

[fossil junco]

cool im totally not gonna brute force through AIME solutions for it

[odd nexus]

anime problem? @digital kettle

oh mb i misread

[sacred junco]

47?

@digital kettle

[digital kettle]

nope

[sacred junco]

shoot

@digital kettle 103?

wait is that prime

ye should be 103 i think

[digital kettle]

ye

[sacred junco]

I think I figured out a way to solve it in general

hmm I got 23 but I didn't really think too hard, maybe I'm wrong if the answer is supposed to be 103

my method is probably flawed

[wild zinc]

hmm I've got an interesting approach maybe

if there's a way to do something nicely

which there may very well not be

yeah on second thoughts it's going to be just as hard :^)

I was thinking consider (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)

but that also doesn't work because it doesn't include any where two or more are the same

[mossy folio]

@sacred junco you're right

it is 103

[sacred junco]

the way I did it was this, tell me where my mistake is

I defined S(n) = p(1)+p(2)+...+p(n)

then I recognized

=tex S(999) = S(99) + 1S(99) + 2S(99) + ... + 9S(99)\ = S(99) (1+910/2)

[sacred junco]

really in general,

=tex S(10^n-1) = S(9)S(10^{n-1}-1)

[sacred junco]

anywho this boils down to then

=tex S(999) = S(9)^3 = (1 + \frac{9*10}{2})^3 = 46^3

[sacred junco]

so that's my reasoning for 23

[wild zinc]

S(999) != S(99)+...

because 101 counts as 0 rather than the 1 it does in the first 100

so I guess it's S(999) = S(99) + (S(99)-S(9)) + (2S(99)-S(9)) + ... = 46S(99)-9S(9)?

idk

wait no

product of non-zero

I'm dumb

I'll have a think and see if I can see why this doesn't work

[sacred junco]

I left out

S(999) = S(99) + p(100) + 1S(99) + p(200) + 2S(99) + ... + p(900) + 9S(99)

I wasn't counting some points

because I was sorta assuming it had started at 0

=tex S(10^n-1) = S(10^{n-1}-1) (1+\frac{910}{2}) + \frac{910}{2}

[sacred junco]

the p function has this really nice property, p(235) = 2*p(35)

so if you imagine like S(999) being chopped up into these runs of 100 parts, specifically,

p(201)+p(202)+... + p(299) =2*[p(1)+p(2)+... + p(99)]

is one such chunk

anyways I guess I should compute my solution now that I found out I was off by 1 lol

but actually I'd like to just find the general solution as much as possible because, like, why not lol

.-. I think ur ovefconplicating it

I've already solved it and now I'm generalizing

don't you enjoy math?

join me

muwahaha

[solar vale]

i actually think this approach is elegant

joining @sacred junco

[sacred junco]

whoooo

hmm it's hard to say though how to generalize this since I was kind of thinking of doing an infinite recursion on this thing but that will end up kinda sorta diverging

[solar vale]

well forget about p(999) if you want to generalize lmao

[sacred junco]

yeah maybe I should at least make sure it does get the right answer now

=tex S(999) = 45+46(45+46*9)

[solar vale]

it should now

[sacred junco]

factor out the 9 and throw it away since I think 3 isn't the biggest prime factor

5 + 46*5 + 46^2

[solar vale]

lmao

[sacred junco]

hmm is there any nice way to solve this other than to just add it

doesn't seem to be coming up to the right answer 2351 is not divisible by 103

I don’t remember but I don’t think your S(999) is right

S(99) = 45 + 45(46) = 45(47)
S(999) = 45(47) + 46(46) + 2(46^2) + 3(46^2) + ... 9(46^2)
= 45(47) + 45(46^2) = 45(47+46^2)

=pup prime factorization of 47+46^2

[stoic basin]

Do more with Wolfram|Alpha pro

Query made by

@hybrid elbow

[sacred junco]

p(100) + p(101) + ... + p(199) = 1 + p(101) + p(102) + ... + p(199) = 1 + S(99) = 1 + 45(47) = 46^2

This was AIME 1994 #5 if anyone’s curious

[wild zinc]

well I figured it out over dinner but I see you also figured it out :^)

[sacred junco]

xd

op

also imo this problem isn’t NT

[wild zinc]

it's a little bit number theory I guess

[glad nacelle]

@left blade we can work here

What questions do you have?

[left blade]

i jsut had

[glad nacelle]

(my number theory is very limited so no guarantees I can actually solve this question)

[left blade]

for part a, i cant see how to show one direction of this iff

in particular

how to show

$$3|g(\alpha)\implies \bar{f}|\bar{g}$$

[left blade]

the other direction i got

at worst i can go to office hours, its nbd

but ive worked on it long enough where i should prolly just move on

oh f is the minimal polynomial of alpha, btw
i think thats just non standard notation

[glad nacelle]

Ah okay

3 divides g(alpha) meaning in O_K, right?

[left blade]

in Z[alpha]

so g(alpha)=3beta where beta is some element in Z[alpha]

[glad nacelle]

Alright, hmm

How does the other way go? Maybe that'll be a hint at the method

[naive idol]

Hi. Let a,b be coprime positive integers, N>ab. Please how to prove there are positive x,y such that ax+by = N ?

(also sorry to interrupt you with trivial stuff)

[glad nacelle]

Hmm, it'd be nice if we could conclude \overline{g}(alpha) = 0 and make some case to the effect of, \overline{f} is the minimal polynomial of alpha in F_3(alpha)?

@left blade my NT is a bit low level for your problem but if I had to guess how an argument could go, it may be along those lines. Sorry that this is probably not that helpful

[left blade]

Nah that’s pretty much what I had been thinking but

No clue how to show it

No worries this is what office hours are for

[glad nacelle]

@naive idol you can get a solution to ax + by = N where x and y might not be positive, and then use a seed, keep adding and subtracting by ab

So if you have a solution, then consider a(x-b) + b(y+a), or some variant thereof

You can show that this'll eventually make things positive

[naive idol]

God I suck so much at elementary n.t. it's unbelievable

Idk man 😦 Let (x,y) be any solution. Let m be smallest integer such that x + mb >= 0

For the love of god can't prove that y-ma >= 0

No idea how to use the fact that N>ab either

AH

fuck me. I'm pretty sure it's not normal for it to take so long to do

(I solved this shit btw)

I've got a question though - what books helped you become good at elementary number theory?

[glad nacelle]

Uh, I dunno if I'm good but

Weil Number Theory for Beginners

It's terse but I read the first few chapters with some friends and did a bunch of problems

[naive idol]

I got the impression that all you have to do in order to be good at elem number theory is to be smart

Nothing can help a dumb person to become good at it, no matter how many exercises he does

[glad nacelle]

And that helped. This problem was there and we were sorta talking it out until we sorta figured this out. Turns out it's a generally important idea, solving diophantine equations with a seed

Uh, there's some creativity involved but I've got negative creativity tbh

You eventually learn a bag of tricks

Then again I'm still new so maybe you do need to be real creative and I'm doomed, idk

[sacred junco]

@naive idol all u need is to thoroughly understand the basics

And solve a shit ton of problems

There isnt anything like a gift here

[autumn swift]

confused on logic

apparently (PvQ) -> ~R is false if all those are true?

how can that be possble?

especially since it's the original statement. How can the original statement be false when all the variables are true?

I know that when R is true it turns to false

But the original statement says (If p or q is true, then r is false)

[ebon frigate]

That's a good interpretation.

So if P and Q are true, and R is also true, then that statement is wrong.

ie, it is false.

[autumn swift]

The statement is

If that's the original statement, how can it be false?

[ebon frigate]

Not sure what 3-connected means.

But a regular graph with two vertices is planar.

A regular graph with 3 vertices is also planar.

[autumn swift]

ok? how does that make the original statement able to be false?

I didn't think an original statement could be false

is r planar or not planar?

[ebon frigate]

The whole statement is false.

I'm not sure I understand your confusion here.

[sacred junco]

Huh how is this NT

[ebon frigate]

An If-Then statement can be false.

Shush, shrom.

[sacred junco]

.>

[autumn swift]

But I don't see how it can be false if it's the statement initially given

[vast vessel]

@autumn swift note that A -> B is true if A is false as well

[autumn swift]

i know that

[left blade]

what happened to denmark

i was gonna tell him the answer to the problem from yesterday

@autumn swift an implication is false when the antecedent is true and the consequent is false

ie T->F

and true otherwise

[glad nacelle]

@left blade hello

[digital kettle]

if

$$n = \prod_{i=1}^k {p_i}^{\alpha_i}$$

[digital kettle]

is it true that

$$\phi(n^2) = n\phi(n)$$

[digital kettle]

or am i just making some silly mistake?

[vast vessel]

That would imply that holds for every perfect square tho

[digital kettle]

um

so?

because

$$\phi(n^2) = n^2 \prod_{m=1}^{k} \left(1- \frac{1}{p_m}\right)$$

[digital kettle]

from the definition

<@&286206848099549185>

[vast vessel]

Did you even try plugging in some values?

e.g. n = 5?

[digital kettle]

lol no

let me

[vast vessel]

[digital kettle]

(it just seemed obvious)

[vast vessel]

🤷

[digital kettle]

works for n=5

[vast vessel]

oh yeah it holds

👌

[digital kettle]

nice

thanks

(for what tho)

[vast vessel]

k

idk either

[digital kettle]

at least im confident now

[vast vessel]

r u tho

[digital kettle]

at least more than before

hmm

if we apply FLT

[vast vessel]

r u tho

[digital kettle]

we can say that

oh wait nvm

im just stupid

I was just rewriting FLT for a special case

[analog kraken]

Let $${a_n}$$ be a sequence of real numbers with $$\lim_{n \to \infty} {a_n} = L$$, and define $${b_n} = \frac{{a_1}+...+{a_n}}{n}$$ Prove that $$\lim_{n \to \infty} {b_n} = L$$

[analog kraken]

can anybody do this?

[sacred junco]

Is this number theory tho

Also look for proof sketches of Cesaro's theorem as mentioned.

[somber rampart]

You bound the a_n and throw away a finite number of terms

Essentially

[sturdy dirge]

$$
\forall \epsilon>0, \exists n_0\in\mathbb{N}, \forall n\geq n_0,\newline
|a_n-l| < \epsilon,\newline
\left(1-\frac{n_0-1}{n}\right)(l-\epsilon)+\frac{1}{n}\sum\limits_{k=1}^{n_0-1}a_k < \frac{1}{n}\sum\limits_{k=1}^na_k<\left(1-\frac{n_0-1}{n}\right)(l+\epsilon)+\frac{1}{n}\sum\limits_{k=1}^{n_0-1}a_k\newline
$$
By limit:
$$
l-\epsilon<\lim\limits_{n\infty}\frac{1}{n}\sum\limits_{k=1}^na_k < l + \epsilon
$$.

[sacred junco]

Nice texing my dude

[dusky egret]

"Show that the product of three consecutive numbers is divisible by 504 if the middle one is a perfect cube", I'm a bit stumped here

504 = 7 * 3^2 * 2^3 , I know that if I have n(n+1)(n+2) at least one of those is divisible by 2 and one should be divisible by 3 but that's as far as I've got

[wild zinc]

look at the possibilities for n^3 when considered mod 7, mod 9

mod 8 there's a nice trick

[dusky egret]

hm alright I'll give it a go

apparently also if (n+1) is a perfect cube then one of n(n+1)(n+2) is divisible by 7

but that's trial and error I have no idea how to show it

[wild zinc]

fermat's little theorem

also

I'd suggest writing the numbers as n^3-1, n^3 and n^3+1

[dusky egret]

ah let me see

[wild zinc]

mod 8, if n is even then n^3 is divisible by 8, otherwise one of {n^3-1,n^3+1} is divisible by 2 but not 4, and the other is divisible by 4

if we consider their product mod 7

we get n^3(n^6-1)

by fermat's little theorem, either n \equiv 0 mod 7, or n^6 \equiv 0 mod 7, and hence n^3(n^6-1) is always divisible by 7

in fact you can do a very similar thing mod 9 using euler's totient theorem

since phi(9) = 6, where phi is euler's totient function, you have

3 | n (giving 9 | n^3), or n^6 \equiv 1 (mod 9)

and hence you always have n^3(n^6-1) is divisible by 9

[dusky egret]

okay so I got the remainders, I can see the 8 trick thing

I don't follow when you say " one of {n^3-1,n^3+1} is divisible by 2 but not 4, and the other is divisible by 4"

if n³ is divisible by 8 then n³ is even so n³+1 and n³-1 should be odd ?

[wild zinc]

that bit is in the otherwise, as in when n is odd

[dusky egret]

oh okay

[wild zinc]

when n is even the n^3 gives you the factor of 8 so you don't need to worry about it

[dusky egret]

oooh okay

okay I just did the mod 7 one that was nifty

[wild zinc]

there's technically a way to do it all without fermat's little theorem or euler's totient theorem

[dusky egret]

oh and for 9 it works as well

yeah it's fine I know those two pretty well

thanks that made sense!

[wild zinc]

nice, that simplifies it a lot 😃

np, if you have any other number theory stuff feel free to send it my way

[dusky egret]

Well since you said so, I need to find the remainder of dividing $$\sum_{i=0}^{2p} i^{p^2 - 1}$$ by p (being p a positive prime)

[dusky egret]

now; I found out that that summatory only gives you either 1s or 0s (terms not result) (depending on whether i is coprime with p or not) so it's just a sum of 1s , howevr I couldn't find out how many 1s I'm adding

[wild zinc]

i^(p^2-1) = (i^(p+1))^(p-1) \equiv 1 (mod p) unless p | i, in which case, 0 (mod p). I'm guessing this is what you did?

[dusky egret]

yes I did fermat

[wild zinc]

so i^(p+1) is coprime with p exactly when i is, so we just have 0 for i = 0,p,2p and 1 for all other i

[dusky egret]

makes sense

hum

[wild zinc]

so the sum is the same as 2p+1-3 = 2p-2 = -2 mod p

[dusky egret]

yeah you're right

you've got 2p+1 terms and only 3 are 0s

although the python script I wrote says the value for p=11 is 4, and it should be 9 :S

maybe I wrote something wrong

[wild zinc]

wolfram says 9 so I think it's fine

[dusky egret]

So I have an excercise which says:
4x ≡ 2 mod 9
3x ≡ 5 mod 14
3x ≡ 1 mod 20
Find x mod 2520

I started by solving each equation in order (so for example from the first one I get x = 9k+5 so now on the second one I have 3(9k+5) ≡ 5 mod 14, and then on the last one I get 3(126j+91) ≡ 1 mod 20 which eventually gets me to x = 2520k + 2107 so x ≡ 2108 mod 2520.

This approach usually works but I have a bad feeling here because 9, 14 and 20 (factors of 2520) are not coprime between each other so Chinese Remainder Theorem might not quite apply and I have to check something else.

Any ideas?

[blissful venture]

Nice!

Chinese Remainder Theorem

Convert them to their prime component equations

I don't exactly remember but there was something about changing to modulo prime exponents

As long as they are co prime you can use Chinese remainder theorem

you have a mod 4 and mod 5 eqn from the third

and a mod 7 from the second

you can combine those to get mod 140

[dusky egret]

yes

[blissful venture]

then combine with mod 9 to get 1260

[dusky egret]

so I'm missing a 2 now right?

[blissful venture]

I feel there are multiple solutions here 🤔 let me see if I did something wrong

[dusky egret]

I think I've seen something similar to this once and you had to divide on two braches and solve for each one but I can't remember how

[blissful venture]

Give me a couple of minutes

Just got some pen and paper

[dusky egret]

okay thanks

[blissful venture]

Ok, so I got x is congruent to 347 mod 1260

So can be either 347 or 1607 mod 2520

Let me check if those work in the original equations

Alright seems to fit

[dusky egret]

why either two of those?

depends on the modulo 2?

[blissful venture]

Because the given equations only guarantee that x is congruent to 347 mod 1260

yeah, we don't have a value modulo 8 to fix x

[dusky egret]

okay then thanks

I'll check if I can get to that point

[blissful venture]

Cool then, I might not be on, but feel free to tag me, and I'll take a look

[dusky egret]

oh crap I arrived at x = 32 mod 315 🤦

so first I did mod 9, then mod 7 then mod 5

and now I have an 8 left

sooo I have to check what happens for x for each modulo 8?

[blissful venture]

Wait, you can't have an 8

you can only have a 4

you don't have an 8 to begin with

That's why you end up with 1260

[dusky egret]

oh I multiplied the 4 and the 2 idk why I thought that'd work

hmmm I got 32 mod 1260 though

oh small typo

Okay I got to 347 mod 1260

Okay I got 347 mod 2520

the case when 3x = 0 mod 2 was absurd

so it had to be 3x = 1 mod 2, then that's the only solution?

but you said that 1260 also fits?

[blissful venture]

@dusky egret

Okay I got to 347 mod 1260
Great! So now you can see that you can only confirm a solution mod 1260, so if you want a solution mod 2520, you can either have 347 or 347 + 1260, since they would both fit the criteria mod 1260.

[dusky egret]

Sorry I don't see why. I get that 1260+347 is 0 mod 1260 but I don't see why that relates to 2520 why did you decide to add those?

[blissful venture]

If you have x = 1 mod 1260, that would mean x = 1 mod 2520 is a solution

It would also mean x = 1261 mod 2520 is a solution

Since x = 1261 mod 2520 => x = 1261 + 2520*k => x = 1 mod 1260

That's basically my reasoning in this

[sacred junco]

@dusky egret the fornicating falcon has solved it

Slide in the dm's ill send u the solution

[sacred junco]

Nvm..horse chan already nailed it.

@blissful venture props to u

[wicked talon]

=pup 16000/11

[stoic basin]

Do more with Wolfram|Alpha pro

Query made by

@wicked talon

[blissful venture]

Thanks!

[sacred junco]

=pup 22/7

[stoic basin]

Do more with Wolfram|Alpha pro

Query made by

@left bobcat

[delicate wyvern]

are there any patterns to finding the divisibility of large integers?
like integers that are 6 or 7 digits?

[royal zodiac]

Yes, there are lots of sources that talk on divisibility

It's usually looking at the last digit or the sum of the digits themselves and studying the divisibility of that

[delicate wyvern]

i understand the divisor rulles for integers 2 through 10

but beyond that, are there any patterns?

like factoring 6886 to find the largest prime factor

or do you literally have to divide mechanically beyond 10 to get to the quotient?

[royal zodiac]

Prime factorization then just knowing the prime which satisfies that by checking divisibility of smaller primes etc. and moving up to find a supremum. Something beyond I would say is modular arithmetic properties/theorems relating to modular arithmetic

[delicate wyvern]

modular arithmetic eh?

okay, cool, thank you

[royal zodiac]

No problem

[delicate wyvern]

i was looking for hours today to figure this out and you solved it in 10 seconds QQ

[royal zodiac]

You can probably do better than me tbh so keep going further

[sacred junco]

is it possible to have a unique minimal element that is not a least element of a set?

[swift shard]

@sacred junco
Take (0, 1)
There is no smallest element, but it's bounded below by 0.

And is not bounded below by any number greater than 0

[blissful venture]

What would be the fastest way to take modular square root?

$$x^2 \equiv 7 \pmod{101}$$

[blissful venture]

Say, something like this

[glad nacelle]

I at least know how to prove existence or lack thereof quickly enough

[wild zinc]

I think I remember reading at some point that this is a hard problem for prime moduli

[sacred junco]

Cant we assume that x =y mod 101

And then sqaure this eqn

And we can subtract the two

Just leaving y

So 0=y^2-7 mod 101

No sorry

[wild zinc]

just leaves you where you started

[sacred junco]

7-y^2

Yess.

But

If we subtract 1 from both sides

We can use wilsons theorem

So 100!=-1 mod 101

Hence y^2-8=100!

?

[wild zinc]

that's true but I don't see how it helps

[sacred junco]

Yeah me neither.

How would u approach this senpai

[wild zinc]

please don't call me that :^)

[sacred junco]

Lol ok.

[wild zinc]

also idk how to do this quickly

[sacred junco]

Well i dont know how to do it at all

[wild zinc]

do you remember the discussion about finding the possible values for f(n) mod p?

[sacred junco]

Sorta

Oh..i see

No but wait.

Its a prime..

Defining a function.

[blissful venture]

I remember something about legrange's symbol, but that seemed slow

If anyone figures something out, please tag!

[sacred junco]

Idk how to do it without the time boundage

@blissful venturehow do u methodically solve this

[blissful venture]

I saw some c++ code that referred to the Toneli-Shanks Algorithm, but not sure what exactly it's doing.

[sacred junco]

Omg..how many theorems do u know

[blissful venture]

not many

[sacred junco]

My x^2 is coming out to be 108

[blissful venture]

Hm, I still have issues understanding this in general

[sacred junco]

Well..in this case..

The factorial of a number is equal to the nunber.

Like 100!=100..all in mod p

So if we say f(x)=x!

I summon u @wild zinc

@blissful venture whats the answer if i may ask

[blissful venture]

I don't know, that's why I asked

[sacred junco]

Does x need to be integer?

[blissful venture]

Yeah

Oh, there are no solutions for the example I gave

=latex x^2 = 13 mod 101

[sacred junco]

Yep

[blissful venture]

This has solutions though

But I don't understand how to actually reach them

[sacred junco]

Ok let me step aside from modulo for a bit

Not my strong point..just got introduced to it 4 days earlier

try out x=1 to 100 :DDD

Omg..didnt come to my mind!!!

[blissful venture]

There is a log(n) solution apparently

And that's where I am completely lost

[sacred junco]

Dude the legendary mudkip is offline

Or else it wudve been sorted till now

The totient of 101 is 100

So

=latex x^ 100=1 mod 101

[blissful venture]

Yeah, obviously if I can solve for primes, I can use CRT to solve in general

[sacred junco]

But u need more than 1 relation for that

[blissful venture]

No, I meant x^2 = a mod c, where c is composite is easy when you have a solution where c is prime

Because of CRT

[sacred junco]

Ah okay

[blissful venture]

But solving for primes still seems beyond me

[sacred junco]

Dude idk..im still doing it.

[blissful venture]

Maybe it involves selectively trying some numbers

[sacred junco]

Whats the answer to this..

[blissful venture]

Yeah, I wish I understood too

[sacred junco]

Aaaaaa

<@&286206848099549185>

[blissful venture]

https://en.wikipedia.org/wiki/Quadratic_residue#Complexity_of_finding_square_roots There is some stuff here,
and here, https://en.wikipedia.org/wiki/Tonelli–Shanks_algorithm

But I still don't get it

[sacred junco]

Oh no oh shit

Quadratic residue!!!

Dude there will be infinitely many solutions

Just find the first solution.

Which im doing right now.

Its basically asking all quadratic residues for mod 101

[blissful venture]

That is horribly inefficient

[sacred junco]

Thats for sure.

[blissful venture]

I'll think more tomorrow, it's late for me

[sacred junco]

Yeah sure dude.

Ill keep trying.

Wth..those wikipedia articles basically give out the full method lol

[blissful venture]

Yeah, but I wasn't able to understand much

Maybe I should read again later

[rancid oasis]

Hello all

I have a question! I have $$n, k, \ell \in \mathbb{N}$$ with $$n = 2^k \frac{2^{\ell +3} - 1}{2^{k+2} + 1}$$

[rancid oasis]

I need to prove this. Last time, I had $$n = 2^k \frac{2^{\ell + 2} -1}{2^{k+2} -1}$$ which I could prove since $$ \frac{2^a - 1}{2^b - 1} = 2^{a-b} + 2^{a-2b} + 2^{a-3b} + \cdots + 2^{a-cb} + \frac{2^{a-cb} - 1}{2^b - 1}$$

[rancid oasis]

However I have no such identity this time, and I'm stuck. Any help would be appreciated.

[sacred junco]

I dont understand he question

[rancid oasis]

Sorry. I need to prove that $$2^k \frac{2^{\ell + 3} - 1}{2^{k+2} + 1}$$ is a natural number for some $$k, \ell \in \mathbb{N}$$

[rancid oasis]

The trouble is proving that it does indeed equate to a natural number. Last time I expressed a similar rational function as a series that ended in 0, proving that it was equal to a series of powers of two - which is, of course, a natural number. The trouble is, this time I can't do that (in the same way) because the addition on the bottom is really busting my behind

[sacred junco]

Omg..busting my behind.

Thats a keeper.

So basically u need to prove that..the fraction will be an integer

Have u tried componendo dividendo?

[rancid oasis]

Erm. No, because I don't believe I know what that is.

[sacred junco]

Its like a very basic concept of fractions..

Might come in handy here..

[rancid oasis]

Perhaps I just call it something different. A brief demonstration, please?

[sacred junco]

If a/b is a fraction

Then a+b/a-b is basically the same thing

[rancid oasis]

I have never seen that before

[sacred junco]

Its a good problem thats for sure.

[rancid oasis]

Last time I could make the following case

[sacred junco]

Ur basically askig to prove a+b/a-b-1

Is always an integer.

[rancid oasis]

$$\frac{2^a - 1}{2^b - 1} = 2^{a-b} + 2^{a-2b} + \cdots + 2^{a-cb} + \frac{2^{a-cb} -1}{2^b - 1}$$

[rancid oasis]

Eventually you reach the point where $$cb < a$$, and either $$a - cb = 0$$ and you're left with a series of powers of two, or you have $$0 < a-cb < b$$ in which case the fraction is a rational number. But that leaves a contradiction since we know it must equate to a natural number

[rancid oasis]

So I guess what I'm asking is how to find the relationship between a and b

The trouble is this time I have $$\frac{2^a - 1}{2^b + 1}$$

[rancid oasis]

And that plus on the bottom screws everything up

[sacred junco]

Well..im trying some stuff out.

[rancid oasis]

Thank you

[sacred junco]

Ill let u know if something good happens

Dont thank me brother..

Im the most stupid person in the server tbh.

[rancid oasis]

No sir that title has already been claimed by yours truly :p

[sacred junco]

Wait it isnt a natural no for k=1,l=1

Do i have to find solutions..

Or do i have to prove that it MUST be equal to an integer for every natural no.whatsoever

Because if its the latter case..

U already have a contradiction

[rancid oasis]

There are solutions, and that fraction always equates to a natural number

I just need to find the relationship between a and b

[sacred junco]

Put l=1,k=1

Aint a natural no.

[rancid oasis]

Correct, not all combinations yield natural numbers

I need to find a relationship between those variables that generates natural numbers

[sacred junco]

Ahhhhh

Ok.

Damn ..makes the question even harder lol

[rancid oasis]

And so you understand my problem 😅

[sacred junco]

What grade are u in?

[rancid oasis]

Third year university

[sacred junco]

Whhaaaa...

Dude im in 10th grade. spare my tiny butt..

Ask this in the higher mathematics server..

They might be able to help u out.

Because all that comes into my mind is somehow forming a quadratic. And then use vieta jumping

[rancid oasis]

You're doing pretty well for someone in 10th grade

[sacred junco]

Im really not that good..

Just struggling to make it to the imo

😥😥

[rancid oasis]

The wot

[sacred junco]

International mathematical olympiad

The most prestigious math competition for highschool students

Ive heard that line so many times..cant live without stating it

[rancid oasis]

I never went because I'm a dumbass

[sacred junco]

Alright here we go

For k=0

There are infinitely many solutions.

Now 0 aint a natural

Fuck lets go again with 1

So for every k.

There will be an infinite number of solutions...

Method coming soon

So for k=1

The denominator is 9

So the first answer is 2^6

And then u keep adding 6 on the powers..

Quite obv

This goes to infinity...hence infinite solutions.

Then for k=2

Denominator is 17

Then l=2

So for every value of k.

There are infinite solutions..

And there are infinite k

So infinite*infinite solutions..

mic drop

[thorny jay]

Anybody know if this is true? It seems extremely likely based on the growth rate of the primes but how do you prove it?

[cobalt birch]

this follows from Bertrand's postulate

proofs can be found online

[thorny jay]

I see, ty

[vapid tree]

Yo guys I have question:

If a, b, c element of N
and
c= a + b/a - 1/b
How do you prove that c is always a square

[cobalt birch]

c = (a^2 b + b^2 - a) / ab

so a divides the numerator, hence it divides b^2

b divides the numerator, so it divides a

[sacred junco]

Not necessary..

They can leave odd remainders.

[cobalt birch]

c is supposed to be a natural number

[sacred junco]

Lets say..it leave it leaves a remainder of ab -1 when it divides b^2

And of 1 when it dividing a

[cobalt birch]

a divides a evenly

[sacred junco]

No no.

Ok let me try to make it more clear.

Do u agree ab>a

When b>1

[cobalt birch]

yes

[sacred junco]

So we can be sure the remainder here is a

So when ab divides b^2

It must leave a remainder which when adds up with a

Leaves a multiple of ab

[cobalt birch]

I'm talking about the fact that (a^2 b + b^2 - a)/a is an integer

so a divides a^2 b + b^2 - a

[sacred junco]

And thats what im trying to say isnt necessary

Alright u demonstraate ur method

Im gonna try solving it for now.

[cobalt birch]

so since b | a | b^2

if for any prime p, we let a_p and b_p be the number of times p divides a and b respectively

then b_p <= a_p <= 2 b_p for all p

now if for any p, a_p < 2b_p

then p will divide a fewer times than a^2 b or b^2

so p will divide the numerator exactly a_p times

whereas it divides the denominator a_p + b_p times, so we won't get an integer

(I'm only looking at the p for which p | b)

so we must have a_p = 2b_p for all p

meaning a = b^2

and then c = b^2

[sacred junco]

@cobalt birch ur smart

[cobalt birch]

thx

[sacred junco]

😃

[cobalt birch]

😛

[sacred junco]

This has a very similar taste to q6 sort of.

If we swap a for b

Its still the same..

The values wont change.

@cobalt birchi didnt understand ur solution..

[cobalt birch]

okay, I'll try to explain in a minute

[sacred junco]

Kk

[sacred junco]

Ok now my turn

Solve.

3x=11mod25

3x=11mod7

3x=11 mod 13

This will obv use CRT

But i cant confirm my answer so help

@cobalt birch

[cobalt birch]

so...

x = 12 mod 25

x = 6 mod 7

x = 8 mod 13

[sacred junco]

Wait how?

[cobalt birch]

just keep adding 25,7, or 13 until you can divide by 3

well, this is the harder way

3x = 11 mod 25713

[sacred junco]

Just learnt that

So thanks.

Now i can do this question.

[cobalt birch]

okay, good

[sacred junco]

Thanks a lot mate.

[cobalt birch]

no prob!

[vapid tree]

nani

wp fine sir

a^2(b/a) + b(b/a) - 1
[(a^2+b)(b)/a]-1

ah so ur saying either b/a or (a^2+b)/a has to be an interger or else this wont be prime?

which ends up being b/a = 0 mod a

wait its 1/b

nvm i confused myself

leme reread the solution

[sacred junco]

Just transpose a from the denominator once

And b the other time

Natural times natural must be a natural number

[vapid tree]

uhuh

[sacred junco]

So u can say a|b|b^2

[vapid tree]

uhuh

[sacred junco]

What follows is understandable

[vapid tree]

tru

[spiral heath]

Not sure this is exactly number theory but in a testament to my inability to use my time wisely, I just managed to spend over two and a half hours approximating an irrational number to 28 decimal digits with a calculator

[blissful venture]

Which irrational number?

[spiral heath]

I've been thinking a little recently about the sequence f(x)=f(x-1)²+f(x-1), f(0)=1, and discovered today that double exponentials can express sequences of that type, but of course usually have irrational or transcendental values in them

so I was trying to find a such that a^(2^x) ≈ f(x) with that prior sequence

[blissful venture]

👀

[spiral heath]

If you're wondering, I know it is in the range 1.5979102180318731783380701182 ± 10^(-28) 😃

All of that... by hand...

sometimes I just... sort of lose track of myself for a while 😄

[blissful venture]

Sounds pretty painful

[spiral heath]

actually I said that wrong before, I'm finding a such that a^(2^x)-0.5 gets arbitrarily close to f(x) as x increases - it's never exactly equal to it though

and no, not painful, just takes a while

I'm prone to obsession, you know

the sequence, btw, is something like 1, 2, 6, 42, 1806, 3263441, 10650050423922...

[sturdy dirge]

$$1=f(0)=a^{2^0}=a^1=a\neq 1$$.

[spiral heath]

yeah I'm using the floor of a^(2^x)

if you take the floor of it then subtract 0.5 it gets closer and closer to the sequence as you go further

[haughty falcon]

you have my attention. I gotta work the basics out on paper real quick (with some excel for the number crunching^^' )

according to open office calc(which I think is lying to me), it approaches this:

$$\sqrt{2}^{(\sqrt{2}^x)}$$

[haughty falcon]

this need another 2 somewhere, though

alright, this is what I got

$$\sqrt{2}^{(\sqrt{2}^{(2x+0.87)})}$$

[haughty falcon]

now the behaviour of the coefficients of the recursive function is remarkable, when you expand it (at least for 0 to 4 so far):
1
1 1
1 2 2 1
1 4 8 12 9 6 3 1
1 7 32 77 178 276 342 338 282 200 122 66 30 12 4 1

maybe we should try to find the function for that and integrate it

[vast vessel]

\👀 big numbres

[sacred junco]

Ok..

I feel dumb.

I dont understand anything going here.

[glass ravine]

It's alright, this is all decently complicated stuff

I don't get some of it either lol

[weak rapids]

Can someone push me in the right direction, not sure what to do after doing extended euclidean algorithm

[celest plinth]

1 = 77(59)-6(757) = 77(59) mod 757

so x=77

mod 757

@weak rapids

[weak rapids]

@celest plinth thanks, how did you get 77(59)mod 757.. I just don't understand that part. Are you just dividing each side. By 59?

[celest plinth]

no

you did that yourself

I just took both sides mod 757

all multiples of 757 get killed off

because 757 = 0 mod 757

[weak rapids]

Sorry I mean how did you get to 77 from that part?

On YouTube most people were subtracting the mod from the other number. Like 757 - 77 , to get x.

[weak rapids]

@celest plinth I guess it's if 77 was negative, we would subtract it from the negative to find a congruent answer?

[celest plinth]

@weak rapids you already calculated
1=77*59 - 6*757

I didn't do anything, I just read it off your paper

you're trying to solve:
1 = 59x (mod 757)

you have on your piece of paper

1 = 59(77) + (-6)(757)

1 (mod 757) = 59(77) + (-6)(757) (mod 757)

757 = 0 (mod 757)

so -6(757) = 0 (mod 757)

therefore
1 = 59(77) + 0 (mod 757)

1 = 59(77) (mod 757)

1 = 59(x) (mod 757)

we conclude that x=77 (mod 757)

I got 77 from your last line

[weak rapids]

oh, so I can exchange 59x and 1, flip sides I mean. Ha way over my head. I do understand when you break it down. Its just weird how 0(mod 757) is just mod(757) in this part: 1 = 59(77) + 0 (mod 757), I would think it would be 757

Can someone guide me on how to use the chinese remainder theorem to solve this. I got m = 280 and m1 = 556, m2 = 40, and m3= 35. The easiest part ha.Not sure how to go on from there.

[sacred junco]

@weak rapids id recommemd watching a video for it.

Because its quite lengthy to explain in plain text.

[blissful venture]

Or you can look at some code 👀

See, the easiest way to do CRT is like,

you know that x = 1 (mod 5)

so x = 1 + 5k

So 1 + 5k = 6 (mod 7)

so k = 1

So x = 6 (mod 35)

So x = 6 + 35k

[celest plinth]

this is so So

[blissful venture]

So 6 + 35k = 3 (mod 8)

Aye, it doesn't do rigorous analysis

But it's fast

[celest plinth]

no I mean

you said "So" a lot

[sacred junco]

Yeah thats what most people do

[blissful venture]

Ayy

[tight agate]

lmao

[elder bobcat]

so what

lelel

[blissful venture]

Oh, one of my s' was lowercase

[wild zinc]

that is the method I know for it

I swear there's a different method that's supposedly better but I might be misremembering

[blissful venture]

Naw, this is the fastest

It's different if you can calculate inverses fast

But as a human that's not as easy as solving these equations

[wild zinc]

how are you planning on solving 6 + 35k = 3 (mod 8) without calculating an inverse 🤔

[blissful venture]

I can show you some code if you are interested 👀

These inverses are easier

But on a machine you can take bigger inverses faster

So those on average are better

[wild zinc]

do you take the inverses modulo the product of the moduli?

[blissful venture]

#computing-software ?

Might be more relevant there

[weak rapids]

I've been watching videos all day ha. I can't grip what they are trying to say after that. I searched different terms "system of linear congruences" and "chinese remainder theorem" . Anything else that could help?

Yes please show some code @blissful venture

[wild zinc]

go to #computing-software because it is there

[weak rapids]

@wild zinc ok

[blissful venture]

@wild zinc When I take inverses by hand I do it like how I do gcd

[wild zinc]

yeah, extended euclidean algorithm

[blissful venture]

Yeah, exactly

But on a machine, I use fermat's little theorem

[wild zinc]

trying to think which would be faster on a machine

FlT certainly easier to code :^)

[blissful venture]

😁

The problem is that pow is implemented with C

So it turns out a lot faster

[weak rapids]

ok yea Extended euclidean algo is what we've been doing. I would rather do it like that

[wild zinc]

right so in python it's going to be faster ye