elementary-number-theory

[obtuse mason]

I mean I'm not looking for a solution more a direction

that's why I dont want to look on the web

[winter bear]

well a^b+1, looking at this form whats the first thing that comes to ur mind

[obtuse mason]

fermat

[winter bear]

hmm, its more direct than that

a^b+1^b, if that helps

[obtuse mason]

thank you for the hint

[jovial hemlock]

can someone explain the Chinese Remainder theorem to me? it keeps coming up and I can’t make sense of the wikipedia article on it

[winter bear]

hmm how much algebra do you know(ring theory etc)

if not we'll go for a more elementery aproach

[jovial hemlock]

I’m in linear algebra (matrix theory) now

[winter bear]

hmm ok try reading this, its well written

https://brilliant.org/wiki/chinese-remainder-theorem/

you can inquire about anything that confuses ya

[jovial hemlock]

okay thanks

brb

I don’t understand this statement about extended euclidean algorithm

[winter bear]

its essentially to find the inverses

[alpine jasper]

an example probably will help

[jovial hemlock]

“It is wellknown that if gcd(a,b) = r then integers p and s exist where p(a) + s(b) = r”

[winter bear]

(bezo'uts)

[jovial hemlock]

but

the gcd of a and b is at most the size of the smallest one

[winter bear]

can be negative

p,s

[jovial hemlock]

how can you multiply a and b by... wait negative numbers nvm

(bezo’uts)

?

[winter bear]

its the name of that

bezo'uts identity

[jovial hemlock]

ah

okay so

the extended euclidean algorithm finds the modular inverse of n

[winter bear]

yep

[jovial hemlock]

but the eea seems to require a lot of iteration

[winter bear]

its relatively efficient, but this is not really the important part. The important part is to get the theory that there are inverses and what follows

(if ur doing problems by hand usually finding inverses will be easy, and if not then computers already have eea coded)

[jovial hemlock]

okay back to CRT

on that page, y1^-1 literally means 1/y1 or no?

[winter bear]

wherever thers ^(-1)s or 1/something, it means modular inverse

[jovial hemlock]

so i’m confused as to what step 3 is saying

[winter bear]

its just asking for the inverses of each y_i

which we defined before hand

[jovial hemlock]

with respect to what

[winter bear]

umm it says mod n_i

[jovial hemlock]

like you can’t say the modular inverse of x

you have to say the modular inverse of x to 26

[winter bear]

i mean it literally says what mod tho

[jovial hemlock]

oh

ohhh

so it’s defining zi as being the modular inverse of yi to ni

[winter bear]

yep

[jovial hemlock]

I don’t get why it’s so complicated

like, intuitively it seems like it’s not that hard to find a number satisfying multiple modulos

[winter bear]

well its not really easy to do so

like think about inverses, even those arent easy to find

the existence itself isnt hard to show with less tedium, but to actually compute it is annoying yeah

[jovial hemlock]

but like

if you were to say a number is 5 mod 12 and 9 mod 20

I can easily think to myself, well, the LCM of those is 60 so it’s X mod 60

and then I think, what numbers are 9 mod 20? well, 9, 29, and 49

[winter bear]

well you dont need to check a lot of things, u only need to check a few here

[jovial hemlock]

and only one of those is 5 mod 12

I’m guessing that the second step is what makes it hard?

[winter bear]

yeah like if u have 3 numbers its p easy to see it right

but if u have bigger numbers i think this is inefficient compared to the other algorithims

[jovial hemlock]

but which step is the hard part?

[winter bear]

step 2

[jovial hemlock]

finding the list of things 9 mod 20 or figuring out which of those is 5 mod 12?

[winter bear]

the latter

(also imagine having more mods to consider, that becomes more inefficient, like instead of 2 if you had 5)

[jovial hemlock]

I only care about 2 at the moment

Say you have a number

and you know it’s a mod c and b mod d

and c is larger

I’d think you could find y = d - c mod d if c mod d =/= 0 and 0 if c mod d = 0

hang on a sec let me

okay I think it’s like this

you have a number X

a = x mod c and b = x mod d
c > d

let y = d - (c mod d)
let z = (a-b) mod d
let n = (z/y) c + a

then n = x mod (LCM(c,d))

I don’t know how to prove it but I think that’s right

no need for c and d to be xp prime btw

oh, and if c mod d = 0 then the LCM is already c so it’s trivial

[obtuse mason]

a^b + 1 is prime, then a even and b is power of two
Okay a has to be even, because otherwise (a^b + 1) would be even and therefore divisible by 2

[jovial hemlock]

why is it prime?

[winter bear]

yeah thats one part

[jovial hemlock]

oh wait you’re a different discussion

John, who are you responding to?

[winter bear]

deekaan

[obtuse mason]

the second part I can't figure out why

[winter bear]

well a^b+1^b

think about factors

also notch im p sure this isjust what the wiki says for 2 primes

[jovial hemlock]

what wiki

[winter bear]

the brilliant one i sent you

[jovial hemlock]

and wdym two primes?

[winter bear]

or sorry 2 mods

(typically CRT is used with primes so i said primes)

[jovial hemlock]

where does it say that?

[winter bear]

i mean i just glanced at it, but urs should be equivalent to the algo they have

for the case of two mods

otherwise its incorrect

[jovial hemlock]

the only algorithms I see there are iterative though and mine isn’t at all

[obtuse mason]

The second part is because of factorization, if b is uneven then the expression (a^b + 1) = (a + 1)(***)

and if its not a factor of 2 then it can be turned into an uneven factor using prime factorization

is this right?

[winter bear]

yep

[obtuse mason]

okay wow thank you very much

[queen locust]

Alright, I am stuck, i got this problem which I need to solve:

[swift shard]

$\frac{x_1+x_2+...+x_n}{n}\geq(x_1x_2...x_n)^{1/n},~ \forall n=2^k,~ k\geq1,~ \text{and}~ x_1,x_2,...,x_n>0.$

[sterile pumice]

Kaynex:

[queen locust]

thank you

[swift shard]

Yaya we just use $ surrounding here

[queen locust]

aight, will keep in mind

I went through the whole method for induction problems

I started with assuming that it's true for a k, and went from there

however i couldnt do anything with the expression for only k, so I thought I'd work with k+1 instead

that didnt help either

so any advice or tip is appreciated, however no solutions please heh

I've managed to get this, dont know if it's useful tho

[craggy solstice]

ok i think i got it

so uh what i did is rewrite the (x1.x2.x3..xN)^1/N where N=2^k as k products with two terms each

like (x1.x2)^1/2^k

now ${(x1.x2)}^{1/2^k}<=(x1.x2)^{0.5}<=0.5*(x1+x2)$

[sterile pumice]

Lionel:

[craggy solstice]

latex is aids but tolerate

no actually i skipped alot of steps

this is what i was trying to get at

you can use the same method for higherpowers

the steps will just be repeated

@queen locust

[jovial hemlock]

so

I don’t want to type out my question again let me screenshot

[sacred junco]

how do you find num of solutions to a polynomial congruence that could look like a quadratic poly with substituion?

i.e $x^{20}+30x^{10}+300=0 \pmod{249}$

[sterile pumice]

Slate:

[sacred junco]

you could rewrite this to $z^2+30z+300=0\pmod{249}$ and then you would use lagranges numbers to find number of solutions to this

[sterile pumice]

Slate:

[sacred junco]

but i dont think this is right because x^20 can have at most 20 solutions while ^2 will have at most 2

[light flicker]

That's not true

The equation x² = 1 (mod 8) has 4 solutions

You need to be careful when working with composite modulus, so your first step should always be to use Chinese remainder theorem to split things apart into prime modulus

[sacred junco]

i must have messsed it up with roots

assuming the mod is already prime?

[light flicker]

It doesn't matter x² - 1 = 0 (mod 8) has 4 roots

[sacred junco]

im wrongly remembering some other fact then

[light flicker]

It's true if the modulus is prime

And that's why you want to use crt to work with prime mod instead

[sacred junco]

the odd number i wrote was supposed to be prime

as mod

i forgot to mention i have a prime modulus

and the 249 i typoed wasnt prime

[light flicker]

Then the problem is that z will have two solutions

[sacred junco]

but x could have up to 20

[light flicker]

But you have that x^10 = z

So there will be ten solutions (counting multiplicities) for x for each solution of z

[sacred junco]

but then you have to rule out the multiplicitites and to find the number of unique solutions you need to enumerate the solutions

[light flicker]

you can use quadratic/5-th power residue things to say how many solutions you have

[sleek cove]

is it necessary to prove that eulers function is strictly nondecreasing

[light flicker]

its uh

not?

phi(7) = 6, phi(8) = 4?

[sleek cove]

wait nvm

sory

[jovial hemlock]

is that right?

and if so how do I prove it

[wild zinc]

I don't think so

let c = d+1, for example. then y = d-1, z = a-b so n = (a-b)(d+1)/(d-1) + a which is not necessarily an integer

[jovial hemlock]

z =/= (a-b) though

it equals (a-b) mod d

let’s try a = 2 c = 20 b = 3 d = 19

y = 19-(20mod19) = 18

z = (2-3) mod 19 = 18

18 * 20/18 + 2 = 22 which is correct

@wild zinc

[wild zinc]

a-b is kinda irrelevant because it's arbitrary

you would have to have (d+1)/(d-1) an integer

[jovial hemlock]

no it isn’t

what a-b does is essentially cycle it

[wild zinc]

your claim is that this holds for all a,b?

[jovial hemlock]

for all a,b and c,d yes, assuming there is a solution

so it wouldn’t hold for 3 mod 18 and 2 mod 16 for example

but if there is a solution with a,b I think this finds it. I’m not sure why though.

[wild zinc]

what about a = 1, b = 0, c = 17, d = 18

[jovial hemlock]

c is the bigger one remember

so it would be a = 0, b = 1, c = 18, d = 17

[wild zinc]

oh right

still shouldn't work

[jovial hemlock]

y = 17-(18mod17) = 16
z = (0-1)mod17 = 16
16/16 * 18 + 0 = 18

[wild zinc]

a=1,b=0 shouldn't work tho

[jovial hemlock]

oh, a = 1, b = 0, c = 18, d = 17?

[wild zinc]

maybe

yeah, I forgot which way round I did the +1

we get 9/8 + 1

I think if you let the division be modular inverse then it holds but

this has the same problem of requiring modular inverse calculation

[jovial hemlock]

hang on

is there an easy way to manually check numbers

like somewhere I can put these equations and change numbers on the fly

[torn escarp]

learn some basic python and program it

easily brute force check for counter examples

[jovial hemlock]

what I don’t know how to do is test only valid a,b pairs

I wouldn’t want it stopping on 0 mod 2 and 1 mod 4 for example

[torn escarp]

what is the criteria for being a valid pair

[jovial hemlock]

I think that A and B have to be equivalent modulo (GCD(c,d))?

[torn escarp]

then only test ones that satisfy that

[jovial hemlock]

is that correct condition?

[torn escarp]

I have no clue, why are you asking me

I didn't read the problem, it didn't interest me

[jovial hemlock]

you were responding to my questions about the problem

[torn escarp]

but I'm just offering to help you set up your program

[jovial hemlock]

ah

[torn escarp]

but you don't know what the criteria is to check so I can't help, good luck

[jovial hemlock]

I assume that’s it

[torn escarp]

sounds sketchy

[jovial hemlock]

just

[torn escarp]

you only want to check the true cases to know what's true...?

[jovial hemlock]

the criteria I want is for one number to be a mod c and b mod d

and I’m testing how to find that number

and I only want to check cases where that number exists

[torn escarp]

I see

[jovial hemlock]

which python version do I get? python 3 or python 2?

[torn escarp]

3

I'd download visual studio code if I were you

[jovial hemlock]

what’s that

[torn escarp]

and then I think it will take care of getting the stuff you need from addons in it

it's an IDE

basically like, spell check for code

and runs code too

like, very roughly speaking lol

might be better to find a kind of tutorial or something to learn some from since you might have to learn a few things before you can write a few for loops to get what you want but it shouldn't be too far off

[jovial hemlock]

no I have basic python experience

It’ll just take me a bit to refresh and check how to use mod operator and that sort of thing

[sacred junco]

if I have p = 3 mod 4 and a =/= 0 mod p, how would I go about showing either -a or a == b^2 mod p where b is some integer?

[light flicker]

this would follow from the fact that -1 isn't a square mod p

[sacred junco]

Could you please explain more?

[jovial hemlock]

I can’t figure out

how to use this

[torn escarp]

visual studio code?

[jovial hemlock]

yeah

it won’t run anything

[torn escarp]

make a new thing like test.py

it might recognize the file name and ask if you wanna add stuff

[jovial hemlock]

I think I’m used to the version 2

gonna try that

[torn escarp]

it'll give you a run button at the top right once you install what it prompts

possible

https://learnxinyminutes.com/

[jovial hemlock]

oh wait do I need to name it .py?

[torn escarp]

yes

[jovial hemlock]

oh okay thanks

I had it as a .txt no wonder

[torn escarp]

lol yeah that would do it

[light flicker]

@sacred junco there's a fact that the product of two things that aren't squares mod p is a square mod p

So like if a and b are both not squares mod p, then ab has to be a square mod p

So if -1 isn't a square mod p and a isn't either, then -a has to be a square mod p

[jovial hemlock]

can you help me?

I probably just messed something up but my code isn’t working and I’m not sure why

[torn escarp]

copy paste it

[jovial hemlock]

hang on let me open discord on my computer

import math
for c in (3,100):
for d in (2,c-1):
lcm = (c*d/math.gcd(c,d))
for x in (0,lcm-1):
a = math.fmod(x,c)
b = math.fmod(x,d)
y = d - math.fmod(c,d)
z = math.fmod((a-b),d)
if z == 0:
n = 0
elif z > y:
n = (z/y)*c + a
else:
n = (y/z)*c + a
if n != math.fmod(x,lcm):
print (a,b,c,d,x)
print ("done")

um, the * symbol for multiplication is making things italic

how do I fix that

[torn escarp]

put three ` before and after

[jovial hemlock]

import math
for c in (3,100):
for d in (2,c-1):
lcm = (c * d/math.gcd(c,d))
for x in (0,lcm-1):
a = math.fmod(x,c)
b = math.fmod(x,d)
y = d - math.fmod(c,d)
z = math.fmod((a-b),d)
if z == 0:
n = 0
elif z > y:
n = (z/y) * c + a
else:
n = (y/z) * c + a
if n != math.fmod(x,lcm):
print (a,b,c,d,x)
print ("done")

[torn escarp]

for i in range(1,100):
  print(i)

[jovial hemlock]

oh did I forget the word range? is that all?

[torn escarp]

I didn't look at your code I was just saying put three ` before and after to format your code

and gave an example

if your code is broken test individual pieces in smaller chunks rather than doing it all in one huge thing

[jovial hemlock]

I did

I got confused

I made typos

[sacred junco]

Solved it nvm

[jovial hemlock]

new question

apparently python interprets “fmod(-2,5)” to be -2

when it should be 3

how do I fix that

k, I just add the 5 to the -2 until it’s above 0

[torn escarp]

you can do mod instead of fmod

I think that fixes it

[alpine jasper]

i know that K is a k vector space but i'm not sure how to show K is finite dimensional

i was thinking of finding a basis, but i have no idea how

i can also try to argue that field containing finite dimensional fields are also finite dimensional, poor wording ik

@winter bear help me papa

[winter bear]

hmm gimme a sec i think i have a proof for this written somewhere

[alpine jasper]

[winter bear]

i mean, the idea is to induct

start of by removing one root, and then extend this new field by removing another

[alpine jasper]

i'll give induction a shot

okay i'm still not sure how

i want to prove that $[k(\alpha_1):k]$ is finite right, and then keep adding on more $\alpha_i$ until it hit $\alpha_n$

[winter bear]

kinda

[sterile pumice]

Publius:

[winter bear]

i mean basically

lemme refine this for you a bit by giving you an inductive hypothesis

[alpine jasper]

but i need that as a base case tho

unless you're talking about a different induction

[winter bear]

$P(n)$: for each poly in any field of degree $n$, theres a splitting field thats a finite degree extension of this field

[sterile pumice]

JohnDoeSmith:

[winter bear]

i mean this is kinda equivalent to your idea of "keep adding a_i"

[alpine jasper]

tru

i see

let me give that a shot

i'm still not sure where to complete the hypothesis

the first line assumed f is degree 1

and green marks the degree of f

[winter bear]

i mean the idea is basically right? try to refine it, for example f(x)=x-a splits in k, no need to say k(a_1)

also i think it would be nicer if you directly use prop 7.2.1

but yeah idea is correct, you reduce the degree by finding a field for one of its roots, and then by induction that splits in a finite extension field

[alpine jasper]

i sort of did by assuming those $K_{\alpha_i}$ exists?

[sterile pumice]

Publius:

[alpine jasper]

but i just don't see where to go with this induction

[winter bear]

oh right ok so

[alpine jasper]

oh wait

g has degree less than n

so it factors

?

[winter bear]

yeah

but ill say, its nicer to consider just one root

rather than all that i roots

[alpine jasper]

i'm not sure how that works

i factor out 1 root at a time?

[winter bear]

basically

[alpine jasper]

but induction tells me to assume it for arbitrary number of roots no..?

[supple furnace]

Just remember that if K is contained in L, [K(a):K]>=[L(a):L]

[winter bear]

i mean yeah idea still works i wont complain about the asthetics ig

[alpine jasper]

hmm

[supple furnace]

You probably also want to verify that adjoining elements maintains field structure

Which can be done inductively as well

[alpine jasper]

i see

thanks john and tree3

[alpine jasper]

okay i'm actually still stuck

what am i doing wrong

am i suppose to say deg f = i?

my small brain just can't make sense of this

[winter bear]

ok so lets say f has deg i+1 right

[alpine jasper]

mhm

[winter bear]

by prop 7.2.1 we can find a finite extension of this such that theres a root, i.e f(a)=0 right

[alpine jasper]

yes

[winter bear]

then do f(x)=(x-a)g(x)

do you see what to do from here

[alpine jasper]

g splits since it is degree i

[winter bear]

mhm, specifically splits in a finite extension

[alpine jasper]

right

hmm, but what field does f splits then?

is that field K + [field that g splits]

[winter bear]

ok so we extended $k$ to lets say $K_a$, and then we extended $K_a$ to lets say $K_a'$, i.e where $g$ splits

[sterile pumice]

JohnDoeSmith:

[winter bear]

so where does f split here

[alpine jasper]

the reason why the second extension from K_a to K'_a is finite is because degree of g is i?

[winter bear]

yeah, by inductive hypothesis we can make a finite extension of g since its deg i

[alpine jasper]

icic

thank you dad

[winter bear]

np son

[sleek cove]

Is it bad that I didn't use the second property (ii) in the proof?

wait, i need it for proving the first direction nvm

[sleek cove]

is there a fast way to show that phi (n) = 2 is only true for n = 3,4,6

nvm

[alpine jasper]

so by euler's theorem, f = phi(n)

i also note that i can factor out x-1 in x^n - 1

but other than that, i'm pretty clueless

hint pls

[winter bear]

so first its not phi(n)

the order itself can be smaller than phi(n)

for example, 2^3=1 (7)

and heres your hint

$n|q^f-1$ which is also the order of the multiplicative group

[sterile pumice]

JohnDoeSmith:

[winter bear]

@alpine jasper

[alpine jasper]

i'm not sure if this makes any sense

i still don't see where i used (q, n) = 1

[winter bear]

hmm i dont see how ker dividing |E^x| told u n|q^f-1

[alpine jasper]

if G a group and H a subgroup, |H| | |G|? is that the lagrange theorem or somethin

[winter bear]

yes but where did n come from

that tells you ker(x^n-1) | q^f-1

[alpine jasper]

kernel has size n since we're basically working with nth root of unity?

i'm not entirely sure

[winter bear]

oh yeah ok

cause it splits it has n roots

[alpine jasper]

uh huh

[winter bear]

you couldnt find a splitting field in the first place if q|n btw

is where this would fall apart

[winter bear]

for the (q,n)=1 condition

[alpine jasper]

is unsolvable if (n, q) != 1?

[winter bear]

also that

[alpine jasper]

hmm my smol brain don't immediately see why

[winter bear]

ill let u work it out

but as to the bigger problem

you have shown if $x^n-1$ splits in extension of degree $f$, then $n|q^f-1$ right.

[sterile pumice]

JohnDoeSmith:

[winter bear]

but unfortunately this doesnt tell you wether f is minal and when n is minal

so what do you think you should do

oof i mispelled minimal twice the same way

[alpine jasper]

hmm yeh good point

[winter bear]

think about it for a bit, its p similar to this to show the other part

[alpine jasper]

alright

ty

[winter bear]

np

[alpine jasper]

i think this shows that gcd must be 1 otherwise the eq. is unsolvable as expected

but i'm stuck on how to argue the minimality of f

[winter bear]

yeah, btw what u did is basically euclids algorithim for gcd

[alpine jasper]

yeah i've noticed

[winter bear]

ok so remember what u showed b4

try showing the converse

[alpine jasper]

that gcd=1 implies solvability?

[winter bear]

no the other one

the splits implies n|q^f-1

[alpine jasper]

i see

i'll give it a shot

ty

idk how

i also don't understand why splits iff n|q^f-1 implies f is minimal

[winter bear]

well remember when u solved x^n=1 in fields right

[alpine jasper]

uh huh

[winter bear]

how many solutions and what was the criterion for solving this

[alpine jasper]

n solutions, (n, q) = 1

[winter bear]

and what else

umm lemme phrase it like this

how many sols does $x^n=1$ have in $\mathbb{F}_{p^k}$

[sterile pumice]

JohnDoeSmith:

[winter bear]

prolly look back in IR if you dont remember

[alpine jasper]

hmm

[winter bear]

wait (n,q) is tru nvm that

[alpine jasper]

either 1 or n?

[winter bear]

yeah, depending on what

[alpine jasper]

since either just trivial soln or a non trivial soln which generates the rest

umm

n | p^k - 1

[winter bear]

there you go

and if it has n solutions, what can you say about the poly x^n-1

[alpine jasper]

it splits

[winter bear]

yeah, so do you see how this works then

[alpine jasper]

not really

gimme 5 mins

[winter bear]

ok

[alpine jasper]

so if n|q^f-1, then x^n = 1 have n solutions in F_(p^f), so it splits in E?

[winter bear]

F_(q^f) but yeah

[alpine jasper]

am i suppose to write F_(q^k) instead?

[winter bear]

umm not really. theres only one field of each order so as long as its clear what the order is its fine

[alpine jasper]

hmm alright

i still don't understand why splits iff n|q^f-1 implies f is minimal tho

[winter bear]

well take the smallest f such that n|q^f-1

what happens if there is some f'<f

such that extension of that degree splits the poly

[alpine jasper]

n | f'?

or contradiction since such extension does not exist?

[winter bear]

contradiction basically

also n|f' is not true

[alpine jasper]

oh right that makes no sense i just realize

i should think about why that extension dne

[winter bear]

another way to look at this that extension degrees where this split is one-to-one corrospendence with f such that n|q^f-1

if that helps

[alpine jasper]

damn

i just don't see how this works

[winter bear]

thats ok

umm so

$f$ deg extension where $x^n-1$ splits $\iff$$n|q^f-1$. Take the smallest $f$ such that $n|q^f-1$, if there is a smaller degree extension $f'<f$ where $x^n-1$ splits, then we know $n|q^{f'}-1$, but this contradicts minimiality of $f$.

[sterile pumice]

JohnDoeSmith:

[alpine jasper]

oh

i thought we can't assume f being minimal

[winter bear]

well it was kind of a contradiction right

[alpine jasper]

Take the smallest f such that n | q^f - 1
this is the important bit right

[winter bear]

yeah

this kind of corrospondence is called an order preserving corrospondence btw

[alpine jasper]

Take the smallest f such that n | q^f - 1

smaller extension that splits
implies
f is not the smallest

yeh.?

[winter bear]

yep

[alpine jasper]

ok i see

that works

do u mind reading a more elementary attempt that i had

[winter bear]

yeah sure

[alpine jasper]

uhh, so the idea is to reduce the degree that q is raised to all the way to 1

idk if that's a contradiction

probably pure garbage

[winter bear]

well so one thing is that u arent garunteed to get 1

just gcd(f,s)

in exponent

(think about why)

and hmm i mean this isnt a contradiction

you'd have to use the corrospondence we used above

i think

[alpine jasper]

yeah

i don't see how that works too

but i'm still not sure why the lowest is (f, s)

[winter bear]

hmm do you see it cannot be any lower than (f,s)

[alpine jasper]

i was thinking we can perform this lowering thing at most f times

i don't see

[winter bear]

ok so the exponent will end up as a combination of f and s right

and gcd divides both

[alpine jasper]

oh.

[winter bear]

to show its exactly gcd is a bit more slick

$q^f=1 (n)$, $q^s=1(n)$ then $q^{(f,s)}=1(n)$

[sterile pumice]

JohnDoeSmith:

[winter bear]

ill give u a hint if u wanna prove this

(btw it can be lower than (f,s), but u wont get that number by just this info, is what i meant)

[alpine jasper]

i will leave this to another day

[winter bear]

i mean its a one liner

[alpine jasper]

i feel like i just have to write out the defn

[winter bear]

in a manner of speaking yes

if u define gcd by ||bezouts||

[alpine jasper]

hmm

still don't see it and i won't bother more

thanks for your help btw

[winter bear]

np

[alpine jasper]

hello

i'm once again asking for your nt support

is this the set i'm considering?\
${x\in\mathbb Z[i]\mid x\equiv p(p)}$

[sterile pumice]

Publius:

[alpine jasper]

doesn't make much sense to me

this is the same as\
${x\in\mathbb Z[i]\mid p\text{ divides } x}$

[sterile pumice]

Publius:

[alpine jasper]

or is the set something else..?

i suppose that set would make sense

it has
p + pi
2p + pi
...
p^2 + pi = 0 + pi
and for each kp + pi, i can "branch out" the imaginary component so like
kp + pi, kp + 2pi, ... kp + p^2 i = kp

for p^2 elements in total

that seems too easy or am i missing something again

[light flicker]

asking you to look at z[i]/(p) basically

[alpine jasper]

this is a quotient ring right

[light flicker]

yes

[alpine jasper]

ok i should try to formally write down what this means my algebra sucks

$\mathbb Z[i]/(p)={a+bi+x\mid x\in(p)}$

[sterile pumice]

Publius:

[alpine jasper]

umm

is this right?

[light flicker]

yeah

[alpine jasper]

i'm not sure how to like, count it tho..

it looks like i can make the real part running from 1 to p

but do i have any control over b?

i'm being dumb again probably

[alpine jasper]

i still can't figure it out

help pls

[light flicker]

oops sorry

one way is to figure out when {a + bi + x} = {c + di + x}

[alpine jasper]

i'd say b = d(p) and a = c(p)

actually

.. yeah?

{a + bi + x} = {c + di + x}
by this do you mean

${a+bi+x\mid x\in(p)}={c+di+x\mid x\in(p)}$

[light flicker]

yes

[sterile pumice]

Publius:

[alpine jasper]

so ok, only b = d(p) is needed

[light flicker]

are you sure? so {1 + x} and {2 + x} are the same set?

[alpine jasper]

is it a single x

or is the first set like... {1 + 1, 1 + (2), ..., 1 + (p-1)}

[light flicker]

Sorry, I'm just writing the shorthand

I mean {1 + x | x \in (p)}

but remember that (p) is an ideal in Z[i]

so it has elements like pi

er p * i

[alpine jasper]

oh wat

i thought its just 1, 2, ..., p-1

let me think about it

ok i don't understand why it'll have stuff like pi

[light flicker]

(p) is the ideal generated by p

[alpine jasper]

this means it contains (a+bi)p for any (a+bi) in Z[i]

right

[light flicker]

and only those

yes

[alpine jasper]

oh. i thought you meant pi as in 3.14...etc

my bad

[light flicker]

lmao

yeah

that's why I said

er p * i

[alpine jasper]

ok gud let's more forward

let me think about it for a sec

${1+x\mid x\in(p)}={1+p(a+bi)\mid a+bi\in\mathbb Z[i]}={(pa+1)+(pbi)\mid a,b\in\mathbb Z}$

[sterile pumice]

Publius:

[alpine jasper]

are you sure? so {1 + x} and {2 + x} are the same set?
so no afterall, if {a + bi + x} = {c + di + x}, then b = d(p) and a = c(p)

[light flicker]

yeah

so can you list out the p^2 sets

[alpine jasper]

that is some gud shit

idk if this notation makes sense

to show that Z[i]/(p) is a field, should i try to show (p) is a maximal ideal or working with the RHS directly?

[light flicker]

the former's easier

[alpine jasper]

wops one small detail

i should add another pair of {}

ok, let me try that

[alpine jasper]

i hope this makes sense

[alpine jasper]

actually, why was the fact that p=3(4) never used? @light flicker

[light flicker]

q is an element of Z[i]

so its not immediately true that since q |p that q = 1

here is where you use the fact that p = 3(4)

if p = 1 (4), for example if you take p = 5, then its not true since 5 = (2 - i)(2 + i)

[alpine jasper]

woow damn that's beautiful

tyty

[alpine jasper]

welp

i tried computing the sum separately but no luck with the first one, and this is what i have so far for the second one

[light flicker]

have you seen the proof of the euler product for the zeta function

[alpine jasper]

this right, yes

but the form is not exactly right, for example, LHS has the base as n, but what i had above has it as p

i looked at it for a while and don't know how to apply it

[light flicker]

Sure, but

the f in the product and f in the sum are over different sets

[alpine jasper]

and that too

umm, so more complicated?

[light flicker]

It's the same idea

Look at the proof for the euler product for the zeta function

and adapt it

[alpine jasper]

hmm alright

so i looked at the proof

and the idea is that i multiply and remove stuff with factors systematically

sort of like treating monic poly as numbers, and monic irred poly as primes

but i'm stuggling to formualize and figure out what to multiple and subtract

bascially i don't see what's the analogous equivalent thing to 1/2^s

[light flicker]

Uh, I'm not sure what proof you looked at

But maybe I'll just say that

[alpine jasper]

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

[light flicker]

$\frac{1}{1 - p^{-s}} = 1 + \frac{1}{p^{-s}} + \frac{1}{p^{-2s}} + \cdots$

[sterile pumice]

Zopherus:

[light flicker]

yeah, you should think of this fact as coming from the fundamental theorem of arithmetic

Basically if you take some number like 12

12 = 2^2 * 3

so to get 12, you'll multiply $\frac{1}{2^{-2s}} \cdot \frac{1}{3^{-s}} = \frac{1}{12^{-s}}$

[sterile pumice]

Zopherus:

[alpine jasper]

ok i'll think about that

[light flicker]

It's kind of the intuitive idea why it works

So like the product is

$\left( 1 + \frac{1}{2^{-s}} + \frac{1}{2^{-2s}} + \cdots \right) \left( 1 + \frac{1}{3^{-s}} + \frac{1}{3^{-2s}} + \cdots \right) \left( 1 + \frac{1}{5^{-s}} + \frac{1}{5^{-2s}} + \cdots \right) \cdots$

[sterile pumice]

Zopherus:

[alpine jasper]

ok gimme some time to digest that

[light flicker]

This is just rewriting the product

[alpine jasper]

uh yeah, but i don't know how to adapt this

[light flicker]

So the idea is that every positive integer can be uniquely written as the product of power of primes

Similarly, every monic polynomial can be unique written as the product of powers of irreducible monic polynomials

[alpine jasper]

god damn

i just don't see what to do really

i'm suppose to multiply this by something in terms of monic irreducible poly right

call this original sum S, then suppose after i multiply, i get AS, then i do S - AS, which equals to S - some stuff that are not irredicuble

that probably made no sense

[light flicker]

Idk, again

$\left(1 - \frac{1}{(Nf)^s} \right)^{-1} = \frac{1}{1 - (Nf)^{-s}} = \left(1 + \frac{1}{(Nf)^{-s}} + \frac{1}{(Nf)^{-2s}} + \cdots \right)$

[sterile pumice]

Zopherus:

[light flicker]

idk, if you want to do it the way that wikipedia does it

Then yeah, take some irreducible monic polynomial g

And if you do what they do

you'll remove all the terms in the sum that were divisible by g

[alpine jasper]

idk where that second equality came from really

and yes i've been trying to do it analogous to the one from wiki with 0 progress

let me think about this more ig

[light flicker]

the second equality is just geometric series

[alpine jasper]

oh

i just don't think i can do it the way wiki did it

because the numerators

if i multiply everything by like, p/p^(2s) then do the same thing

the terms wouldn't like, line up

[light flicker]

where are those p's on top even coming on top

[alpine jasper]

i count by degree of f

but suppose degree f = 7, then i have x^7 + ax^6 + bx^5 + ...

oh wait did i counted it wrong

[light flicker]

no its right

but you don't need to write it like this

[alpine jasper]

there are p^6 choices of degree 7 monic poly

[light flicker]

and I'm not sure writing it like this will help you

[alpine jasper]

i turned it into this and that's not a nice form too

[light flicker]

$\zeta_R(s) = \sum_{f \in R, f \text{ monic}} \frac{1}{(Nf)^s}$

[sterile pumice]

Zopherus:

[light flicker]

then take some irreducible monic polynomial g

you have that $\frac{1}{(Ng)^s} \zeta_R(s) = \sum_{f \in R, f \text{ monic}} \frac{1}{(Nfg)^s}$

Since $(Nf)^s (Ng)^s = (Nfg)^s $

[sterile pumice]

Zopherus:

Zopherus:

[alpine jasper]

ok let me see if i can continue

ok i think i can finish it

just to make sure, 1 is not a monic irreducible polynomial by definition right? since it can't be factored into two non-constant poly

[light flicker]

yeah

irreducibles have to be not units

they have to be non-invertible

[alpine jasper]

i see

[light flicker]

This is by definition

[alpine jasper]

thanks for your immense amount of patience my dude

[light flicker]

You're lucky I've been grading and have nothing better to do

[alpine jasper]

lmao

[alpine jasper]

i'm stuck once again

i completed 2 successfully but 3 is giving me a hard time

@light flicker

[alpine jasper]

<@&286206848099549185>

[alpine jasper]

ok i figured it out, i was digging the wrong rabbit hole smh

[alpine jasper]

@winter bear hey dad i'm online now

are you available

[winter bear]

yeah im online now

[supple furnace]

Me too

[winter bear]

@alpine jasper

[alpine jasper]

noice

[winter bear]

oof

[alpine jasper]

do you want me to repost the problem again

[winter bear]

yeah

[supple furnace]

Did you figure it out

[alpine jasper]

no i did not

[winter bear]

wait ill let tree3 help u out here, cause i kinda have some hw to do

[alpine jasper]

ok nice

N(n) is the number of monic irreducible polynomial in Z/pZ[x] where p is a prime

[supple furnace]

log(1-x) taylor series

[alpine jasper]

i did exactly that actually

didn't really work, let me post my work

[supple furnace]

You want to compare coefficients: the 1/p^(ks) coefficients should match up for each k.

[alpine jasper]

let me try again i guess, i'm less tired than yesterday

mostly stuff written in blue

[supple furnace]

Yeah

Start by finding the 1/p^(ns) coefficients of both sides

Do LHS and RHS individually

[alpine jasper]

gimme a moment i'll report back

so this is precisely where i'm stuck

i don't know how to deal with 1 sum on the LHS and 2 sum on the RHS

i tried to write it in form of\
$\sum_{\beta\geq1}\sum_{n\geq1}(\textit{stuff})$

[sterile pumice]

Publius:

[supple furnace]

We want to find the coefficient of 1/p^(ns)

Which acts like an x^n coefficient for x=p^s

In -log(1-p^(1-s)), we get (p^(1-s))^n/n

=p^n/n(p^(-ns))

So the coefficient is p^n/n

[alpine jasper]

let me digest that

gimme a moment

In $-\log(1-p^{1-s})$, we get $(p^{1-s})^{n/n}=\frac{p^n}{n(p^{-ns})}$

[sterile pumice]

Publius:

[alpine jasper]

umm i'm just texing it so easier to read

hmm

[supple furnace]

In $-\log(1-p^{1-s})$, we get $\frac{(p^{1-s})^n}{n}=\frac{p^n}{n}(p^{-ns})$

[sterile pumice]

tree3:

[supple furnace]

This is just from the Taylor series expansion

[alpine jasper]

yeh gimme a moment

so stuff in purple must match?

[supple furnace]

Um no because the stuff circled in purple can’t have an s in it

[alpine jasper]

idk how to manipulate the RHS really :/

i also dont get the idea of matching terms, since i can have two different series summing to the same thing but the nth term need not be the same right?

[supple furnace]

On the RHS, we have terms in the form $-N(k)\log(1-p^{-ks})$, which when expanded give terms in the form $\frac{N(k)p^{-krs}}{r}}$. We just need to identify which of these terms correspond to $p^{-ns}$. This occurs if and only if $kr=n$, so to account for all possibilities, we just select $k$ to be any divisor of $n$ and $r$ to be $\frac{n}{k}$. This gives terms of the form $\frac{N(k)p^{-ns}}{\frac{n}{k}}=\frac{kN(k)p^{-ns}}{n}$, where $k$ ranges over the divisors of $n$. Summing these and comparing coefficients gives the result.

[sterile pumice]

tree3:
Compile Error! Click the reaction for details. (You may edit your message)

[alpine jasper]

is this what you mean?

[supple furnace]

Um you lost a sum

We’re just extracting coefficients

[alpine jasper]

i don't think i'm getting the idea here

let me reread what you said above

so like this...?

i'm a big dumb

bc i'm only looking for coefficient of p^(-ns)

[supple furnace]

It should still be a double sum

[alpine jasper]

i don't know what i'm summing over when i write kr=n

[supple furnace]

Divisors of n

[alpine jasper]

i know that, i mean like, where i should put my second sum

[supple furnace]

Inner sum is over k|n

Outer sum is n>=1

[alpine jasper]

okay this must be it

[supple furnace]

There you go

[alpine jasper]

so removing the 1/n term gives us the result

[supple furnace]

Yes

[alpine jasper]

i don't fully understand how you turn the sum into that form, let me think about it more

thanks btw

[supple furnace]

The idea is that you want to find all terms corresponding to p^(-ns), so you set kr=n and solve for r in terms of k.

The new constraint is just k|n because k can be any divisor of n

[alpine jasper]

ok i think i get it now

thanks

[supple furnace]

Np

[sleek cove]

did i miss smth obv, why is the second equality true

[sleek cove]

i literally cant find anything saying that (p/q) = (q/p)

wait sry im dumb

[karmic edge]

Let p be an odd prime. Use Theorem 7.5 to prove that there is exactly one positive primitive
Pythagorean triple (x, y, z) with y = p, and find formulas for x and z in terms of p.

Theorem 7.5.
If (x;y;z) is a primitive Pythagorean triple, where x is even and x, y, and z are positive, then x= 2st; y=s2-t2; and z=s2+t2; for some relatively prime positive integers s and t. Conversely, if s and t are relatively prime, s > t >0, and s or t is even, then (2st; s2-t2; s2+t2) is a primitive Pythagorean triple.

I'm having some difficulty with how to go about this proof. I set y=p=s^2-t^2 but I see no way to write x and z in terms of just p.

[light flicker]

Think about factoring s² - t²

[karmic edge]

Right, so p = (s-t)(s+t). Ohhhh -- but p is prime! So s-t = 1.

I did that earlier and i don't know how i missed that lol

Thank you! I got it now

[obtuse mason]

let a be in N and p be a primefactor of n = (2a)^2 + 1

show p is congruent 1 modulo 4

I can eliminate 2 because (2a)^2 + 1 is uneven so 2 can't be a prime factor

but don't get much farther than that

[upbeat wren]

is n an element of N?

(in the problem)

[light flicker]

@obtuse mason p dividing (2a)^2 + 1 means that (2a)^2 + 1 is 0 mod p

Then think about quadratic residue things

[obtuse mason]

oh

@light flicker thank you so much 🙂

[dreamy rain]

i dont really get where this um +uv= g comes from?

for this to be true, u and v would have to be non-integer?

because g is smaller than both u and v

[fervent crest]

No

u and v can be negative

That’s the whole point

@dreamy rain

[dreamy rain]

oh are they using bezouts identity?

if so thats strange since bezouts hasnt been covered in the book yet

but thatd make the most sense yea

[fervent crest]

Lmao yes

It's the bezout's identity

[torn escarp]

,calc xgcd(13,17)

[sterile pumice]

Result:

[1, 4, -3]

[fervent crest]

,calc 413-173

[sterile pumice]

Result:

1

[stuck lintel]

Determine all positive integers $n$ such that $n^2$ divides $2^n+1$.

[sterile pumice]

Plum:

[winter bear]

consider the ||2 smallest prime factors of n||

[stuck lintel]

hmm

[sacred junco]

That's an imo p3 lel

[stuck lintel]

really?

[sacred junco]

Yes

[winter bear]

yeah lol

[stuck lintel]

its part of the amsp N2 pset bruh

[winter bear]

i solved this b4

which is why i managed to give u a hint in a second lol

[stuck lintel]

can u give me another hint lol

[winter bear]

umm ok start by considering smallest possible prime factor p of n

what can you deduce about p from this

[sacred junco]

Are you doing awesome math?

[stuck lintel]

no the deadline passed but im trying some sample problems for fun

hm yeah im not sure

[winter bear]

ok try arguing on order

[winter bear]

so first definations

[sterile pumice]

JohnDoeSmith:

JohnDoeSmith:

JohnDoeSmith:

JohnDoeSmith:

JohnDoeSmith:

[stuck lintel]

o lord

[sterile pumice]

JohnDoeSmith:

[winter bear]

yeah try doing these to also absorb the defination

tell me when you are done

[stuck lintel]

wait (n/d) is the legendre symbol right?

[winter bear]

oh no its just division lol

[stuck lintel]

oh 🤦‍♂️

oh its associative commutative since ${ d | d\text{ divides }n}$ is the same as ${n/d | d\text{ divides }n}$

[winter bear]

commutative* but yes

[sterile pumice]

Plum:

[supple furnace]

The sol to n^2|2^n+1 is very short tbh

[stuck lintel]

can u give a solution sketch?

also im not sure why its associative john

[supple furnace]

For associativity, think about convolution as sum ab=n f(a)g(b)

[winter bear]

yeah^

[supple furnace]

||Clearly n=1 works. Suppose n>1 and let p be the least prime factor of n. Then p|2^n+1 and so p|2^(2n)-1. Therefore p|2^(gcd(2n),p-1))-1. p-1 must be relatively prime to n since it is less than each prime factor of n, so p|2^2-1=3. Hence n=3k. If k=1, then n=3 works. Now suppose k>1 and let q be the least prime factor of k. We have q|2^(3k)+1 and so q|2^6-1 for similar reasons. 2^a+1 is never divisible by 7, so q=3 again. However, since n is odd, 1+v_3(n)=v_3(2^n+1)>=2v_3(n), contradiction for v_3(n)>1.||

[winter bear]

progress plum?

[stuck lintel]

i think i have an idea

[supple furnace]

John was anything different in your sol

[winter bear]

i think it was p much exactly the same, with minor divergences in the end

[supple furnace]

Forgot to say n is odd before lte oops

[winter bear]

right my order of logic was a bit different then urs, but ultimately same thing ||i said power of 3 must be 1 before doing the 2nd smallest prime factor||

[leaden compass]

yeah my sol was the same as yours john iirc

[supple furnace]

I like getting contradictions lol

[stuck lintel]

$f * (gh) = \sum_{ab = n} f(a)(gh)(b) = \sum_{ab=n} f(a) \sum_{cd=b} g(c)h(d) = \sum_{acd = n} f(a) g(c) h(d) = \sum_{ed = n} \sum_{ac = e} (fg)(e) h(d) = (fg)*h$

[sacred junco]

Tree is anti constructivist

[sterile pumice]

Plum:

[supple furnace]

Yeah nice

@stuck lintel

[stuck lintel]

hold on i need to work on my hw but ill brb

[leaden compass]

Personally I prefer intuitionist proofs

By this I do not mean following intuitionist logic, but rather by saying "intuitively it seems true, qed"

[winter bear]

alright we'll continue it after u return

[supple furnace]

Anyone have any problems

[leaden compass]

Yeah

[sacred junco]

What's 2+2

[leaden compass]

My problem is I have overdue homework

i do actually have one though

[winter bear]

show spectrum of a ring is disconnected iff it contains a non-trivial idempotent

[leaden compass]

a) Find a nice way of writing 1 + 2x + ... + (n+1)x^n

b) Find a nice way of writing 1 + 3x + ... + T_(n+1)x^n

c) Find a nice way of writing 1 + (T^k)_2 + ... + (T^k)_(n+1)x^n

Where (T^k)_n is the n-th k-th order triangular number (the sum of the 1st n (k-1)th order triangular numbers)

(for example (T^2)_3 = T_1 + T_2 + T_3 = 1 + 1 + 2 + 1 + 2 + 3

[supple furnace]

By Hockey stick this reduces to 1/k! times some kth derivative

Is that considered “nice”

[leaden compass]

Hint: (this idea applies to all of them so be careful before deciding to read) ||let's look at a: denote the sum by S, S - Sx = -(n+1)x^(n+1) + x^n + ... + x + 1 = (n+1)x^(n+1) + (x^(n+1) - 1)/(x-1) and so forth||

Not sure what you mean tree

[winter bear]

ew lol

derivative is much better

[supple furnace]

How is this “nice”

[leaden compass]

Nice enough

Care to explain the other method?

[stuck lintel]

Just use finite differences for all of them

[leaden compass]

Don't know that

[winter bear]

$$f(x)=\dfrac{1}{1-x} = \sum x^n$$ $$f^{(k)}(x) = \sum n(n-1)\dots (n-k+1) x^{n-k}$$ $$\dfrac{x^k}{k!} f^{(k)}(x) = \sum \binom{n}{k} x^n$$

[leaden compass]

Can you illustrate it

[sterile pumice]

JohnDoeSmith:

[winter bear]

i assume he meant

oh plum you are back

ready for some more?

[stuck lintel]

im procrastinating xdd

[winter bear]

oh oof

[stuck lintel]

i think ill be finished at around 9 est if ur available then

[supple furnace]

T^k(n)=(n+k+1)C(k+1)

[winter bear]

oh oops, i mean same logic tho

i prolly will plum

[supple furnace]

1/(k+1)!(d^(k+1)/dx^(k+1)((x^(n+k+1)-1)/(x-1)))

[leaden compass]

Now find all integral solutions to $n^420 - 1 | 420^n - 1$

[sterile pumice]

Archsys:

[supple furnace]

My formula’s fairly succinct and leads to a closed form with number of terms independent of n by product rule expansion.

[jovial hemlock]

Is there a way for me to prove that if A and B are coprime, then there is a k in 0 to (B-1) such that kA = X mod B for any arbitrary X in 0, B-1?

[stuck lintel]

do you know how to prove that the multiplicative inverse to A exists given (A,B)=1?

[jovial hemlock]

um

I’m not sure off the top of my head from an analytical level, but I can easily take your word for it

I can take that as a granted in what I need this for

Actually, is the thing I said I needed to prove enough of an established fact that I can use it in a proof of something else without having to prove it first?

[fervent crest]

Yes

But it really depends on what you are doing

If you're taking a number theory course, then your professor probably wants you to prove it

But if you're talking to people who know even a little number theory, then they know this fact

[jovial hemlock]

No, not a course

[winter bear]

i mean ig proving this needs ||bezouts|| and the only proofs i know for that use WOP

[fervent crest]

Or || use Euclidean algorithm and argue backward ||

[jovial hemlock]

Okay second year ion

question*

if you have a prime A, and the next prime is B, is B guaranteed to be less than 2A? I’m pretty sure it is

but I don’t know for sure if it’s been proven

[winter bear]

yes, its proven theres always a prime between p and 2p

[brittle patrol]

It's called Bertrand's Postulate

[jovial hemlock]

okay good

I think... I think I’m lined up really well

[winter bear]

its very high powered though, compared to what you are doing

[jovial hemlock]

I don’t think so

it’s more of a case of I cba to try proving those right now as I don’t want to get distracted from what I’m doing overall and lose my train of thought

[brittle patrol]

BTW an elementary proof that multiplicative inverses exist can be done by ||showing that no two elements in {0, a, 2a, ... , (b-1)a} give the same residue mod b||

[jovial hemlock]

“I don’t think so” was bad phrasing and arrogant, sorry

it was more “I’m not sure”

[winter bear]

oh thats smart gonzo

[jovial hemlock]

Gonzo, showing that no two elements have the same residue mod B is what I wanted to know if it was proven lol

[brittle patrol]

Yeah, but formulating it like that is like half the solution

think about what if ax=ay mod b

[remote sinew]

Quick question.

I'm not sure I understand why, about 2/3 of the way down, each of a,b,c must be even.

Why couldn't, say, a be even and b,c be odd?

...is it because then you'd end up with a number congruent to 2 mod 4?

[light flicker]

Yes

[remote sinew]

K.

I realized that as soon as I posted it :P

[fervent crest]

[sacred junco]

lmao

[exotic scarab]

did i do smthn

[sacred junco]

What

[jovial hemlock]

so

I don’t want to sound stupid or like a crank here

I have in front of me what I believe to be a proof of the twin prime conjecture. I am also fairly certain that such a proof must be wrong because of how long people looked for one and didn’t find one, but I personally can not find any errors in it. Where can I put it to have someone look over it and see if/where there are error?

[light flicker]

right here

[jovial hemlock]

like just type it out in the channel?

it’s a couple pages long in my word doc, is that okay?

[stoic bear]

Link it or upload

[jovial hemlock]

Um okay let me open discord on my computer

I'm sorry it isn't written formally I don't know how to do that

[light flicker]

should probably remove your name from the doucment

[jovial hemlock]

oops

In the third line I refer to L- this is a typo it should be O in that line

[light flicker]

I don't see how your earlier analysis, which seems correct to me, applies to the situation with Q, C and R

Earlier, when you were assuming you had a pair of twin primes, there was exactly one number N inbetween the twin primes

but here, there is no single number between Q and R, so you can't really apply your earlier analysis

[jovial hemlock]

The earlier analysis describes the property a number must have, the Q/C/R bit is about proving a number with those properties exists

Q and R aren't twin primes, they're the primes that are the equivalent of A and B in the earlier analysis

[light flicker]

Then, I guess I'm confused what the "list of last digits" is

[jovial hemlock]

like

[light flicker]

the last digits of what number

earlier, it was the possible last digits of N, but now there's no N

[jovial hemlock]

well there is an N

what you're doing is you're saying "If an N exists, it must have these properties; and if these properties are satisfied on a number then that number is an N"

and so the list for Q is saying "If an N exists, it must have one of these last digits when written in base C"

and obviously you can't immediately say "if these properties are satisfied on a number then that number is N" because the N could be 1 mod a prime above Q but below sqrt(N), which is what the section about G and F is dedicated to showing that a number has to exist without that problem

the list of last digits is basically saying "N = Y mod C" and the list is all possible values of Y

I'm just using the last digits phrasing because it's more intuitive to my brain so by using this phrasing I could avoid confusing myself when writing it up

what may help you is G is the equivalent of N

[light flicker]

Okay, I'm also confused because, you're not assuming such a G exists, you're trying to show that such a G exists right

[jovial hemlock]

right

[light flicker]

Then how can you let F be the largest prime below G - 1, if you don't know if G exists yet?