elementary-number-theory

[sterile pumice]

Publius:

[alpine jasper]

i stared at it for like 2, 3 mins and i got nothing

[light flicker]

what happens when you get to

[alpine jasper]

is there something useful when looking at element raised to the divisor of its order?

[light flicker]

$a^{k \frac{p-1}{k}}$?

[sterile pumice]

Zopherus:

[alpine jasper]

oh

i reached 1

[light flicker]

indeed

[alpine jasper]

and so everything after repeats

..

[light flicker]

yep

[alpine jasper]

and since a has order p-1

everything before k does not repeat

hmm did i do something stupid with my examples?

,rotate

[alpine jasper]

like, for 17 and 19, something repeats before kth element

oh

i didn't start with a^((p-1)/k)

thank you zoph

[light flicker]

any time

[azure pendant]

i have a pretty baby question but i cant see how to make any progress on it. i'm supposed to find the least residue of $10^{2019}$ mod 1009. it's pretty easy to factor out a 3 and show that this is the same as $(-9)^{673}$ mod 1009, but i'm not sure where to go from here

[azure pendant]

the problem is written in a way that makes it seem like 1009 being prime is important, but im not sure how to use that information

[fervent crest]

Fermat's little theorem

[azure pendant]

oh

we havent learned that but i can probably work through the proof in my solution

[fervent crest]

So Fermat's little theorem states that if $p$ is prime and if $\gcd(a,p)=1$, then $a^{p-1}\equiv1\pmod{p}$

[sterile pumice]

Whoever:

[sacred junco]

Yeah there’s no way to do this w/o fermat

[fervent crest]

There is always another way

[winter bear]

indeed there is

u manually calculate (-9)^673 first

in Z

and then take the mod

ez

[torn escarp]

the only alternative I'd consider legitimately is writing 673 in base 2 and then work out by repeated squaring but even then 🤢

[light flicker]

This is more #discrete-math

[sacred junco]

sorry

[sacred junco]

if we have f(x) = 1988x^2 + bx + 8891, and g(x) = 8891x^2 + bx + 1988, why is it that any factor of g or f must be a factor of g(x) - f(x)

i think i may be missing the obvious here

[light flicker]

its not obvious

you have to use the fact that the polynomials are kind of symmetric to one another

[sacred junco]

so if the quadratic and constant coefficients were different, this would be not true?

[light flicker]

it says or

whoever

[sacred junco]

why did you delete your messages @fervent crest

?

[light flicker]

yes you're right

its only because you get the polynomials from switching the constnat and quadratic terms that this works

if it said that any factor of g and f must be a factor of g - f

then this is always true, regardless of g and f

[sacred junco]

what if it said some (or more) factors of g must be factors of f

but not necessarily all?

[light flicker]

uh

idk, that's a weird statement to make

[sacred junco]

well what about what @fervent crest said

[light flicker]

I guess its always true since 1 is a factor of g, and is always a factor of f

[sacred junco]

before they deleted it

it was something like

f(x) = q(x) * h(x) and g(x) = p(x) * h(x)

[fervent crest]

lol i deleted because it wasn't the question

[sacred junco]

then g(x) - f(x) = h(x)(p(x) - q(x))

[light flicker]

I mean, this is what I said

in this case, h divides both g and f

[sacred junco]

okay so the original polynomial difference was g(x) - f(x) = 6903x^2 - 6903 = 6903(x^2 - 1)

so i suppose i'm a little bit unclear on why this means x^2 - 1 is a factor or something?

[fervent crest]

Now I don't think the statement is true tbh

[hasty jacinth]

If it is a factor of f or g, and a factor of g-f, then it is a factor of the other

Implies f and g has the same roots?

But they dont have the same roots..

[sacred junco]

i am just really lost right now ngl

[light flicker]

yeah, this isn't true

[sacred junco]

what is not true?

[light flicker]

your original statement

"that any factor of g or f must be a factor of g(x) - f(x)"

[sacred junco]

sorry

it's missing 1 important word

"any common factor of g and f must be a factor of g(x) - f(x)"

[light flicker]

oh

[winter bear]

g and f

not or

[sacred junco]

so now it's true?

[light flicker]

yes

[sacred junco]

okay. could you explain why?

[light flicker]

this is true in general for any ring

just write out the definition of divides

[sacred junco]

g(x) - f(x) = h(x)(p(x) - q(x)), so is 6903 h(x) in this case?

or is it (x^2 - 1)

[light flicker]

h is anything that is a common factor of g and f

[olive remnant]

anyone know how to solve this its driving me crazy

[sacred junco]

alright thanks

[olive remnant]

welp

[light flicker]

what do you know about these types of problems

[olive remnant]

that's not what i thought that was going to happen

[light flicker]

{ }

[olive remnant]

i know euler's function

3^(71)^82

and i guess i know like how to find inverses with like

[light flicker]

euler's function?

[olive remnant]

like phi

[hasty jacinth]

Phi

The totient one

[olive remnant]

where it gives like the # of coprime numbers less than it

[light flicker]

And why do you think that helps you here?

[olive remnant]

maybe like

reduce down the 82 first

by doing (3^71)^82 (mod phi(23))

where im modding the 82

idk if that's a word

[light flicker]

And why would this help?

[olive remnant]

bc maybe then i would get a smaller number to work with so i could just brute force it lol?

[hasty jacinth]

23 is already prime

[light flicker]

Does this preserve your original number though?

[olive remnant]

i think?

[light flicker]

how do you know that $3^{72^{82}} \pmod{ 23} = 3^{72^{82}} \pmod {\phi(23)}$?

[sterile pumice]

Zopherus:

[olive remnant]

well the exponent gets modded by phi(23) but the number sa a whole is modded by 23 still

[light flicker]

by the way, $\phi(23) = 22$ since 23 is prime

[sterile pumice]

Zopherus:

[olive remnant]

yeah

[quick furnace]

[light flicker]

Yeah okay, that's the right idea

So we can reduce down the exponent, which is $72^{82} \pmod{\phi(23)}$

[sterile pumice]

Zopherus:

[light flicker]

This is called Euler's theorem by the way

[olive remnant]

oh okay

[light flicker]

Okay, now what

(you should also try to understand why this is even allowed)

[olive remnant]

tbh idk after that point :/

[light flicker]

well okay, we know that $\phi(23) = 22$ so its just

[olive remnant]

ive been trying just random brute force stuff for an hour lol 😅

[sterile pumice]

Zopherus:

[light flicker]

$72^{82} \pmod{22}$

[sterile pumice]

Zopherus:

[light flicker]

what's one way we can simplify this? How would you simplify $72 \pmod{22}$?

[sterile pumice]

Zopherus:

[olive remnant]

6 mod 22

i think

so 6^82

?

[hasty jacinth]

You can do better than that

[light flicker]

And why can we do that?

[olive remnant]

6^16 mod 22?

uh not really sure

i've kinda just been doin it

lol

[light flicker]

Can we reduce down the exponent mod 22? Are we allowed to do that?

[olive remnant]

yes bc of euler's theorem?

[light flicker]

What does Euler's theorem say

I.e., what did we do earlier?

Did we reduce the exponent mod 23?

[olive remnant]

phi (23)

we reduced it by phi 23

so it becomes mod 22

[light flicker]

So what do we need to do here, to reduce the exponent

[olive remnant]

and thats what we're reducing the exponent by?

[light flicker]

But the exponent is $72^{82}$

[sterile pumice]

Zopherus:

[light flicker]

so we're trying to find $72^{82} \pmod{22}$

[sterile pumice]

Zopherus:

[olive remnant]

oh

[light flicker]

and we need to reduce this somehow

[olive remnant]

ohhhh

[light flicker]

and it'd be nice if we could reduce the exponent of this

[olive remnant]

im not entirely sure how to do that ik the bottom is 6^82

but now i see that we can't just do the euler's stuff to the power

[light flicker]

and why not?

[olive remnant]

bc 22 isn't prime

or coprime at least?

[light flicker]

Does that matter?

[olive remnant]

coprime to uh

22 and 6 aren't coprime

so we can't use euler's theorem i think

[light flicker]

Yeah

So we want to split this into things so it is coprime

Do you know of anything that could help us with that

[olive remnant]

can't we just do 3^82 * 2^82

[light flicker]

but are those both coprime to 22?

[olive remnant]

no i thought we'd just do one of them and worry about the other one later lol

in that case i have no clue :/

[light flicker]

Then sure, we can

so which one can we do

[olive remnant]

3

[light flicker]

and what do we get when we do it

[olive remnant]

3^16

which is 5

mod 22

[light flicker]

uh, how'd you get that?

[olive remnant]

uh i did (3^4)^2

and i just did 3^4 mod 22

[light flicker]

how'd you reduce it down to 3^16?

[olive remnant]

and then squared it and put it through mod 22 again

oh

oh right

phi(22) is 10

3^2

so its 9 mod 22

mb

[light flicker]

yeah

now for the second part

any ideas?

[olive remnant]

uhh

[light flicker]

if we write that $x = 2^{82} \pmod{22}$

[sterile pumice]

Zopherus:

[light flicker]

then we know that $x = 2^{82} + 22k$ for some integer $k$ right?

[sterile pumice]

Zopherus:

[olive remnant]

yeah

so that's euclidean algorithm right

[light flicker]

uh

not really, this is just the definition of mod

[olive remnant]

o

i thoguht maybe we had to use euclidean algorithm at this point

oops

[light flicker]

well okay, is there anything you notice about x here?

Like some condition x has to satisfy?

[olive remnant]

it has to be even ig?

[light flicker]

yeah exactly

so we can write that $x = 2y$ for some integer $y$

[sterile pumice]

Zopherus:

[light flicker]

so that $2y = 2^{82} + 22k$

now what

[sterile pumice]

Zopherus:

[olive remnant]

now y= 2^81 + 11k

so that's uh y = 2^81 (mod 11)

and then its like uh 2

bc 81 mod phi(11) (which is 10)

[light flicker]

yeah

exactly

now bring everything together

[olive remnant]

wait but how i have a 9 mod 22 and a 2 mod 11

[light flicker]

but $y = 2$ so what is $x$

[sterile pumice]

Zopherus:

[olive remnant]

oh so x = 4

wait

no

im st0pid

x = 1

so its just 9 in the exponent?

[light flicker]

$x = 2y$

[sterile pumice]

Zopherus:

[olive remnant]

o

so its 4

[light flicker]

(x had to be even remember)

[olive remnant]

yeah smh im stupid mb lmao

so 36 in the exponent which reduces down to 14

bc the exponent was mod 22

[light flicker]

yes

still a bit of calculation to do but its a bit easier

[olive remnant]

so now we have 3^14

and im good from there i think

thanks!

sorry for being a bit slow lol

[light flicker]

you're fine

[olive remnant]

i haven't had the time to focus on number theory bc my discrete and multivar classes were harder than i thought they would be B(

but thank u very much for the help

[light flicker]

any time

[sacred junco]

hey, I have a doubt regarding definitions: does a prime element always has to be irreducible as well?

I know that in an integral domain that's always the case, but it is not strictly required by definition, right??

[light flicker]

Yes

[sacred junco]

okay, thanks!!

[sacred junco]

I solved the a part by proving b and c part

I wanted to know if there is a way to directly prove the a part

(also because that's what the author wanted us to do)

[light flicker]

how did you do part b and c?

[sacred junco]

for part b, as any field with 0 characteristic has subring isomorphic to Z, and so will have a subfield isomorphic Z's field of fractions, Q

for c I used the map :

$\phi (x) = 1.x$

[sterile pumice]

mega:

[sacred junco]

where x element of $Z_p$

[sterile pumice]

mega:

[light flicker]

hm, what's the definition of prime subfield?

[sacred junco]

a prime field is one which has no proper subfields

a prime subfield is one which has no proper sub-subfields

[light flicker]

ah okay

Well I guess one problem would be that you haven't done part (a) in showing that this is the unique prime subfield

[sacred junco]

ah okay, you mean I have shown only existence, not uniqueness?

[light flicker]

yeah

[sacred junco]

I think it can be proved that's it's unique in the sense that all other subfields will have to contain the subfield formed from the image of the mapping

I mean all subfields will have 1, and thus will have 1, 1+1, 1+1+1...

[light flicker]

yeah

[sacred junco]

does this help in anyway in proving part a) ?

[light flicker]

wdym?

[sacred junco]

I wanted to know anyway to directly prove part a) without first proving part b) and c)

I proved only existence for part c) but now also got to uniqueness

does this help in part a)?

[alpine jasper]

i tried it with x^4=-1(p) but i don't know how to argue that (p-1, 4) = 4, i tried to assume that it equals to 2 and get a counter example but that didn't really worked out

@light flicker my nt papa help pls

[light flicker]

okay, well if (p-1,4) = 2, then you know that p is 1 mod 4 right

[alpine jasper]

yes

[light flicker]

so you just have to show its not 5 mod 8

[alpine jasper]

actually, if i pick p=7 then that breaks it?

[light flicker]

wdym breaks it

[alpine jasper]

the implication (p-1, 4) = 2 implies p = 1(4)

logic is not my strong suit

so you just have to show its not 5 mod 8
p != 5(8)?

i don't understand how you arrived at that

[light flicker]

the statement (p-1,4) = 2 is the same as (p-1,2) = 2, and so it just becomes the first part

if p is 1 mod 4, then it is either 1 mod 8 or 5 mod 8

[alpine jasper]

oh

hmm

alright let me see if i can prove that

uhh idk man,
if i assume (p-1, 4) = 2
then that's iff p = 1(4)
and i'm suppose to show that this implies(iff?) to p = 1(8)

[light flicker]

indeed

Again, p = 1(4) iff p = 1 (8) or p = 5(8)

can a prime p satisfy both p = 5(8) and (p-1,4) = 2?

[alpine jasper]

hmm probably not, let me try to prove that

4 does not divide p-1 hence 4 does not divide p-5, hence 8 does not divide p-5

sounds ok..?

wops

[light flicker]

uh

[alpine jasper]

i'm eating rn, sorry but maybe i should finish eating first

[light flicker]

yeah, that's right

[alpine jasper]

ty

[alpine jasper]

how can i figure out $x^7\equiv2(49)$ has any solution without solving it?

[sterile pumice]

Publius:

[alpine jasper]

@light flicker

so i was trying to use this but p = 7, and 7 divides 7

i don't know if there are any other theorems out there

[light flicker]

i dont think its easy

[winter bear]

i mean i'd just say

that x^42=1 so 2^6 must =1 which is contradiction, but this doesnt generalize that nicely i think

2 and 6 are small numbers here

[alpine jasper]

hmm

[light flicker]

oh no thats nice

[alpine jasper]

wow

that's a pro fucking gamer move

wait actually i don't quite understand why x^42 = 1

oh wait phi(49)=42

nice

[alpine jasper]

actually, how do you know x^42 = 1? i thought i need prime to apply FLT @winter bear

wops

eulers theorem

[winter bear]

x^42=1 for all x in Z/49Z*

yeah

[alpine jasper]

am retarded

[torn escarp]

looks like Hensel's lemma

[alpine jasper]

yeah, my next question is apply hensel's lemma to calculate some stuff

anyway... new question

so i'm trying to do the same for $x^7\equiv18(49)$\
i rise both sides to 6th power and since $18^6\equiv1(49)$, i get $x^{42}\equiv1(49)$, which is true by euler's theorem, this means i can pick any $x$ that is co prime to 49 and it'll work

[sterile pumice]

Publius:

[alpine jasper]

but it doesn't

the pattern seems to be +7, which is the power

what is going on

[winter bear]

this garuntees the existence of a solution

so basically use the fact that Z/49Z* is cyclic

[alpine jasper]

so i find a generator/primitive root

?

[winter bear]

i mean if you want the explicit solutions sure

to show existence/number of sols u dont need that

[alpine jasper]

i know, but know more is better than know less

can you explain why this won't give me solution?

did i alter the congruence when i rise both sides by 6

i guess i surely did

[winter bear]

um suppose g is a generator of the multiplicative group

[alpine jasper]

uh huh

[light flicker]

your implications are basically going backwards

[winter bear]

we are saying that since 18^6= 1(49), that this thing has a solution right, cause 18=g^n for some n and g^(6n) = 1 which means 7|n

[light flicker]

just because x^42 = (18)^6 doesn't mean you can just take 6th roots

and say that x^7 = 18 for all x

[winter bear]

yeah

[alpine jasper]

okay let me digest john's message, i'm big dumb at algebra

[winter bear]

thats proof of existence btw

[alpine jasper]

gotcha

hmm alright

how does the logic work here if i'm going backwards?

i showed that that congruence leads to euler's theorem

uhh...? but F => T is still true

[winter bear]

yeah that wont work here, which is why i wrote that thing

[alpine jasper]

[winter bear]

if 18=g^n and since 18^6=1, then g^6n = 1 which means 7|n as order of g is 42. this means 18 is a 7th power

the fact that these groups are cyclic are very powerful for precisely this kind of manipulaitons

[alpine jasper]

i don't see how that implies solvability

can you explain more please

[winter bear]

ok erm

do you agree with 18 is 7th power

[alpine jasper]

yes, since 18 \equiv x^7 right

[winter bear]

no using the generator argument i used

forget about the x^7 for now

[alpine jasper]

yes

i agree that 18 is the 7th power

w/ the generator method

[winter bear]

so if 18 is a 7th power

then does x^7=18 have a solution

[alpine jasper]

i suppose that allows me to take 7th root on both sides?

oh wait yes duh

[winter bear]

the root isnt going to be unique

but it shows there is atleast 1 sol

[alpine jasper]

if 18 is the 7th power then there exist some x such that x^7 = 18

oh then that's it?!

there exists that x

[winter bear]

yup

show a general statement using what i showed u now

[alpine jasper]

like generalize it to $x^n\equiv a(p^e)$

[sterile pumice]

Publius:

[alpine jasper]

so pick a generator of Z/p^eZ*

[winter bear]

suppose $d|\phi(p^k)$, show $x^d= a (p^k)$ is solvable iff $a^{\phi(p^k)/d}=1 (p^k)$

[alpine jasper]

oh hmm

[sterile pumice]

JohnDoeSmith:

[winter bear]

just generalize both arguments we made

[alpine jasper]

i'm lost in all those symbols lmao

gimme a moment please let me write it down

[winter bear]

yeah take ur time

[olive remnant]

I'm trying to figure out how many solutions there will be to a^2 \equiv 1 (mod 195) and I know the solutions to this will be the same as the solutions to the system of congruences of a^2 \equiv 1 (mod all the prime factors of 195) and I know that each of these prime factors will have two answers a= plus or minus 1 (mod some prime factor) but idk how to account for double counting solutions solutions?

like in the case of 35 i have the system of congruences a^2 \equiv 1 (mod 5) and a^2 \equiv 1 (mod 7)

and i'll double count 6 as a solution

idk if that makes any sense

[winter bear]

you would not double count 6 as a solution

infact there would never be any double counting, courtesy of chinese remainder theorem

[olive remnant]

fuck

you're right

im stupid

idk how i thought 6 was the solution to two things

two system of congruences

[winter bear]

-6 is a solution too

[olive remnant]

yeah

[winter bear]

prolly messed it up there

[olive remnant]

yeah

that makes sense

[alpine jasper]

is what i wrote full of garbage or

i know we just went over an example but i'm dumb

[winter bear]

so a few things

[olive remnant]

thanks by the way

i think forgot to say that

[winter bear]

a^x=1 means ord(a)|x

np

not the other way around which is what u seem to have there

also umm yeah the other implicition needs to be done

[alpine jasper]

i can't even do one implication how can i do both lmao

[winter bear]

well do this implication first then

look back at what i did for the example

[alpine jasper]

i sort of did but couldn't connect the dots

let me try harder

,rotate

[alpine jasper]

this makes sense?

[winter bear]

yeah

do you get an intuition for this now?

[alpine jasper]

yes, but i wanna think about this part a bit more

[winter bear]

aight go ahead do that, keeping in mind that a^x=1 means ord(a)|x

[alpine jasper]

thank you so so much, and thank zoph also

[winter bear]

np

[winter bear]

(btw try continuing this to when d doesnt divide n, use gcd)

[olive remnant]

so the question im trying to answer is how many solutions to a^2 \equiv 1 (mod 2^e) where e is any natural number (not zero)

from test cases it seems that a \equiv plus or minus 1 (mod 2^e) and (mod 2^(e-1))

so when e > 2 theres 4 solutions, else there are 2 solutions

i started proving this but uh i realized i my proof would also show that any odd number should also work

so i have no clue how to solve this or if my conjecture(?) is even right

nvm got it

[torn escarp]

yep it's true if you want me to check your proof or whatever

[olive remnant]

i think i got it

its just not very rigorous bc its 5:50 am

and im too lazy

but i just said a^2 \equiv 1 (mod 2^e) equivalent to the system of congruences a^2 \equiv 1 (mod 2^(e-1)) .... a^2 \equiv 1 (mod 2)

which is the same as just doing a^2 \equiv 1 (mod 2^(e-1)) (which i proved in words lol)

and then i just showed that applying this to different cases with different values for e give me what my conjecture(?) was

[torn escarp]

hmm

what I had in mind was more like, take $a=1+2^bx$ and square it

[sterile pumice]

Merosity:

[torn escarp]

then look at which e this could be equal to 1

and then go back and determine what xs were allowed

[olive remnant]

ughhh

im sure that's better

i just don't know if its worth it at this point

im just so drained

does what im saying have a semblance of truth

so i can justify putting what i just said down

also i have no idea where u got that from

[torn escarp]

well you said some stuff and said "which you proved in words" and I didn't see what you said so

I really have nothing to go on

[olive remnant]

lol yeah :(

[torn escarp]

you sound like you just need to sleep and come back when you're fresh

I was terse in what I said

[olive remnant]

lol

aint no rest for the wicked or something like that

but yeah

thanks i appreciate the help

[torn escarp]

just try to reason why every a that's a solution is of the form 1+(2^b)*x for some b, x

then by squaring it you end up with some more concrete restrictions mod 2^e

[olive remnant]

well that's just cuz you can just represent it as 1 + 2^(e-1)*(2k)

so x = 2k and b = e-1

oh wait im dumb

still thinking

yeah i can't figure it out

ill just turn in the homework as is i guess

[torn escarp]

no you restricted x too much

that's not the reason why you're sorta working the opposite direction

$a=1+2^bx \mod 2^e$

[sterile pumice]

Merosity:

[torn escarp]

we're supposing we know nothing about b and x except that b>0 and x is an integer

but we know that squaring gets us 1

$1 = 1 + 2*2^bx +2^{2b}x^2 \mod 2^e$

[sterile pumice]

Merosity:

[olive remnant]

can't we get something larger than 1 leftover since we don't know if b is divisible by e

[torn escarp]

$0 = 2^{b+1}x ( 1+2^{b-1}x) \mod 2^e$

[sterile pumice]

Merosity:

[torn escarp]

yeah so you can start to see what are the minimum constraints to force this to be 0 mod 2^e

b+1=e would do it but that doesn't force x=2k

you just have to be careful about what b is when it's small which is where it switches over, right?

is this clear or do you not see where to take it from here

[olive remnant]

oh

i see

i don't understand how you started with 1 + 2^b*x

tho

[torn escarp]

because it's the most generic choice I could make

it has to be either a=0+2^b*x or a=1+2^b*x

for b>0

either it's even or odd, that's really all I'm saying

the 2^b is just saying there are some number of 0s (possibly none) up until this value times x

err written in base 2

sorry I implicitly assume I'm working in base 2 when thinking about stuff mod 2^e

since each e basically chops off a further digit, it's just cleaner

it's not too serious if that's confusing to you

either a=0+2^b*x or a=1+2^b*x

this is just a generic representation for some b>=1 and integer x

and we can assume x=1 mod 2

[olive remnant]

yeah i get that

[torn escarp]

it's just a way to force it more cleanly

[olive remnant]

ive just never seen that representation of even and odd numbers

[torn escarp]

$a= a_0+a_12+a_22^2+a_32^3+\cdots$

[sterile pumice]

Merosity:

[torn escarp]

basically looking at its base 2 expansion and then saying, there's some number of 0s

yeah yeah you get it I guess I forget this isn't so common to see

specifically working in the 2-adics is working with numbers mod 2^e for all e simultaneously

so I got used to that, which is where I'm kind of coming from

[olive remnant]

aight

sorry just thinking

thanks for the help

[torn escarp]

take your time I've got nothing better to do right now

[olive remnant]

im sorry i just can't

i wanna cry

i've pulled an all nighter

and i just can't figure it out

[torn escarp]

I'm just here to talk about math ideas, time limits and morale problems are out of my sphere sorry

[alpine jasper]

$\left(\frac{\cdot}{p}\right):(\mathbb Z/p\mathbb Z)^\times \to {\pm1}$

[sterile pumice]

Publius:

[alpine jasper]

how would i show that this is a homomorphism..?

kinda lost

this is legendre symbol btw

[light flicker]

by proving it

[alpine jasper]

thanks i'm cured

..

[light flicker]

np

always here for you

[alpine jasper]

love you

[light flicker]

so prove it

[alpine jasper]

i guess i can try harder, i'll report back if i fail

actually

could i just map the quadratic residues to 1 and non residue to -1... uhh

i know that there are equal amount of them..

[light flicker]

I mean

That's the definition of the legendre symbol yes

The legendre symbol is a function from (Z/pZ)* to {\pm 1}

[alpine jasper]

$\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$

this proves it..?

for a and b in (Z/pZ)*

[sterile pumice]

Publius:

[alpine jasper]

is it this simple..?

or am i big wrong again

[light flicker]

That would prove it

if you know that statement is true

[alpine jasper]

ah neat

ireland and rosen proved that fact

[light flicker]

true

[alpine jasper]

very cool, thank you

[light flicker]

any time

[sacred junco]

yeah this one I dont think is easy to prove

[light flicker]

its not particularly difficult, Ireland Rosen does it in like 3 lines

[sacred junco]

(ab)^n = a^n b^n

[sacred junco]

lol

[light flicker]

I mean, he's not wrong

At least, the way Ireland Rosen proves it is to show that

(a/p) = a^((p-1)/2) (mod p)

And then you can use the fact that exponents split

[wild zinc]

most of it can be done just fine

but showing product of two non-residues is a residue is not as simple as the rest

the rest are like one line contradiction or construction

[sacred junco]

Hello

Can anyone help me prove Brocards theorem using induction?

Will be thankful to anyone who helps

[torn escarp]

,rotate

[torn escarp]

the problem just wants you to just check a few cases by hand

proving the general statement is an unsolved problem

[plain fable]

ya it's just 8 cases (if we include 0)

[sacred junco]

Oh so i have to just put the values and check?

[stoic bear]

Yes

The open problem is called Brocard's problem if curious

[fervent crest]

When does the equation $$x^k\equiv a\pmod{m}$$ have a solution in $x$?

[sterile pumice]

Whoever:

[torn escarp]

when you can find $x$ for every $p|m$ such that $|x^k-a|_p < |k x^{k-1}|_p^2$

[sterile pumice]

Merosity:

[fervent crest]

[torn escarp]

once you have that, you can hensel lift up to the appropriate power of p and combine with the chinese remainder theorem

;P

[fervent crest]

Way too hardcore of a condition

😔

[torn escarp]

I'm partly teasing

well two things

the p-adic absolute value part can be reworked into just modular arithmetic if you want to see that part,

$$f(x)\equiv 0 \mod p^{2a+1}$$ $$f'(x) \equiv 0 \mod p^a$$ $$f'(x) \not \equiv 0 \mod p^{a+1}$$

[sterile pumice]

Merosity:

[torn escarp]

usually people just show the a=0 case for Hensel lifting but this is the more general circumstance when f'(x)=0 mod p

idk hopefully that's not too scary as a requirement

but in some sense I didn't really answer the original question either, I just pushed the answer back to looking at x^k=a mod individual small powers of primes

[fervent crest]

🙃 Take in mind that I don't know a lot about p-adic

[torn escarp]

ok ok back up have you seen this condition for solving for x for f(x)=0 mod p^k

if you satisfy $f(x) \equiv 0 \mod p$ and $f'(x) \not \equiv 0 \mod p$ then you can lift?

[sterile pumice]

Merosity:

[fervent crest]

Not the |x^k-a|_p condition you posted above

Oh

Yeah you've showed me this

I think

[torn escarp]

yeah

[fervent crest]

It was in the hensel's lemma

[torn escarp]

I didn't want to scare you too bad so I only showed you this case when f'(x) != 0

https://cdn.discordapp.com/attachments/418981682117345288/691692663593762916/622029249053982720.png

plug in a=0 and you see this second condition is f'(x)=0 mod 1 which is trivially true

it collapses back down to that, this is just so that when you look at your f(x)=x^k-a (different a from my powers whoops bad choice)

your f'(x) = kx^{k-1} might have k divisible by some power of p in common with m

which can mess things up

take gcd(k,m)=1 and your problem becomes simpler

now you really are just looking at x^k=a mod p in all circumstances

[fervent crest]

Oh

Well if gcd(k,m)=1 then I have an elementary method to solve this equation

Idk if this guarantee existence tho

[torn escarp]

yeah you need to start talking about the phi function / fermat's little theorem or something

[fervent crest]

Oh yeah

The condition isn't gcd(k,m)=1

my bad

it was gcd(phi(m),k)=1

Then you write u*k-v*phi(m)=1

somehow make u,v≥0

then

$a^u\equiv\br{x^k}^u\equiv x^{1+v\cdot\varphi(m)}\equiv x\pmod{m}$

[sterile pumice]

Whoever:

[fervent crest]

But like

I don't think this shows the solution exist

Oh wait nvm it does

I'm dumb xD

But still what if gcd(phi(m),k)≠1

[torn escarp]

I stepped away for a bit

hmm what you wrote looks backwards to me, seems like you're assuming a=x^k

[fervent crest]

?

So I assumed gcd(phi(m),k)=1, then write u*k-v*phi(m)=1 where u,v≥1, then you have that x=a^u is a solution

[torn escarp]

your first equals sign, how do you make this substitution:

$a^u\equiv (x^k)^u \pmod{m}$

[sterile pumice]

Merosity:

[torn escarp]

aren't you assuming a=x^k here?

[fervent crest]

yeah

i realized that

but it's also really easy to show that a^u is a solution

$\br{a^u}^k \equiv a^{uk}\equiv a^{1+v\cdot\varphi(m)}\equiv a\pmod{m}$

[sterile pumice]

Whoever:

[torn escarp]

oh ok

[torn escarp]

I guess the next thing is if gcd(k,phi(m))>1 to split off the offending powers of primes from m so that we have the simple cases taken care of

and then specifically look at gcd(k, p^{n-1}(p-1)) > 1

[fervent crest]

Hm ok

[torn escarp]

kind of distracted but I got a handful of results for some cases

if k isn't divisible by p, then you just need to have x^k=a mod p

i k is divisible by p, and p>2 then you need the numerator of (a-1)/k divisible by p once

if k is divisible by p and p=2 then you need the numerator of (a-1)/k divisible by p twice

that's enough to guarantee you have a solution but there could be some weaker conditions I'm pretty sure

@fervent crest tell me if these are obvious or not I have tunnel vision

[fervent crest]

I'm not sure what you meant by offending powers of prime

[torn escarp]

oh

gcd(k, phi(m)) > 1

so you factor m into two parts m=r*s such that gcd(k, phi(r))=1 and gcd(k, phi(s))>1

then we look only at the powers of primes in s

and look at just these cases for the sake of gluing them back together with the chinese remainder theorem

because phi is multiplicative and I'm picking gcd(r,s)=1 for m=r*s

[fervent crest]

Ok

[torn escarp]

gcd(k, phi(m)) = gcd(k, phi(r)*phi(s)) but

the idea was to pick them such that this splits as well, that's what I meant by offending

[fervent crest]

Hmmm ok

[inland hawk]

has anyone used the chinese remainder theorem before

to solve a system of congruence

[light flicker]

What are you confused about?

[inland hawk]

trying to solve this system

i got an answer but it doesnt check out with the second congruence 3(mod 5)

[light flicker]

Maybe show your work?

[inland hawk]

sorry couldn’t get a good picture

I got 662

[light flicker]

Ah yeah

check your b2

[inland hawk]

yeah I get 2(mod 5) when I check it so ik that’s wrong

[light flicker]

Yeah, so check your calculations for b2 again

[inland hawk]

i think I’m blind or just missing something lol

I keep getting b2=2

[light flicker]

what's 252 * 2 (mod 5)?

[inland hawk]

4

[light flicker]

but what do you need it to be

[inland hawk]

1

[light flicker]

so what do you need b2 to be instead

[inland hawk]

3

[light flicker]

yes

[inland hawk]

oh ok lol

[light flicker]

And I mean, you even wrote it down

that 2 * 3 = 1 (mod 5)

[inland hawk]

thx im a silly goose

yeah I was just kinda following another example without even thinking

[sacred junco]

I noticed that
15/154 = 9.7%
10/149 = 6.7%
is that a coincidence?

15-1/3 * 15 = 10
9-1/3 * 9 = 6

[stoic bear]

?

[torn escarp]

there's no pattern here

[sacred junco]

likewise 20/159 = 12.6%

[torn escarp]

what

[sacred junco]

10/149 = 6.7%
15/154 = 9.7%
20/159 = 12.6%
25/164 = 15.2%
30/169 = 17.7%
35/174 = 20.1%

it seems to me like you can roughly approximate the %

10/149 = 6.7
12/151 = 7.9
14/153 = 9.1
16/155 = 10.3

obv few jumps like 15.2 to 17.7 and you'd be off by quite a bit

but it's still better than nothing

[winter bear]

im guessing your claim is that the approximate increase is by 3% at each turn?

[sacred junco]

@winter bear with 5 yeah

with 2 it seems to be 1.2

[winter bear]

$\dfrac{n}{x+n} = \dfrac{n}{x} \dfrac{1}{1+\dfrac{n}{x}} = \dfrac{n}{x}-\dfrac{n^2}{x^2}+\mathcal{O}((n/x)^3)$

[sterile pumice]

JohnDoeSmith:

[winter bear]

with x>>n

this approximation should show why its increasing at about a 3% rate when ur numbers are small enough

[sacred junco]

what's that letter at the end

[winter bear]

big O notation

[sacred junco]

when do you learn it

[winter bear]

just google it, the defination is p easy to understand

[sacred junco]

also what does the >> mean

[winter bear]

much greater than

[sacred junco]

thanks

[winter bear]

im using some taylor series approximation, if your familiar with those

cause thats what your problem essentially breaks down to

you are seeing the linear term in the expansion dominate for small values of n

[sacred junco]

thanks for that, I'll try to read some wiki pages on it

[rigid loom]

Any books u would recommend for number theory?

[light flicker]

How much math have you studied?

@rigid loom

[rigid loom]

I had maths in my alevels but now am in first year of uni

Having trouble with modular arithmetic and stuff

[light flicker]

hm, you could look at Silverman's A Friendly Introduction to Number Theory

But, if you just want to understand the stuff you're struggling with, you can just ask here, or there are videos online

[rigid loom]

More of a book guy tbh

[light flicker]

Yeah I mean, trying to read a book just to figure out modular arithmetic probably won't be the most efficient

As in, this book goes over some unrelated stuff before talking about modular arithmetic and you might not be able to skip that stuff

[rigid loom]

Ahh i see alright I’ll look into this stuff tbh and if encounter any problem will get back to u

[light flicker]

I could always just try explaining stuff to you too

[rigid loom]

Thanks bud

[alpine jasper]

why is the flipping rule allowed with (1001/9907) even tho 1001 is not a prime?

i know that the Legendre smbol is defined for (a/p) a need not be a prime

but what confuses me is why can i flip without a being a prime too since the law says q and p both ood prime

pls come to rescue zoph papa @light flicker

[supple furnace]

The situation for the Jacobi symbol is a bit different. Jacobi symbols don’t always detect QRs properly, but they will always detect QNRs (if the Jacobi symbol is -1, you have a nonresidue).

[alpine jasper]

right, i understand that, but not what i'm asking, sorry for my shit wording above

the one on the left is a Legendre symbol, yes?

[light flicker]

Jacobi symbols are equal to legendre symbols when the "denominator" is prime

[alpine jasper]

hmm alright

but how does that allow the flip.. in the first step?

[light flicker]

quadratic reciprocity

[alpine jasper]

but that requires both top and bottom to be primes tho...?

[light flicker]

Look at property 6 on that wiki page

[supple furnace]

Since the Jacobi symbol matches with Legendre symbol in prime case, it’s always right

It has the do with the number of QRs modulo n in general

[alpine jasper]

oh huh, so since the LHR is both Legendre and Jacobi, i just treat it like Jacobi and get 1 in the end

[supple furnace]

The Jacobi symbol takes permutation signs so it is half +1 and half -1 always while quadratic residues mod composite moduli are much rarer in general

Yeah

[alpine jasper]

i see

cool stuff

[rigid loom]

Pub is that u???

[alpine jasper]

ain't that a coincidence lmao

[rigid loom]

Wtf

Oh well

[sleek cove]

fermats little theorem is basically useless with p=2 right

[torn escarp]

lol

yeah I guess so

[fervent crest]

I mean if you have to state it in that way

[sleek cove]

im just trying to solve a problem using flt but it would rely on 2 actually doing something

now i have to do extra work

[light flicker]

happens very often in nt that 2 is a special cas

[alpine jasper]

let's kick 2 out of the prime club

fundamental theorem of arithmetic is only for chad odd numbers

[sacred junco]

228

[rigid loom]

Can u explain

I am@weak at these

[random hill]

What remainder is left after dividing by that number

[alpine jasper]

$232\equiv 4(\text{mod }x)$ is same as saying\
$232=4+xn$ for some integer $n$

[sterile pumice]

Publius:

[alpine jasper]

@rigid loom

[sterile pumice]

JohnDoeSmith:

JohnDoeSmith:

[rigid loom]

Ohhh

[alpine jasper]

~~yeah alright throw him some ideals lmao ~~

[winter bear]

always generalize stuff in Z to rings smh

[alpine jasper]

i'm not far enough in nt to see why that'll be very useful

nice new pfp btw, it was one of my favs

[winter bear]

thx

[sleek cove]

in proving a^(p^k) = a mod p, induction is fine, right

[winter bear]

mhm

[sleek cove]

[sleek cove]

can someone tell me if this logic is flawed

i dont want to have to do induction again, but this just seems kinda cheeky

[somber drum]

I got an assignment to solve a problem using newtons method for multiple variables

sat all day coding in matlab and i thought it would work but didnt

and when I debug I notice that for some reason f1 and f2 become the same value after the first iteration has passed

function NewtMultiVar
        function newtMulti = newtMulti(xk)
        u = xk(1);
        v = xk(2);
        function f1 = f1()
           f1 = ((93-u)^2)-((63-v)^2)-(55.1^2);
           disp(['f1 : ',num2str(f1)])
        end
        function f2 = f2()
           f2 = ((6-u)^2)-((16-v)^2)-(46.2^2);
           disp(['f2 : ',num2str(f2)])
        end
        function df1u = df1u()
           df1u = -2*(93-u);
        end
        function df1v = df1v()
           df1v = 2*(63-v);
        end
        function df2u = df2u()
           df2u = -2*(6-u);
        end
        function df2v = df2v()
           df2v = 2*(16-v);
        end
        function S = S()
            s1 = sym('s1');
            s2 = sym('s2');
            
            E = [(df1u()*s1)+(df1v()*s2) == -f1(), (df2u()*s1)+(df2v()*s2)==-f2()];

            X = solve(E,s1,s2);
            S = [X.s1 X.s2];
            S = double(S);
            
        end
        newtMulti = xk + S(); 
    end

The disp is just to debug

I dont get what im doing wrong, (in the body i call this iteratively but thats irrelevant, this is the function that calculates)

After one iteration f1 and f2 turn out the same

and my results just spasm out of control

I'm going insane, spent entire day on this and Im pulling my hair out, defeated

Ill give my first born child to anyone who can help, i'm at a total loss..

<@&286206848099549185> if possible

[sleek cove]

can someone help me with this problem, i've broken it into modulo 2 and 5, giving

x = 1 mod 2, which is obvious, but idk what to do for modulo 5

i know that (x^4)^250 = x^1000, but dont know how to apply FLT

do i assume that x has an inverse and is a unit? or what...

nvm

[fervent crest]

I got 1, 3, 7, 9

[sleek cove]

yes

x=0 mod 5 gives contradiction, then just have to solve the 4 other cong. when x = 1, 2, 3, 4 mod 5

[fervent crest]

You can do it with euler’s theorem

i.e. x^(φ(n))=1 (mod n) if x is coprime with n, where φ(n) is the euler totient function

Euler totient of 10 is 4, therefore automatically 1, 3, 7, 9 are solutions

Then for 0, 2, 4, 6, 8, when you raise them to arbitrary power it’s always even therefore never 1

For 5, when you raise that to any power it’s always congruent to 0 or 5 (actually it should always be 5), so it’s also never 1

[sleek cove]

we havent got to that yet

looks fancy

[sacred junco]

can someone try deciphering this word salad

[stoic bear]

What is a word salad?

[sacred junco]

I imagine the lecturer was wanking himself off while writing this thinking hes a genius

it's clearly not parsable

[stoic bear]

I think the lecturer just wanted to direct you to other related topics to explore

[sacred junco]

this is supposed to be a solvable question

[stoic bear]

Also, it isn't really unreadable, at least it isn't a mess of symbols

[sacred junco]

okay please enlighten me

[stoic bear]

On?

[sacred junco]

what is the question asking?

[stoic bear]

The first sentence?

[sacred junco]

yes, the question

[stoic bear]

The 1st sentence is the question

"show that..."

[sacred junco]

can you put it in words that humans understand?

how is it "producing" said sequence

[winter bear]

the wording is p bad on this

but i think he means to show that Z/pZ* is cyclic

[sacred junco]

he's a fucking charlatan

[stoic bear]

I don't get it upon further examination, sorry downtown

[random sand]

Idk man

[winter bear]

i.e show that the sequence {a^n} for some generator is lenght n-1?

p-1

no

[random sand]

Probably equivalent to that anyway

[winter bear]

p-1

Z/pZ* is p-1

actually

i think?

[random sand]

This is weird

[winter bear]

i mean the proof is p involved anyhow

the one i know uses mobius inversions

[random sand]

Is it showing that a^n is periodic (as a function of n)?

[sacred junco]

such a fucking fraud

[random sand]

I’m out I can’t read it

[winter bear]

yeah idk this is hard af to parse

its like he forgot a sentence

or smth lol

[stoic bear]

Sorry, downtown

[random sand]

I can’t give a definitive answer as to what he wants

[winter bear]

can we see the whole problemset @sacred junco

maybe theres some context in other problems?

[sacred junco]

there isn't

it's all unrelated

[winter bear]

but in any event, your best bet is prolly to just email him

rip

[sacred junco]

I'd rather just call it out publicly

[winter bear]

eh bad move, man controls your grade, just give him the feedback that he made a mistake

everyone makes mistakes lol, doesnt mean they should be shamed for it right

[sacred junco]

no it's pretty consistent

he's a charlatan who thinks he knows what he's talking about

[light flicker]

wait what do you guys not understand lmao

[sacred junco]

he's writing a book that is riddled with mistakes

[light flicker]

its just asking you to show there's always a primitive root

and not even for F_{p^n}, its just F_p

[winter bear]

i mean thats what i guessed, but it really doesnt say that?

[sacred junco]

I hate this guy, he's a total pseudointellectual

it's a computer science course and this is his attempt to teach mathematics

but he's failing since he doesn't know any

[light flicker]

I mean, it says to show that if you take a primitive root, then it'll produce such a sequence

I mean, its pretty clear he does know mathematics lmao, this is fine

I mean, he might be a terrible teacher, but thats different

[winter bear]

nah the wording is still p bad, most of us guessed he meant show its cyclic but still

[sacred junco]

you think this is good evidence that he knows mathematics? a word salad with words he doesn't even understand?

[random sand]

I can’t read this right, but I wouldn’t go so far as to say he’s a charlatan

[winter bear]

i mean he doesnt use any particularly big words

[light flicker]

It's not a word salad lmao, it makes mathematical sense

And it's correct math as well

[winter bear]

my problem is that he just didnt phrase it well enough

[light flicker]

I mean, maybe he just copied it from somewhere and he doesn't know the stuff, but its correct mathematically

[sacred junco]

his construction of the real numbers was $\mathbb{R} = \overline{\mathbb{Q}} \cup \mathbb{Q}$

[sterile pumice]

downtown:

[random sand]

Doesn’t seem like he hides behind terminology, it’s just I can’t read it

[sacred junco]

in his great book that we all get to read

[winter bear]

yeah rifts, its like missing a sentence i feel like

[sacred junco]

I'm a math major taking his course

[winter bear]

i mean i think by that he means like the completion of Q?

[serene crag]

Lol it sounds like he is confused. But maybe the theorem he's trying to get at is that for every prime, there exists a primitive root.

[sacred junco]

no, he was saying that the real numbers are the rational numbers with the rationals

which is blatantly wrong

[light flicker]

You mean the irrationals?

[sacred junco]

yeah

[light flicker]

Because that's correct?

[sacred junco]

...

[light flicker]

lmao

Irrationals are literally defined as the real numbers that aren't rational

So maybe this might be circular, but its definitely not wrong

And there are plenty of other ways to define the irrationals

[random sand]

I don’t see what exactly would make that wrong per se, tho idk what kinda closure that overline refers to

[serene crag]

$\mathbb{Q}$ is the algebraic numbers, not the irrationals.

[sterile pumice]

4c:

[light flicker]

if you mean \bar{\bQ}

[serene crag]

sorry. meant to put a bar.

[light flicker]

this notation is used for both

just because algebraic numbers come up more, you see that used more frequently

Anyways, its clear that he meant irrational here and thats not the confusion

[random sand]

Like, is the bar for a metric closure, algebraic, etc?

it’s annoying hoe stuff gets reused

[light flicker]

set complement

[random sand]

ew who tf uses bar for that

disgusting

[light flicker]

@sacred junco so uh, have you figured out why this is right yet

[random sand]

It’s right but wow I don’t like it

[light flicker]

I mean, its probably the most common notation for denoting irrationals

not that people do that often but

[winter bear]

bar for irrational is ew

or complement*

[sacred junco]

@light flicker nah

I haven't tried

[light flicker]

Anyways, your professor is right, it's not "blatantly wrong"

Whether its a good way of defining the reals is another question

[sacred junco]

I still think he's a charlatan

[light flicker]

Idk, you just seem mad at this guy and are trying to blame him for everything, including you not understanding any of this

[random sand]

I don’t know enough to reliably talk about his skill & knowledge, but I certainly can’t say that he is an imbecile

[sacred junco]

nah I just don't have time to decipher this stupid question

[random sand]

Clearly he knows something

[winter bear]

yeah, as i said he prolly just forgot to put a sentence into that question

[light flicker]

I mean again, its not stupid just because you can't understand it

[sacred junco]

@light flicker that's not the reason I think it's stupid

[light flicker]

Then why do you think its stupid

It's a fine exercise that introduces important number theoretical ideas

[sacred junco]

because it's phrased in such a way that no reasonable person should be expected to be capable of being able to decipher it

[light flicker]

I mean, I understood it fine

If you understand what primitive roots are and fermat's little theorem is, you should be able to understand this

[random sand]

I couldn’t decipher it but I am an imbecile

[sacred junco]

@light flicker could you rephrase the question?

[light flicker]

Do you understand what Fermat's little thoerem says?

[sleek cove]

theres a lot of smart ppl here

[sacred junco]

yes

[light flicker]

And what does it say

[sacred junco]

if $a$ is not divisible by prime $p$ then $a^{p-1} = 1 \pmod p$

[sterile pumice]

downtown:

[light flicker]

Okay, and do you understand what primitive roots are

[sacred junco]

no

[light flicker]

Then you should try to understand that?

Not knowing one of the terms in the problem will obviously make the problem confusing

[sacred junco]

where is "primitive root" a term in the problem

for the record there is no expectation of understanding primitive roots before attempting this question

[light flicker]

"The \alpha here is known as a primitive root"

[sacred junco]

ok i got that

[serene crag]

Why does he say there's a "unique" sequence? Different primitive roots can give different sequences.

[light flicker]

Yeah, it means unique as in choosing different alpha will give you different sequences

[sacred junco]

okay so I choose alpha = 5, what is the sequence given

[serene crag]

As an example, if p = 5, 2 and 3 are primitive roots because they give 2, 4, 3, 1 and 3, 4, 2, 1 resp., but 4 is not because it goes 4, 1, 4, 1, etc.

[light flicker]

The sequence here is the powers of \alpha

for when p = 5, \alpha = 2, the sequence is 2, 2^2, 2^3, 2^4

[sacred junco]

up to p-1

[light flicker]

which you can see reduces (mod 5) to be what he described

[sacred junco]

so how do I prove the uniqueness?

[serene crag]

This is also a pretty hard problem to do without any hints. An outline is, you define the order of an element z to be the least d for which $z^d = 1$. A primitive root is one whose order is p-1. Now you can show that the order must divide p-1 by FLiT. Then you show that for each d, there are at most phi(d) guys with order d. Then you use the fact that every element has an order and the fact that $\sum_{d|p-1}\phi(d) = p-1$ to show that this implies that in fact, there are $\phi(d)$ guys with order $d$. In particular, there are $\phi(p-1)$ primitive roots.

[sterile pumice]

4c:

[serene crag]

I still don't understand what "uniqueness" really means here.

[sacred junco]

I'm guessing it is uniqueness for the same p

@light flicker could you state the claim in precise terms?

[light flicker]

Sorry, I was in the shower but

I'd write it as

Show that for any prime $p$, there exists primitive roots mod $p$. In other words, there exist numbers $\alpha$ such that $\alpha^{p - 1} = 1 \pmod p$ and $\alpha^n \neq 1 \pmod p$ whenever $0 < n < p-1$.