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So far I know that $1 \bmod p$ will always be $1$ as the lowest $p$ can be is $2$, meaning that $mn \bmod p$ will have to $= 1$. This means either $mn < p$ or $mn = p + 1$.

I've also figured out that to prove the above statement I would need to show the following

  • $n$ exists (obviously)
  • $n$ is unique
  • $0 < n < p$

This is were I get stuck. I can see that this is the case (and show that there is an n such that this is the case,) however, I am having trouble proving that n is unique.

Bernard
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ChuChugga
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3 Answers3

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Suppose you have 0 < n < p and 0 < n' < p such that mn = mn' mod p = 1 mod p. Then you would get that m(n-n') = 0 mod p. But multiply this by n and you get n-n' = 0 mod p. So n - n' is a multiple of p, but -p < n-n' < p so n-n' can only be 0. Thus, we have n = n'

Mickaël M
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  • not quite sure I follow this... are you trying to say that n can only be 0? – ChuChugga Apr 05 '20 at 10:04
  • I edited the answer to clarify that it is n-n' = 0 and not n = 0. – Mickaël M Apr 05 '20 at 10:08
  • i understand what you are saying, but i don't see how this covers all the cases specified in the original post for n – ChuChugga Apr 05 '20 at 10:15
  • Well, if you have two numbers n and n' between 0 and p that verify mn = 1 mod p and mn' = 1 mod p, it shows that we must have n = n'.

    Edit : I think I might have identified our misunderstanding. Are you trying to prove that if there is a n such that mn = 1 mod p, then necessarily 0 < n < p ? Because this claim is false. Take for instance p = 5, m = 2 and n = 8.

     – Mickaël M Apr 05 '20 at 10:17
  • yes I am trying to prove that is an n such that mn = 1 mod p, where 0 < n < p. no point considering n = 8 if p = 5, because of the constraint that n < p – ChuChugga Apr 06 '20 at 02:50
  • Well then I don't see how my argument doesn't answer your question. It shows that if you have two numbers such that nm = 1 mod p and n'm = 1 mod p, and if 0 < n < p and 0 < n' < p, then n = n'. This is the unicity of n that you asked for, is it not? – Mickaël M Apr 06 '20 at 11:46
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Hint:

Since $p$ is prime and $0<m<p$, $m$ and $p$ are coprime, hence there exists a Bézout's relation $um+vp=1\quad(u,v\in\mathbf Z)$.

Furthermore, all integer solutions of the equation $xm+yp=1$ satisfy the relations $$\begin{cases} x=u+kp\\ y=v-km \end{cases}\quad(k\in\mathbf Z). $$ There remains to show you can find an $x$ s.t. $0<x<p$.

Bernard
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  • any chance you could 'dumb this down a little'.. i have never heard of Bézout's relation. also what does x s.t. 0 mean? – ChuChugga Apr 05 '20 at 10:11
  • Sorry for the typo. 'Tis fixed. Don't tell me you've never seen a relation like $xm+yp=1$ for coprime numbers. Maybe your teachers didn't mention its name. – Bernard Apr 05 '20 at 10:14
  • i have not. i have only done an introductory course to discrete math. i understand what you are saying in your hint but i don't really see how to go forward from there because as i mentioned i have never seen any of this before – ChuChugga Apr 05 '20 at 10:18
  • No arithmetic in high school? – Bernard Apr 05 '20 at 10:20
  • so if i understand correctly, gcd(m,p) = 1 because they are coprime, and then xm + yp = 1 because of 'Bézout's relation'. okay that makes sense, but how does that relate to n – ChuChugga Apr 05 '20 at 10:26
  • The Bézout's relation becomes modulo $p$: $xm\equiv 1$, so by definition, the congruence class of $x$ is an inverse of the congruence class of $m$. The problem comes down to finding a representative of this congruence class which satisfies the inequality condition.s – Bernard Apr 05 '20 at 10:32
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Hint: Consider the map $x \mapsto mx$ in the classes mod $p$. Prove that it is injective. Conclude that it is bijective.

lhf
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