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Been working on it for a couple of hours with no success. My intuition tells me that it has to do something about $\gcd(n,x) = 1$ I have managed to prove that if $\gcd(x,n) = 1$ the claim holds, now I try to prove the other way around. Any hint or a solution will help. thanks!

Bill Dubuque
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alon1230
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  • The condition $\gcd(x,n)=1$ is indeed a sufficient condition, since it means $x$ is a unit mod. $n$, but it is not necessary. You have the same conclusion $(a\equiv b \mod n$) if $x$ is a non-zero divisor modulo $n$, which is a more complex problem. – Bernard Nov 22 '20 at 12:38
  • What do you think of my solution @Bernard? – Contestosis Nov 22 '20 at 12:50
  • The implication is equivalent to the statement that the map $, c\mapsto cx,$ is $1$-$1$ modulo $N$ (i.e. $,x,$ is cancellable), which is equivalent to $,\gcd(x,N)=1, $ by the well-known Theorem in the dupe. Btw, $x$ is not good notation for a constant. Better notation would be $\ ax\equiv ay\Rightarrow,x\equiv y,$ for all $,x,y\ \ $ – Bill Dubuque Nov 22 '20 at 13:28

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You're right, this holds if and only if $\gcd(n, x) =1$.

So, as you've already proved the first part, let's show the second one.

Assume for all $a, b \in \mathbb{Z}, ax\equiv bx \mod n \implies a \equiv b \mod n$.

Let $d := \gcd(x, n)$ and $(x', n') := (\frac{x}{d}, \frac{n}{d}) \in \mathbb{Z}^2$.

Take $(a, b) = (n', 0)$. One has $xn' = x'n'd = x'n \equiv 0 = x 0 \mod n$.

Then one should have $n' \equiv 0 \mod n$, which means $n | n'$. As one also has $n' | n$, this implies $|n'| = |n|$ thus $d = \gcd(n, x) = 1$.

Contestosis
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