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According to this answer (and the modular arithmetic theory), $ax\equiv ay \pmod{n} \iff x\equiv y \pmod{n}$, if $a$ and $n$ are relatively prime. I tried to prove the forward implication but reached a contradiction:

$ ax\equiv ay \pmod{n} \rightarrow x\equiv y \pmod{n} \iff ax \in n \mathbb{Z}+\{ay\} \rightarrow x \in n \mathbb{Z}+ \{y\} $ $\quad$ ("$+$" here denotes Minkowski addition)

$x \in n \mathbb{Z}+\{y\} \iff ax \in an \mathbb{Z}+\{ay\}^{(*)} $, therefore $ ax \in n \mathbb{Z}+\{ay\} \rightarrow x \in n \mathbb{Z}+ \{y\} \rightarrow ax \in an \mathbb{Z}+\{ay\}^{(**)} $ in particular.

For simplicity, we will assign the following names to our formulae: $ A:=ax \in n \mathbb{Z}+\{ay\} \quad B:=x \in n \mathbb{Z}+ \{y\} \quad C:=ax \in an \mathbb{Z}+\{ay\}$

If $ A \rightarrow B \rightarrow C $ (see $^{(**)})$ then $ A \rightarrow C $:

$ ax \in n \mathbb{Z}+\{ay\} \rightarrow ax \in an \mathbb{Z}+\{ay\} \iff (n \mathbb{Z}+\{ay\}) \subset (an \mathbb{Z}+\{ay\}) \iff n \mathbb{Z} \subset an \mathbb{Z} \iff \mathbb{Z} \subset a \mathbb{Z} $ which is clearly false, therefore $ A \rightarrow B \rightarrow C $ didn't pass the necessary condition and since we've deduced $ B \rightarrow C $ to be true (see $^{(*)})$ it follows that $A \not\rightarrow B$

What am I doing wrong?

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Alex B.
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    This looks quite complicated. I do not see, where you use the condition $\gcd(a,n)=1$. And without this, the forward implication is false, take e.g. $a=0$. If $\gcd(a,n)=1$ then $a^{-1}$ exists and $ax\equiv ay \Longrightarrow a^{-1}ax\equiv a^{-1}ay \Longrightarrow x\equiv y$. The other direction is trivial because $x\equiv y \Longrightarrow ax\equiv ay,$ for all $a.$ – gammatester Jul 01 '16 at 10:21
  • Thanks, this helped me for the modular arithmetic part of the question. Actually it all makes sense now. – Alex B. Jul 01 '16 at 10:34

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the error is the application $ ax \in n \mathbb{Z}+\{ay\} \rightarrow x \in n \mathbb{Z}+ \{y\}$ is not true ; it is true only if $a$ invertible modulo $n$, that is only if $gcd(a,n)=1$

m.idaya
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