# If $m\in\Bbb{Z}_n$ and $q\in\Bbb{Z}_n$, show that $(m + q) \bmod n$ uniquely determines $q$.

Suppose $m$ belongs to $\Bbb{Z}_n$ and $q$ belongs to $\Bbb{Z}_n$, show $q$ in $c=(m + q) \bmod n$ is unique.

i.e.

Suppose you are given $m$ and $c$, show that $q$ is unique.

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Could you talk about what you've attempted? From a certain perspective, $q$ is equal to $c - m$ and there's really no choice at that point. –  Dylan Moreland Mar 4 '12 at 1:09
Yes, i do have the idea of why it is true - but can't really seem to figure out how to approach the proof :( –  user996522 Mar 4 '12 at 1:37

It's because $\mathbb{Z}/n$ is a group under addition. In particular every element has an additive inverse: if $[a]$ denotes the congruence class (modulo $n$) of the integer $a$, then $[a]+[-a]=[0]$. If $[c]=[m]+[q]$ (where $c,m$ and $q$ are integers) adding $[-m]$ to both sides gives

$[c]+[-m]=[q]+[m]+[-m]=[q]+[0]=[q]$

So the congruence class $[q]$ is uniquely determined in $\mathbb{Z}/n$, given by $[q]=[c-m]$

If you meant that the representative integer $q$ is unique, then that is not true. If we assume that $c\equiv m+q\pmod n$, then $c\equiv m+(q+kn) \pmod n$ for any $k\in\mathbb{Z}$

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