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I want to show that , in a given pair, there exists a pair where $x-y \equiv 0 \pmod n$ and in order to do that I can write an algorithm that looks for two different instances where we have the same remainder in $\pmod n$. So that we would have $0 \pmod n$, correct?

So can I then just turn this question into this: Prove that $a \equiv b \pmod n$ if and only if $a$ and $b$ leave the same remainder when divided by $n$? And how can I prove this?

Robert Shore
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nich
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  • I assume you are talking about $a,b\in \mathbb{Z}$? If so you have that $a=a_1n+r_1$ and $b=b_1n+r_2$ where $r_i<n$ by Euclidean division. Then $r_1\equiv r_2$ mod $n$ but as $r_i<n$ they were equal to begin with. – H U Jan 21 '22 at 21:39
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    What is your definition of $a \equiv b \pmod n$? – Robert Shore Jan 21 '22 at 21:46

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