<s>[INST] Simplifying a proof by contradiction: if $a\equiv 1\bmod 5$, then $a^2\equiv 1\bmod5$ Prove the following either by Direct Proof or by Contraposition:
Suppose $a\in\mathbb{Z}$, if $a\equiv 1\pmod 5$, then $a^2\equiv 1\pmod5$

Suppose $a\equiv 1\pmod 5$

Then $5|\left(a-1\right)$, therefore $a-1=5k$

$a^2-1=\left(a-1\right)\left(a+1\right)=5k\left(a+1\right)$

Thus, $a^2-1=5m$ where $m=k\left(a+1\right)$

Hence $a^2\equiv1\pmod5$

So, that is basically a proof by Contraposition, right?
 [/INST] If
$a\equiv 1\pmod 5$,
$a = 5n+1$
for some integer $n$.
Therefore
$a^2
= 25n^2+10n+1
= 5(5n+2)+1
$
so
$a^2\equiv 1\pmod 5$.
</s>