Simplifying a proof by contradiction: if $a\equiv 1\bmod 5$, then $a^2\equiv 1\bmod5$

Question

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?

Answer 1

If $ a \equiv 1 (mod 5) $, then multiplying both sides by $a$ we get $ a^2 \equiv a (mod 5) $.

So $a^2$ is congruent to $a$, and by hypothesis $a$ is congruent to 1.

Answer 2

In proving this relationship, we need only work with one variable. There is no need to introduce the $b$ variable in your proof. Additionally, since we are working in modulus 5, I think it would be best if we avoided talking about odd numbers and stuck to talking about mod 5. Here is a way to start a proof:

Assume that $a \equiv 1$ (mod $5$). Then we can write $a= 5k+1$ for some integer $k$.

Now, from here, is there any way we can express $a^2$ as a multiple of $5$ plus one?

Answer 3

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 $a^2\equiv 1\pmod 5$.