How do I prove this congruence?

Question

let $a \in \mathbb Z$. Prove that if $a \equiv 1 \mod 6$ then $a^2 \equiv 1 \mod 12$

I started with $6 \mid a-1$ so there is some k $\in\mathbb Z$ where $6k = a-1$. That means for the second equation we can write $12 \mid a^2 -1$ which will lead to $2 \times 6 \mid (a-1)(a+1)$ which then leads to $(2)(6) \mid (6k)(a+1)$. I'm having trouble extracting a 2 from the right side of the equation to fully state that 12 can divide the right hand side. By placing the $6$ in front, I can say that $6 \mid 6(k)(a+1) $ but I can't say that $12 \mid 6k(a+1) $ yet. Can someone please help me.

Answer 1

We can write $a=6b+1$.

Now $a^2=(6b+1)^2=36b^2+12b+1\equiv 1\mod 12$.

The last equivalence is true since $36b^2+12b+1=12\cdot(3b^2+b)+1$.

Answer 2

$a\equiv 1 \mod 6\implies $

$a = 6k+1$ for some integer k.

So $a^2 = (6k+1)^2= 36k^2+12k+1\implies... $

Answer 3

Note $\,a\equiv 1\pmod{\!2n}\,\Rightarrow\, a^{\large 2}\equiv 1\pmod{\!4n}$

because $\ a=1+2nk\,\ \Rightarrow\ a^{\large 2} = 1+4n(k\!+\!nk^2).\ \, $ Yours is case $\,n=3$.