Let $n$ be an odd integer not divisible by $3$. Prove that $n^2 \equiv 1 \ (\mathrm{mod}\ 24)$. We just started congruences in class so I'm a little stuck as to how to solve this.
So far, I've done: $24 \mid n^2 - 1$ and since $n$ is odd, I defined $n = 2m + 1$ for some integer $m$. Plugging that in gives me $24 \mid 4m(m+1)$, and now I'm not sure what to do with this. Where did I go wrong/any tips on how to solve this?
You're almost done! Note that either $m$ or $m+1$ is even. And either $m$ or $m+1$ is divisible by $3$. (Why? Try modulo $3$.)
I'd like to suggest an even better approach: Note that $n^2-1=(n-1)(n+1)$. Since $n$ is odd, $n-1$ and $n+1$ are even and in specific one of these is divisible by $4$. Also, one of $n-1,n,n+1$ is divisible by $3$. By assumption, $n$ is not divisible by $3$, so one of the others must be. Combining all of these gives us the divisibility by $24$.
$n=6m\pm 1$ so $n^2 =36m^2\pm 12m+1 =12m(3m\pm 1)+1 $ and $m(3m\pm 1)$ is even since if $m$ is odd then $3m\pm 1$ is even.
a\mod bgives $$a\mod b$$a\pmod bgives $$a\pmod b$$ anda\bmod bgives $$a\bmod b$$ – gen-ℤ ready to perish Oct 27 '17 at 16:06