$24$ divides $a^2+23$ 
Prove that $24$ divides $a^2 + 23$ if $a$ is not divisible by $2$ or $3$.

Well $a^2=8k+1$
for some $k$.
So, $$a^2+23 = (8k+1)+23=8(k+3)$$
So, $8$ divides $a^2+23$.
Now can $3$ divide $a^2+ 23$ so that in the end $24$ divides $a^2+23$?

If $a$ is not a multiple of $2$ or $3$, then $a=6m\pm 1$ and then
$$a^2+23=36m^2\pm12m+1+23=12(3m^2\pm m+2)$$
so it suffices to show that $3m^2\pm m$ is even. And indeed it is, because $3m^2\pm m=m(3m\pm 1)$, and one of these two factors is even.
