<s>[INST] $24$ 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 $a^2+23$?
 [/INST] $a^2+23=(a-1)(a+1)+24$
Since $a$ is odd; $a-1$ and $a+1$ are even. 
Moreover, since $4$ does not divide $a$, 4 divides either $a-1$ or $a+1$.
Since $3$ does not divide $a$, $3$ divides either $a-1$ or $a+1$.
Hence $2\times 4\times 3$ divides $(a-1)(a+1)$.
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