<s>[INST] Let $p$ be prime not equal to 2 or 5. Show $p^2+1$ or $p^2-1$ is divisible by 10. I can do half the proof but can not think of a way to finish. If $p$ is prime then both $p-1$ and $p+1$ are even. In cases where either $p-1$ or $p+1$ is divisible by 5 that implies $(p-1)(p+1)=p^2-1=10n$ for some positive integer $n$. I am not sure if I am on the right track with this. 
 [/INST] Since $p \neq 5$ and $p$ is prime, $p$ cannot be divisible by $5$, so $p \neq 0 \mod 5$. Now,
$1^2 = 1 \mod 5$
$2^2 = 4 \mod 5$
$3^2 = 4 \mod 5$
$4^2 = 1 \mod 5$
That is, $p^2 \mod 5$ can be either $1$ or $4$. If $p^2=1\mod 5$, then $p^2-1=0\mod 5$. If $p^2=4\mod 5$,then $p^2+1=0\mod 5$. That is, either $p^2-1$ or $p^2+1$ is divisible by $5$. Since both are even, it implies that one of them that is divisible by $5$ is actually divisible by $10$.
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