<s>[INST] None of $3,5,7$ can divide $r^4+1$ Let $n=r^4+1$ for some $r$. Show that none of $3,5,$ and $7$ can divide $n$.
I am thinking to use a corollary that "each prime divisor p of an integer of the form $(2m)^4+1$ has the form $8k+1$", but I failed. Anyone can give some hint?
 [/INST] Use Fermat little theorem.  If $r $ isn't divisible by 3, $r^2 \equiv 1 \mod 3$.  If $r$ is divisible by 3 then $r^2 \equiv 0 \mod 3$.  So $r^4 \not \equiv -1 \mod 3$ so $r^4 + 1$ is not divisible by 3.
Similarly $r^4 \equiv 1,0 \mod 5$ so $r^4 + 1$ is not divisible by 5.
If $r$ is not divisible by 7 then $r^6 \equiv 1 \mod 7$.  So if $r^4 \equiv -1 \mod 7$ then $r^8 \equiv r^2 \equiv 1 \mod 7$.  So, contradictory, $r^4 \equiv 1 \mod 7$.  So $r^4 \not  \equiv -1 \mod 7$.  So $r^4 + 1$ is never disible by 7.
</s>