<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] I'm going to try
to make
the most elementary possible proof.
$\bmod 3$,
$(0, \pm 1)^4
\equiv (0, 1)
$
so
$(0, \pm 1)^4+1
\equiv (1, 2)
$
so
$3 \not \mid r^4+1$.
$\bmod 5$,
$(0, \pm 1, \pm 2)^4
\equiv (0, 1, 1)
$
so
$(0, \pm 1, \pm 2)^4+1
\equiv (1, 2, 2)
$
so
$5 \not \mid r^4+1$.
$\bmod 7$,
$(0, \pm 1, \pm 2, \pm 3)^4
\equiv (0, 1, 2, 4)
$
so
$(0, \pm 1, \pm 2, \pm 3)^4+1
\equiv (1, 2, 3, 5)
$
so
$7 \not \mid r^4+1$.
</s>