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?
 A: 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.
A: If $p\mid(r^4+1)$ for one of $p=3,5,7$ then $-1$ is a fourth power mod $p$, hence $(\mathbb{Z}/p\mathbb{Z})^{\times}$ has an element of order 8. 
But this is impossible because $(\mathbb{Z}/p\mathbb{Z})^{\times}$ has order $p-1\leq 6$.
A: Here is a variant of @carmichael561's proof.
As before, we have an element of order $8$. By Lagrange's theorem, $8$ divides $p-1$ and so $p$ is of the form $8k+1$. But $3,5,7$ are not of the form $8k+1$.
This argument then allows us to extend the results to $p=11, 13, 19, 23, 29, \dots$, for which the original argument fails.
A: Check for $r=3k,3k+1,3k+2$
Then check for $r=5k,5k+1,5k+2,5k+3,5k+4$
Then check for $r=7k,7k+1,7k+2,7k+3,7k+4,7k+5,7k+6$
It is the easiest way but lengthy.
A: 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
$7 \not \mid r^4+1$.
