<s>[INST] Prove that each divisor of $5^{(4n+1)}+5^{(3n+1)}+1 =$ $1 \pmod {10}$ Prove that if $d$ divides $5^{(4n+1)}+5^{(3n+1)}+1$ for any $n$, then $d =$ $1 \pmod {10}$. 
A similar statement can be proven, where $d$ divides $(3^{(2n+1)}+1)/4$, for any $n$, then $d =$ $1 \pmod {6}$ by first showing that if $-3$ is a quadratic residue $\pmod p$, then $p = 1 \pmod 3$. For the base $5$ expression I would start by showing that if $-5$ is a quartic residue $\pmod p$, then $p = 1 \pmod {10}$. Any  help on proving the original statement with base $5$ expression? Thanks!!! 
 [/INST] We can restrict to prime divisors $p$. These must be odd, so
all we need to prove is $p\equiv1\pmod5$. We can't have $p=5$ either.
Now
$$5a^4+5a^3+1\equiv0\pmod p$$
where $a=5^n$ and so
$$b^4+5b+5\equiv0\pmod p$$
where $b$ is the inverse of $a$ modulo $p$.
Consider $f(x)=x^4+5x+5$ and let $\zeta$ be a primitive fifth root
of unity. Let $\eta=\zeta^2-\zeta$. Then
\begin{align}
\eta^4&=\zeta^3-4\zeta^2+6\zeta-4+\zeta^4\\
&=-5\zeta^2+5\zeta-5=-5-5\eta.
\end{align}
Thus $f(\eta)=0$ and any zero of $f$ generates the cyclotomic
field $K=\Bbb Q(\zeta)$.
If $f(a)\equiv0\pmod p$ then $p$ splits in $K$. As the primes that
split in $K$ are those congruent to $1$ modulo $5$ then $p\equiv1\pmod5$.
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