<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] Each divisor of $5^{4n+1}+5^{3n+1}+1$ is odd, hence it is enough to show that any odd prime divisor of $5^{4n+1}+5^{3n+1}+1$ is $\equiv 1\pmod{5}$. Now we may consider the splitting field of $f(x)=5x^4+5x^3+1$ over $\mathbb{F}_p$ for any prime $p>5$. The complex roots of $f(x)$ are given by
$$ \frac{1}{4}\left(-1\color{red}{\pm}\sqrt{5}\color{blue}{\pm} i\sqrt{2\left(1\color{red}{\pm}\frac{1}{\sqrt{5}}\right)}\right) $$
hence if $p$ is a divisor of $5^{4n+1}+5^{3n+1}+1$ then $5$ is a quadratic residue $\!\!\pmod{p}$ and $p\mod{5}\in\{-1,1\}$. On the other hand, if $p\equiv -1\pmod{5}$ then $5x^4+5x^3+1$ factors in $\mathbb{F}_p$ as $5$ times the product of two quadratic irreducible polynomials. Long story short,

$$ p>5,\;\; p\mid 5^{4n+1}+5^{3n+1}+1\quad \Longrightarrow \quad p\equiv 1\pmod{5}$$

and the claim easily follows.
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