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!!!
On
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.
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 $p\equiv1\pmod5$.