Primes modulo which an algebraic equation admits a solution Let $p$ be an odd prime. Denoted by $\mathbb{F}_p$ the finite field of $p$ elements. If $x^4+10x^2+5\equiv0\pmod p$ has a solution, what about the prime $p$? I think $p$ should satisfy $ 5\mid(p−1)$ but I don't know how to prove it. Please help me.
Thanks!
 A: Case (a): $p=5$. Then $x\equiv0\pmod{5}$ is the only (four-fold) solution.
Case (b): $p\not\in\{2,5\}$: Then $x\equiv0\pmod{p}$ is not a solution, and we have
$$\begin{align}
x^4+10x^2+5 &= \frac{(1+x)^5-(1-x)^5}{2x}
\\\therefore\quad
x^4+10x^2+5\equiv0\pmod{p}&\iff (1+x)^5\equiv(1-x)^5\pmod{p}
\\&\qquad\text{ for some}\quad x\not\equiv0\pmod{p}
\end{align}$$
Case (ba): $5\not\mid (p-1)$. Then every element in $\mathbb{F}_p$ has a unique $5^{\text{th}}$ root,
which implies
$$\begin{align}
1+x&\equiv1-x\pmod{p}
\\\therefore\quad
x&\equiv0\pmod{p}
\end{align}$$
which contradicts the implication $x\not\equiv0\pmod{p}$ of the parent case (b).
Therefore case (ba) excludes the existence of a solution.
Case (bb): $5\mid (p-1)$. This is the case that remains, so you have guessed that right, except for the case (a) above. As a bonus, we can identify the solutions now.
Note that we can easily exclude $x\equiv\pm1\pmod{p}$ by substituing the values in the original degree-$4$ polynomial.
For $i=0,\ldots,4$, let $\zeta_i$ enumerate the $5^{\text{th}}$ roots of unity in $\mathbb{F}_p$ with $\zeta_0=1$ and the other $\zeta_i$ being primitive.
Then solution candidates $x_i$ can be found by setting
$$\begin{align}
\frac{1+x_i}{1-x_i}&\equiv \zeta_i\pmod{p}
\\\therefore\quad
x_i&\equiv\frac{\zeta_i-1}{\zeta_i+1}\pmod{p}
\end{align}$$
where the fraction involves multiplicative inversion modulo $p$.
Since we know that $x\not\equiv0\pmod{p}$, this excludes use of $i=0$, so the solution set is $\{x_1,\ldots,x_4\}$.
Getting rid of the denominator using
$$2\equiv\zeta_i^5+1\equiv(\zeta_i+1)
(\zeta_i^4-\zeta_i^3+\zeta_i^2-\zeta_i+1)\pmod{p}$$
we finally get
$$x_i\equiv-\zeta_i(\zeta_i-1)(\zeta_i^2+1)\pmod{p}
\quad\text{for}\quad i=1,\ldots,4$$
