Prove that element is a square Trying to post this question again that got downvoted.
Suppose that $x^5 + ax + b \in \mathbb{F}_p[x]$ is irreducible over $\mathbb{F}_p$. Is it true that $25b^4 + 16a^5$ is a square in $\mathbb{F}^{\times}_p$?
My idea: the Galois group of $f$ is a subgroup of $A_n$ iff its discriminant is a square. The discriminant of $f$ is $256a^5 + 3125b^4$ and while this kind of looks like $25b^4 + 16a^5$, its not quite it. So not sure if I even should try to prove the galois group is a subgroup of $A_n$. Any other ideas?

Here's a slightly simpler example, in my opinion.
Let $p=11$ and $f(x)=x^5+x+3$.
We have $25\cdot3^4+16\equiv6\pmod{11}$ and $6$ is not squared modulo $11$.
Check the irreducibility of $f(x)$. This polynomial has no roots in $\mathbb{Z}_{11}$ can be checked using Horner's method.
This will require $10$ lines of computation.
Let us check that it has no divisors of degree two.
If
$$
f(x)=(x^2+\alpha'x+\beta)(x^3+\alpha'x^2+\beta'x+\gamma'),
$$
then $\alpha'=-\alpha$, $\beta'=\alpha^2-\beta$, $\gamma'=3/\beta$ and
\begin{align*}
  2\alpha\beta^2-\alpha^3\beta-3& =0, \\
  \alpha^2\beta^2+3\alpha & =\beta^3+\beta.
\end{align*}
To prove that this system has no solutions over $\mathbb{Z}_{11}$
we can do the following.
Assume alternately $\alpha=0,1,\ldots,10$, then the first equation is a quadratic equation with respect to $\beta$. Calculate $\beta$ and substitute it into the second equation.
All of these calculations can be done manually, but you can use a computer.
I have done all these calculations, I leave it to the author of the question to do them again.
Note.
For $p=3,5,7$ there are no such polynomials;
for $p=11,13,17,19$ there are $4,16,24,34$, respectively.
