I asked the following question and got some awesome answers:
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 $\mathbb{F}^{\times}_p$?
While it turned out to be false, I was thinking to slightly alter the element. Let's suppose that we look at the discriminant instead, i.e $3125b^4 + 256a^5$. It's not obvious to me that this is a square, indeed I know its enough to prove that the Galois group of $x^5 + ax +b$ is contained in $A_5$ but I cant seem to show this. Is there any other way to conclude that the discriminant must be a square?
One way that uses much less Galois theory. Instead I use the fact that $5$ is an odd integer (necessary actually).
If $f(x)=x^5+ax+b\in\Bbb{F}_p[x]$ is irreducible, then its splitting field is isomorphic to $K=\Bbb{F}_{p^5}$. Let $a_1,a_2,\ldots,a_5\in K$ be the roots. The discriminant $$ \Delta=\prod_{1\le i<j\le 5}(a_i-a_j)^2 $$ is obviously a square of an element of $K$. By more basic Galois theory, or a calculation we also know that $\Delta\in\Bbb{F}_p$.
But the index of the subgroup $\Bbb{F}_p^*$ in $K^*$ is $$ m:=[K^*:\Bbb{F}_p^*]=\frac{p^5-1}{p-1}=1+p+p^2+p^3+p^4. $$ Here the powers of $p$ all have the same parity. Taking the first term $1$ into account, we can conclude that $m$ is odd.
Lemma. If an element $d\in\Bbb{F}_p$ is the square of an element of $K$, $d=a^2$, then $a\in\Bbb{F}_p$.
Proof 1. The case $d=0$ is trivial, so we can assume $d\neq0$. Let's move to the quotient group $G=K^*/\Bbb{F}_p^*$. As $d\in\Bbb{F}_p^*$ the coset $d \Bbb{F}_p^*=1_G$. Therefore we have $$ (a \Bbb{F}_p^*)^2=a^2\Bbb{F}_p^*=d\Bbb{F}_p^*=1_G. $$ So the order of the coset $a \Bbb{F}_p^*$ in $G$ is a factor of two. But $m=|G|$ is odd, and by Lagrange, the order must be a factor of $m$. We can thus conclude that the order must be equal to $1$. Hence $a\Bbb{F}_p^*=1_G$, and it follows that $a\in\Bbb{F}_p$.
Proof 2. Assume contrariwise that $a\notin \Bbb{F}_p$. As $a^2=d\in\Bbb{F}_p$, it follows that the degree of the field extension $[\Bbb{F}_p(a):\Bbb{F}_p]=2$. But, by the tower law we must have $$[\Bbb{F}_p(a):\Bbb{F}_p]\mid [K:\Bbb{F}_p]=5.$$ This is a contradiction.
The main claim follows from all of the above.