In an [article](https://www.semanticscholar.org/paper/A-potential-analogue-of-Schinzel%27s-hypothesis-for-Bender-Wittenberg/cc4b8b7082c3ac64809adf5d333ed1c9a6d83f01), I've found the following result. Unfortunately, it was derived from a general, somewhat complicated theory, that would be cumbersome for this result alone.



> 
> Assume that $\mathbb F\_p$ is the field with $p$ elements, and that we choose an integer $a$ with $1< a < 4p$, such that $a$ be coprime to $p(p-1)$ (for example, $a = 2p - 1$ is suitable). Let $t$ be transcendental over $\mathbb F\_p$, and set $K = \mathbb F\_p(t)$. Consider the polynomial $$P(X) = X^{4p} +t^a$$ in $K[X]$.
> 
> 
> Then for every $f\in \mathbb F\_p[t]$, $P(f)$ is reducible in $\mathbb F\_p[t]$.
> 
> 
> 


I would like a direct proof of this result. Any idea?



Proving that polynomials belonging to a certain family are reducible
It follows from [this](https://mathscinet.ams.org/leavingmsn?url=http://projecteuclid.org/euclid.pjm/1103036322) result of Swan: If $t$ divides $f(t)$ there is nothing to do. So assume that this is not he case. Then with $F(t)=P(f(t))=f(t)^{4p}+t^a$, $F'(t)=at^{a-1}$ and $F(t)$ is separable. From this one easily gets that the discriminant of $F(t)$ is a square in $\mathbb F\_p$ (see the alternative formula for the discriminant at the beginning of the article of Swan). By Swan's result, the number of irreducible factors of $F(t)$ is congruent to $\deg F(t)=4p\deg f$ mod $2$, hence it is even and therefore $>1$.


*Note added later:* In fact the special case needed here is already due to Dickson, as pointed out by Swan in his paper. The argument is easy: If $F(t)$ is separable of degree $n$ and irreducible, then its Galois group contains an $n$-cycle. If $n$ is even, then this $n$-cycle is an odd permutation, so the discriminant of $F(t)$ is not a square. (Assuming that we are working over a finite field of odd characteristic.)


