Are there infinitely many polynomials $f \in \mathbb{F}\_3[X]$ for which $f(X^2) + X$ is irreducible?



Infinitely many irreducible polynomials of the form f(X^2) + X mod 3?
$\def\FF\mathbb{F}$**Lemma** The following are equivalent:


(1) $f(x^2)+x$ is prime in $\FF\_3[x]$ 


(2) $f(x^2)^2-x^2$ is prime in $\FF\_3[x^2]$


(3) $f(y)^2-y$ is prime in $\FF\_3[y]$.


**Proof** (2) and (3) simply differ by the change of variable $y=x^2$, and are thus equivalent. Set $g=f(x^2)+x$, $\overline{g} = f(x^2)-x$ and $h=g \overline{g}=f(x^2)^2-x^2$. We see that $h$ is the norm of $g$ from $\FF\_3[x]$ to $\FF\_3[x^2]$. An element with prime norm must be prime, so (2) implies (1).


Conversely, suppose (1). So $h=g \overline{g}$ is the prime factorization of $h$ in the PID $\FF\_3[x]$. It is easy to see that $g$ and $\overline{g}$ are not associate to elements of $\FF\_3[x^2]$, so $h$ is irreducible (and hence prime) in $\FF\_3[x^2]$. $\square$


[Bunyakovsky's conjecture](https://en.wikipedia.org/wiki/Bunyakovsky_conjecture) states that, if $a(T)$ is a polynomial obeying certain natural conditions, then $a(T)$ is prime for infinitely integers $T$. This is widely believed, but has not been proved for any $a$ of degree $\geq 2$.


Similarly, you want to know whether $T^2-y$ is prime for infinitely many $T \in \FF\_3[y]$. 
People have made conjectures analogous to Bunyakovsky in function fields (see [Conrad, Conrad and Gross](http://math.stanford.edu/~conrad/papers/genuszerofinal.pdf), conjecture 6.2). However, Conrad, Conrad and Gross (first page) also write 



> 
> The only proved instance of this asymptotic conjecture is the case where $\deg\_T a=1$, just as in the classical case.
> 
> 
> 


Some brief literature searches suggest that little has changed since that statement was written in 2008. So you are probably out of luck to prove this result.


