Given $F$ finite field, how to prove that there is $f \in F[x]$ such that $\deg(f) =2$ and $f$ is irreducible, meaning there is no two polynomial $g_1,g_2 \in F[x]$ such that $\deg(g_1) = 1 , \deg(g_2)=1$ and $f = g_1 g_2$ ?!
I thought that $x^2+1,x^2+2,\cdots , x^2+n-1$ given that the number of elements in $F$ is $n$, then at least one them is quadratic irreducible (but this way of thinking is more number theory than algebra)
So i also thought that if i could prove that there is more number of 2 degree polynomials than polynomial made of the product of 2 polynomial of degree 1, but could not proceed either.
Could any one provide an algebraic proof.