I want to prove this claim below as an exercise for exam:
Let $F(x) \in GF(p)[x]$ a prime polynomial deg $n>0$, if $gcd(F,x^{q^i}-x)=1$ for $i=1,2,...,[n/2]$ $\iff$ $F$ is irreducible
Here a trace of my proof:
Suppose $F$ reducible.
Then $F(x)=G(x)H(x)$ product of at least two irreducible polynomial, and since $deg F=n \Rightarrow degf \geq degG \times degH$. Suppose then $degG=degH=[n/2]$.
We know that for $q=p^k \Rightarrow x^{q^i}-x=$ {product of all irreducible polynomial of $GF(q)$}. But from the product of thos polynomial, must emerge a polynomial with degree at least $[n/2]$, so $gcd(F,x^{q^i}-x) \not =1$. So in order to have $gcd(F,x^{q^i}-x) =1$, $F$ must be irreducible.
I have difficulties in translating my ideas in mathematical format so thanks for any kind of hints or corrections, thanks.