$\gcd$ of polynomials over a finite field

Question

Let $p$ be an odd prime number and $\mathbb{K}:=\mathbb{F}_{p^{t}}$ be a field of characteristic $p$. Let $u\in \mathbb{K}$ such that $T^2 -u$ is irreducible over $\mathbb{K}$. Prove that for all $n\geq 1$ $\gcd(T^{3.(2^{n-1}-1)} +T^{3.(2^{n-1}-2)} +T^{3.(2^{n-1}-4)} +\cdots+T^{3.(2^{n-1}-2^{n-2})} +1, T^{2}-u)=1$ ?

Answer 1

Write $$P(T):=T^{3.(2^{n-1}-1)} +T^{3.(2^{n-1}-2)} +T^{3.(2^{n-1}-4)} +\cdots+T^{3.(2^{n-1}-2^{n-2})} +1.$$

Since $T^2-u$ is irreducible, $\gcd(P,T^2-u)$ is either $1$ or $T^2-u$. Suppose it's not one, so $T^2-u$ divides $P$.

Let $\alpha\in \overline{\mathbb{K}}$ be a root of $T^2-u$, so $\alpha^2=u$ and $\alpha$ is a root of $P$. Because all the powers of $T$ in $P$ are even except the leading term, the evaluation of $P$ at $\alpha$ gives us $0=P(\alpha)=a+\alpha b$ for some $a,b\in \mathbb{K}$ with $b\neq 0$. But this means that $\alpha=-ab^{-1}\in \mathbb{K}$, which contradicts the fact that $T^2-u$ is irreducible.

Recall that if a polynomial is irreducible over some field, then it has no roots over that field.