Let $F$ be a field and $g(x)$ in $F[x]$. Prove if $1+g(x)^2$ has an irreducible factor of odd degree then there exists $a$ in $F$ such that $a^2 =-1$.
I didn't get too far on this problem. It doesn't seem like it's a Galois Theory or field extension problem since we don't know much about our field.
I first started looking at roots of $1+g(x)^2$ since then $g(r)^2 =-1$ for root $r$ as we want. I didn't really get anywhere with that.
We know $F[x]$ is a Euclidean Domain, and it seems like the Euclidean Algorithm might be useful. We know $1+g(x)^2 =f(x)p(x)$ for irreducible $f(x)$ and so $1=f(x)p(x)+(-g(x))g(x)$.