TITLE: Squares which are not 1 + a square in finite fields of odd characteristic?
QUESTION [8 upvotes]: Given a finite field $F$ of odd characteristic, does there always exist an element $x$ of $F$ such that $1 + x^2$ is not a square in $F$? If so, can one even find a natural description of such an element, e.g. as a function of $\lvert F \rvert$ and a generator of $F^\times$?

REPLY [17 votes]: Short answer I found to the existence part of my question.
Let $\lvert F \rvert = p^n$ for some odd prime $p$ and a natural number $n$. The number of squares in $F$ is given by $\frac{p^n + 1}{2}$, a number not divisible by $p$. If we would have that $1 + x^2$ is always a square, then any orbit of the action of $+1$ would either contain no squares or consist entirely of squares. That is, for any square $x^2$, also $x^2 + 1, x^2 + 2, \ldots, x^2 + (p - 1)$ would be squares.
In particular, the number of squares would be a multiple of $p$, as every orbit of this action contains $p$ elements. Contradiction.