TITLE: Can every element of a finite field be written as a sum of two non-squares?
QUESTION [7 upvotes]: We know that any element of a finite field $\mathbb{F_{q}}$ ($q$ odd prime power) can be written as a sum of two squares - is the same true for non-squares? Can any element of a (sufficiently large) finite field be written as a sum of two non-squares?
I know that the above is not true in general, e.g. in $\mathbb{F_{3}}$, the only non-square is $2$ and so $2$ itself cannot be written as a sum of two non-squares. However, if $q$ is large enough could the above be true?
If it is true, then could anyone provide any hints on I could prove it?
Many thanks!

REPLY [3 votes]: For prime fields its true for $p=4k+1>5$, and for all elements except 0 for $p=4k+3$.

If $p=4k+3$, then $0$ cannot be written in this way.
Now for non-zero elements. It's equivalent to prove that each element is the sum of two non-zero squares.

2a. For squares use  such formula
$$
a^2=\left(\frac35a\right)^2+\left(\frac45a\right)^2.
$$
2b. Let $x$ be non-square without such decomposition. Then, of course, all non-square don't have such decomposition.
Since $x-1$ is non-square (if not, then $x=1+a^2$), then $x-1$, $x-2$, $\dots$ $2$, $1$ are non-squares. Contradiction.