Sum of two $n$th powers in finite fields

Question

Let $q$ be a prime power, let $n$ be a positive integer and let $\mathbb{F}_q$ be the finite field of cardinality $q$. I have some computational evidence that the set $$\{x^n+(-1)^nay^n:x,y\in\mathbb{F}_q\}$$ is the whole of $\mathbb{F}_q$, unless $q$ is small (with respect to $n$).

This does make sense because if, for instance, $q-1=n$, then the set $\{ x^n+(-1)^nay^n:x,y\in\mathbb{F}_q \}$ only consists of $\{0,1,a,1+a\}$ because each element of $\mathbb{F}_q$ raised to the $n$ equals either $0$ or $1$.

Can anyone justify the computational evidence?

Answer 1

Yes, this is true, and is proved e.g. as Corollary 3 of Small's "Diagonal equations over large finite fields" (Can. J. Math. 1984).

Small actually gives explicit bounds on how large $q$ needs to be in terms of $n$ — in particular the equation $ax^n+by^n$ generates all of $\mathbb{F}_q$ whenever $a,b\in\mathbb{F}_q\setminus\{0\}$ and $q>(\delta-1)^4$, where $\delta=(n,q-1)$.

Answer 2

I would like to give some comments and context, but I am not going to cite any improvement to the result of Small quoted by Thomas Bloom. (I do not know if there is any.)

Since the image of $f(x)=x^n$ is the same as that of $g(x)=x^{\gcd(q-1,n)}$, and $\gcd(q-1,n)$ divides $q-1$, we may restrict from now on to $n$ being a divisor of $q-1$.

You ask whether $f(x)+af(y)=z$ is solvable for each $z$. Since $f(0)=0$ we may assume $z\neq 0$.

Since the multiplicative group of a finite field is cyclic, the image of $f$ has size $1+((q-1)/n)$. This leads to two immediate results:

1. If $(1+((q-1)/n))^2<q$, which translates to $n>\sqrt{q}+1$, the answer is negative by a trivial counting argument (bounding the possible values of $z$ by the Cartesian product of the images of $f$).
2. If $2(1+((q-1)/n)>q$, which translates to $n=1,2$, for any given choice of $z$ the coset $z+\operatorname{Im}f$ intersects the subgroup $-\operatorname{Im}f$, hence the answer is positive in this case.

Gauss was highly interested in the number of solutions to $f(x)+f(y)=1$ for $n=3,4$. You can read about it in Chapter 8 of Ireland and Rosen's book "A classical introduction to modern number theory", where Theorem 2 is due to Gauss. One can use Gauss sums and Jacobi sums to derive the estimate $\#\{x,y: f(x)+af(y)=z\}=q+E$ for $\lvert E\rvert\le n-1+\sqrt{q}(n-1)(n-2)$, which ensures a solution for each $z$ if $\sqrt[4]{q}+1>n$, which matches Small's result. See Proposition 8.4.1 for a special case (corresponding to $a=z=1$ and $q$ a prime) of this which contains all the ideas already. See also Theorem 5 in the aforementioned chapter for a general theorem of Weil from 1949 about diagonal hypersurfaces and the number of points on them; this inspired his famous Weil Conjectures.

See also the paper "Sums of Powers in Large Finite Fields: A Mix of Methods" by Vitaly Bergelson, Andrew Best and Alex Iosevich, that appeared in 2021 in the American Mathematical Monthly. It discusses all of the above and more, including Small's work and Schur's work mentioned by Gerry Myerson.