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Is there a method to count the number of solutions to the equation $$ x^{q+1}+y^{q+1} = 1 $$ for $x,y \in \mathbb{F}_{q^2}$, $q=p^k$?

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    We may notice that $x^{\frac{q^2-1}{2}}$ is Legendre symbol in $\mathbb{F}_{q^2}=\mathbb{F}_{p^{2k}}$, hence $x^{q+1}$ is always a $(q-1)$-th root of unity, and not many $n$-th roots of unity add up to one. – 2017-02-05
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    @JackD'Aurizio In $\Bbb{F}_q$ all the non-zero elements are $(q-1)$th roots of unity, and there are plenty of ways for two of them to add up to one. – 2017-02-06
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    @JyrkiLahtonen: all right, I should have written *it is not difficult to count in how many ways $\ldots$*. Nice solution btw (+1). – 2017-02-06

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Yes. This is not too difficult. The key is that $$ N:\Bbb{F}_{q^2}\to\Bbb{F}_q, x\mapsto x\cdot x^q= x^{q+1} $$ is the relative norm map. This implies that the elements $x^{q+1}=N(x)$ and $y^{q+1}=N(y)$ are always in the subfield $\Bbb{F}_q$.

Because $\Bbb{F}_{q^2}^*$ is cyclic of order $q^2-1$, and $q+1\mid q^2-1$, it follows that every non-zero element of $\Bbb{F}_q$ occurs as the image of exactly $q+1$ distinct element of the bigger field. Of course, $N(x)=0$ if and only if $x=0$.

Consider the equation $$ u+v=1 $$ with $u,v$ ranging over the field $\Bbb{F}_q$. There are

  1. $q-2$ solutions $(u,v=1-u)$ such that neither $u$ nor $v$ is zero,
  2. a single solution $u=0, v=1$, and
  3. a single solution $u=1, v=0$.

The idea is to think of these variables as $u=N(x)$ and $v=N(y)$. As $N(x),N(y)\in\Bbb{F}_q$ we get all the solutions to the original equation by pulling back the related solution $(u,v)$ of $u+v=1$. In view of the listed mapping properties of $N$ these give rise to

  1. $(q-2)(q+1)^2$ solutions $(x,y)$ such that $x\neq0$ and $y\neq0$,
  2. $q+1$ solutions $(0,y)$ such that $v=1$, and
  3. $q+1$ solutions $(x,0)$ such that $u=1$.

So we have a total of $$ \begin{aligned} (q-2)(q+1)^2+2(q+1)&=(q+1)[(q-2)(q+1)+2]\\ &=(q+1)(q^2-q)\\ &=q(q-1)(q+1) \end{aligned} $$ solutions $(x,y)$.

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    Thanks, that was very helpful! – 2017-02-09