$N_p := \text{card}\{(x, y, z, t) \in (\textbf{F}_p)^4 : ax^4 + by^4 + z^2 + t^2 = 0\}?$ Assume that $ab \neq 0$. What is $$N_p := \text{card}\{(x, y, z, t) \in (\textbf{F}_p)^4 : ax^4 + by^4 + z^2 + t^2 = 0\}?$$I need this result, but unfortunately I am not a number theorist. Could anyone provide me a reference/supply a computation? Thanks!
 A: We need to distinguish according to whether $p$ is congruent to $1$ or $3$ modulo $4$, and whether $-b/a$ is or is not a fourth power modulo $p$. (Note that the case $p=2$ is trivial since $ax^4+by^4+z^2+t^2 \equiv ax+by+z+t \pmod{2}$, so this gives exactly $2^3=8$ solutions modulo $2$.)
We need to use the following
Fact: if $t \in \mathbb{F}_p^{\times}$, then $x^2+y^2=t$ has $p+1$ solutions over $\mathbb{F}_p$ if $p \equiv 3 \pmod{4}$, and $p-1$ solutions over $\mathbb{F}_p$ if $p \equiv 1 \pmod{4}$. (The proof of this is easy and quite standard.)
Now if $p \equiv 3 \pmod{4}$, and $-b/a$ is non-square modulo $p$, then $ax^4+by^4$ is never zero, so to each of the $p^2-1$ pairs $(x_0,y_0) \in \mathbb{F}_p^2$ with $x_0,y_0$ not both zero, will correspond exactly $p+1$ solutions. This gives $(p+1)(p^2-1)+1$, or
$$p^3+p^2-p$$
solutions, since $x=y=0$ gives only the all-zero solution (because $p \equiv 3\pmod{4}$). If $-b/a$ is a square modulo $p$, then it is a fourth power modulo $p$ since $\mathbb{F}_p^{\times 2} = \mathbb{F}_p^{\times 4}$, and to each non-zero value of $x_0 \in \mathbb{F}_p$ will correspond two values of $y_0$ such that $ax_0^4+by_0^4$ equals zero. So there will be exactly $(p-1)^2$ pairs $(x_0,y_0)$ yielding a non-zero value of $ax_0^4+by_0^4$, giving a total of $(p+1)(p-1)^2+(2p-1)$, or
$$p^3-p^2+p$$
solutions.
On the other hand, if $p \equiv 1 \pmod{4}$ and $-b/a$ is not a fourth power modulo $p$, then by the same reasoning as above (and the Fact) we get $(p-1)(p^2-1)+(2p-1)$, or
$$p^3-p^2+p$$
solutions (keeping in mind that $z^2+t^2=0$ has $2p-1$ solutions this time). Finally, if $-b/a$ is a fourth power, then for each non-zero value of $x_0 \in \mathbb{F}_p$ we have four values of $y_0$ such that $ax_0^4+by_0^4$ equals zero, so there are precisely $p^2-4p+3$ pairs $(x_0,y_0)$ yielding a non-zero value of $ax_0^4+by_0^4$, which gives a total of $(p-1)(p^2-4p+3)+(4p-3)(2p-1)$, or 
$$p^3+3p^2-3p$$
solutions. That is, if I have not made any mistakes.
