We consider a prime number $p\equiv 3 \pmod4$ and $\mathcal{D}(1,p):=\# \{(x,y,z)\in \mathbb{F}_p^3 \mid x^2+y^2+z^2\equiv 1 \pmod p\}$
By a powerful formula I found that $\mathcal{D}(1,p)=p^2-p.$
Now I have to deduce $\mathcal{F}(1,p):=\# \{(x,y,z)\in \mathbb{F}_p^3 \mid x^4+y^4+z^4\equiv 1 \pmod p\}$ . Maybe I can start with writing $x^2=u,y^2=v,z^2=t$. So the equation becomes : $u^2+v^2+t^2\equiv 1 \pmod p$ but I think there are other cases because I did not use $p\equiv 3\pmod 4$ enough.
Thanks in advance !
From the fact that $p\equiv 3 \bmod 4 $ you can in fact identify that there are exactly the same values in the same proportions for $\{a_i^2\}$ and $\{a_i^4\}$. Note that for some primitive root $g$ of $p$, the quadratic residues are represent by $g^{2k}$ and the fourth powers by $g^{4k}$, which collects the same group of numbers over a double range of powers, eg. $g^2 \equiv \left(g^{(p+1)/4}\right )^4$
Therefore $\mathcal{F}(1,p)=\mathcal{D}(1,p)$