Find the numbers of ordered array $(a,b,c,d)$ such $a^2+b^2\equiv c^3+d^3\pmod p$ Let $p$ be prime number,and such $p\equiv 1\pmod {12}$,Find the numbers of ordered array $(a,b,c,d)$ that satisfies the following conditions:
(1):$a,b,c,d\in \{0,1,2,\cdots,p-1\}$
(2):$a^2+b^2\equiv c^3+d^3\pmod p$
maybe can use this problem methods?:But I can't  it ,Thanks  2018 TST 
 A: Outlining the first approach that occured to me by giving steps only. Ask, if you get stuck at some point.
This is about the number of solutions of the equation
$$a^2+b^2=c^3+d^3$$
in the field $\Bbb{F}_p$, $p\equiv1\pmod{12}$.


*

*Because $p\equiv1\pmod4$ there exists an element, call it $i\in\Bbb{F}_p$, such that $i^2\equiv-1\pmod p$.

*Using $i$ we can factor $a^2+b^2=(a+bi)(a-bi)$.

*The linear system
$$\begin{cases}a+bi=u,\\a-bi=v,\end{cases}$$
has a unique solution $a,b\in\Bbb{F}_p$ to any given pair of elements $u,v\in\Bbb{F}_p$.

*In light of item 3, the equation $a^2+b^2=e$ has $p-1$ solutions $(a,b)$ when $e\neq 0$, and $2p-1$ solutions $(a,b)$ when $e=0$. Hint: How many ways to write $e=uv$?

*Similarly, because $p\equiv1\pmod 3$, there is an element $\omega\in\Bbb{F}_p$ such that $\omega\neq1$, $\omega^3=1$.

*$c^3+d^3=0$ if and only if $c=-d$, $c=-d\omega$ or $c=-d\omega^2$. 

*So $c^3+d^3=0$ for $3(p-1)+1$ pairs $(c,d)$, and $c^3+d^3\neq0$ for the remaining pairs $(c,d)$.

*Putting items 4 and 7 together gives you the answer.



There may be a solution requiring less knowledge about finite fields. I was simply pretty much on autopilot here, the congruence $p\equiv1\pmod{12}$ being kind of a give-away.
