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Prove that the number of ordered triples $(x, y, z)$  in the set of residues of $p$ such that $(x+y+z)^2 \equiv axyz \mod{p}$, where $gcd(a, p) = 1$ and $p$ is prime is $p^2 + 1$.
Number Theory
Medium
Suggested Book
Elementary Number Theory by David Burton

Do you really need a hint? Try it first!

First, assume that at least one of $x,y,z$ is zero. Show that there are $3p-2$ solutions in this case. Next we need to consider the case where none of them is zero.

Let us denote the residue class of $p$ by $\mathbb{Z}_p$. Show that, there exist non-zero $b,c$ in $\mathbb{Z}_p$ such that $y\equiv bx\;\text{mod}\; p$ and $z\equiv cy\;\text{mod}\; p$.

From hint 2, show that $(1+b+bc)^2\equiv ab^2cx\;\text{mod}\;p$. This means that $x\equiv a^{-1}c^{-1}(1+b^{-1}+c)^2\;\text{mod}\;p$ (you need to convince yourself that the inverses exist).  Now it becomes a matter of simply choosing $b,c$.

Note that, $1+b^{-1}+c$ cannot be zero. Given any $b\neq p-1$, there exists exactly one non-zero $c$ such that $1+b^{-1}+c$ is 0 modulo $p$. Hence, in this case there are $(p-2)^2$ choices. For $b=p-1$, this special $c$ is actually 0. Hence in this case there are $p-1$ choices. Thus, the total number of choices is $(p-2)^2+(p-1)=p^2-4p+4+(p-1)=p^2-3p+3$. Adding to this the $3p-1$ cases considered in hint 1, we get $p^2+1$ as the answer.

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