Understand the problem

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 $p^2 + 1$.
Source of the problem
Brazilian Olympiad Revenge 2010
Number Theory
Difficulty Level
Suggested Book
Elementary Number Theory by David Burton

Start with hints

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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