Let $a,b,c$ belong to $\{0,1,....p-1\}$, where $p$ is an odd prime. We need to find the number of triplets $(a,b,c)$ such that $a^2-bc$ is divisible by $p$, but $a$ isn't.
The solution to this problem claims that there are $(p-1)$ ways to choose $a$, $(p-1)$ ways for $b$, and for each $(a,b)$, there is one value of $c$. So, a total of $(p-1)^2$ cases.
Now the first two statements make perfect sense: $a$ can be anything except $0$, since $a$ is not divisible by $p$. $b$ cannot be zero, since then $a^2-bc=a$, which can't be divisible by $p$. However, I have no idea why for each $(a,b)$ there will always be a value for $c$, and not more than $1$.
I tried to prove this, but couldn't get far. I could only establish that $a^2$ can be congruent to $\{1,4,9,\ldots,\frac{1}{4}(p-1)^2\}$ mod $p$. Combining this with the fact that $b$ can be $\{1,2,3,\ldots,p-1\}$ mod $p$, I can't think of any reasoning that forces $c$ to take only a single value for each $(a,b)$.
If $a^2-bc$ is divisible by $p$, then $a^2-bc\equiv0\pmod p$, so $a^2\equiv bc\pmod p$.
Given $b\in\{1,...,p-1\}$, there is a unique $t\in \{1,...,p-1\}$ such that $bt\equiv1\bmod p$;
$t$ is the multiplicative inverse modulo $p$ of $b$.
Then $ta^2\equiv tbc\equiv1c\equiv c\pmod p$, so $c$ is uniquely determined.