Establishing divisibility by a prime

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 $(a,b)$.