For how many values of $a,b,c\in(1,2\ldots,p-1)$ does $p$ | $({a^2}-bc)$ where $p$ is an odd prime number

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In a mock test for an entrance exam I am preparing for came the following question:

Let $p$ be an odd prime number and $T_p$ be the following set of matrices

$$ T_p= \left( A=\begin{bmatrix} a & b \\ c & a\\ \end{bmatrix}:a,b,c\in(0,1,2\ldots,p-1) \right) $$

The number of $A$ in $T_p$ such that the trace of A is not divisible by $p$ but $det(A)$ is divisible by $p$ is, (the answer is in terms of p)

Here's what I have done so far:

Trace of $A=2a$.

Clearly, trace of A will be divisible by p iff $a=0$

$\Rightarrow \qquad a\in(1,2,\ldots,p-1)$

We have $det(A)={a^2}-bc$

So $\quad p\;|\;det(A)\Rightarrow \quad p\;|\;{a^2}-bc \qquad (1) $

Now how do I find out the number of ordered pairs $(b,c)$ which will satisfy condition $(1)$ for each value of $a\in(1,2,\ldots,p-1)$

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For each choice of $a$ and $b$, there is a unique choice of $c$ such that $p \mid a^2-bc$ (that is, $c = a^2 b^{-1}$). It follows that there are $(p-1)^2$ such triplets $(a,b,c)$ and $p-1$ pairs $(b,c)$ for a given choice of $a$.