No. Of ordered triplet if $a^{2}-b c$ is divisible by 13.

Question

Let a,b,c $\in\{0,1,2,3, \ldots \ldots, 12\}$ such that $a^{2}-b c$ is divisible by $13,$ then number of ordered triplet $(a, b, c)$ is..

If I start by counting all the combination of {a,b,c} for which $a^{2}-b c$ is a multiple of 13 then it will take too much time. Is there an easy approach to this question.

Answer 1

There are $13$ triples with $b=0$.

We can obviously work modulo $13$. If $b\ne0$, then it has an inverse $u$ and we can set $a=bx$, $c=by$, so the equation becomes $x^2=y$.

Answer 2

To save on notation bloat, lets say here that $x \equiv y$ is a shorthand notation for $x \equiv y \pmod{13}$.

These are the possible values for $a^2 \pmod{13}$.

\begin{array}{|c|c|} \hline a & a^2 \\ \hline 0 & 0 \\ 1, 12 & 1 \\ 2, 11 & 4 \\ 3, 10 & 9 \\ 4, 9 & 3 \\ 5, 8 & 12 \\ 6, 7 & 10 \\ \hline \end{array}

Note that $13$ divides $a^2-bc$ is equivalent to $bc \equiv a^2$

If $a \not \equiv 0$, then we find $c \equiv a^2 b^{-1}$.

So, for each nonzero value of $a^2$, there are $12$ pairs $(b,c)$ such that $bc \equiv a^2$.

So, for example, there are $24$ triples $(a,b,c$) such that $a^2 \equiv 1$ and $bc \equiv a^2$.

You should be able to work out the rest.