# Establishing divisibility by a prime

Mathematics Asked on January 3, 2022

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 $$a^2-bcequiv0pmod p$$, so $$a^2equiv bcpmod p$$.

Given $$bin{1,...,p-1}$$, there is a unique $$tin {1,...,p-1}$$ such that $$btequiv1bmod p$$;

$$t$$ is the multiplicative inverse modulo $$p$$ of $$b$$.

Then $$ta^2equiv tbcequiv1cequiv cpmod p$$, so $$c$$ is uniquely determined.

Answered by J. W. Tanner on January 3, 2022