[math-fun] Difference of Squares Modulo p?

Hello, I've been trying to prove the following statement, which I believe to be true: Let $p$ be a prime number, and let $n$ and $m$ be two distinct positive integers such that $0\le n , m \le (p-1)$. Then, $m^2 - n^2 = a (mod p)$ has exactly (p-1) solutions for each $0\le b \le (p-1)$. I've been just writing out a table of $m$ and $n$, and I tried to use some kind of counting argument, but it has gotten me nowhere. Could someone give me a hint? Best, Ki -- If I'm not working, then I must be depressed.