Oops. I had changed the second "a" into a "b" without changing the first "a". I meant to say: 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, $m^2 - n^2 = b (mod p)$ has exactly (p-1) solutions for each $0\le b \le (p-1)$. On Mon, Aug 25, 2008 at 10:57 PM, Ki Song <kiwisquash@math.sunysb.edu>wrote:
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.
-- If I'm not working, then I must be depressed.