Number Theory - non-zero squares?

Question

For any prime p greater than or equal to 3, prove that the number of non-zero squares in Z (subscript p) = {0, 1, 2, ...p-1} is (p-1)/2

Thanks

Don

Answer 1

If n is in Zp and 0 (p - n)^2 = p^2 -2np + n^2 = n^2 (because p=0 in Zp)

Now, if n=0 n^2 = 0.

For the non-zero numbers in Zp, let's see what happens if m,n are in Zp and m^2 = n^2

m^2 - n^2 = (m - n)(m + n)

m^2 = n^2 -> m-n=0 or m+n=0 -> m=n or m=-n

there are (p - 1) unordered pairs of such that
m^2 = n^2

Answer 2

There are (p-1) numbers from 1 to p-1 inclusive.

In Z sub p, the square of any one of them, say n, is equal to the square of (p-n), and different from the square of any other one m (because if it wasn't, we would have n^2 - m^2 = 0, and a non-trivial factorisation of p, which is impossible).

So, the (p-1)/2 numbers from 1 to (p-1)/2 give that many distinct squares, and there are no other squares.

Answer 3

For there to be some number n to fit your problem, there would wind up being a pair of consecutive perfect powers. But if the Catalan Conjecture is true, the only consecutive perfect powers are 8 and 9.