Hi,

this is an old exam question dealing with some trivial number theory. I'm not sure if this question should go in the "Number Theory" section. I don't feel this is "university level" material though. Please forgive me if this is the wrong section.

Give a proof that any odd prime number can be written as the difference between two squares. E.g. 5 = 3^2 - 2^2

Attempt at proof:

Let p be a prime number > 2.

We wish to show that p = a^2 - b^2, where a and b are integers.

p = a^2 - b^2 = (a + b)(a - b)

I know that p should only have the factors p and 1. Therefore I can reduce the expression to be p = a + b?

From simply a Google search I see that I should also be able to conclude that a - b = 1. How do I reach that conclusion?

Therefore, we have that a = (p + 1)/2 and b = (p - 1)/2

p = (p + 1)/2 - (p - 1)/2 = a^2 - b^2

In addition to the questions above in the attempt it is not clear to me that this is a proof. Could someone explain it to me?

Thanks for any help.