prove if p is an ordinary prime then p=a^2+b^2+c^2+d^2?

Answer 1

You do not indicate any restrictons on a,b,c,d. Can they all be the same? Is 0 allowed?
I believe 2 is an ordinary prime.
We could have a=b =1 and c=d=0
41 = 5^2 +4^2 +0^2 +0^2
43 = 5^2 +4^2 +1^2+1^2
47 = 5^2 +4^2 +2^2 +(sqrt(2))^2
53= 5^2+4^2 +3^2 +(sqrt(3))^2

I think if there are no restrictions on a,b,c,d the statement is true. But if there are restrictions, it is probably not true.

Answer 2

I believe this is Lagrange's four-square theorem. Here's a link with its proof:

http://planetmath.org/encyclopedia/ProofOfLagrangesFourSquareTheorem.html