no. of integeral solutions of x²+y²=2008

x²+y² = 8X251 and 251 is a prime of the form 4k+3.

There is a result that states that if x²+y² is divisible by a prime p of the form 4k+3, then both x and y are divisible by p.

But that means LHS is divisible by 251² whereas RHS is divisible at most by 251 which is a contradiction.

Hence no integral solutions exist