Integer solutions for $x^2+y^2=208$ Which steps I can follow to find the integer solutions for the equation $x^2 + y^2 = 208$?

A positive integer is the sum of two squares iff primes $\equiv 3\pmod 4$ (following the Fermat's theorem on sums of two squares) of its factorization are raised to even exponents.
$$208=2^4\cdot 13$$
So, $208$ meets the condition.
Now, $13=3^2+2^2$, so $208=4^2(3^2+2^2)=12^2+8^2$
