Integer solutions for $x^2+y^2=208$

Question

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

Answer 1

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 $208=4^2(3^2+2^2)=12^2+8^2$

Answer 2

Since $208$ is a relatively small number, it is not too hard to check $208-x^2$ for all numbers $0\leq x\leq 14$ to see if the answer is a square number. The negatives of whatever solutions you find are also solutions.

Answer 3

This represents a circle centered at the origin of radius $\sqrt{208}$. You only need to check integer $x$-coordinates between $-14$ and $14$.