Pythagoras Wouldn't Like This

Relevant wiki: Quadratic Diophantine Equations - Modular Arithmetic Considerations

x and y are odd, so they can be written as:
$x = 2n-1$ ,
$y = 2m-1$
where n and m are positive integers.
$x^2 + y^2 = 4n^2+4m^2-4n-4m+2$ .
This is equal to 2 mod 4 and is clearly an even number.
Let's assume it is a perfect square - call it $z^2$ .
Since this is even, then z is even also and can be written as $z = 2p$ where p is a positive integer.
$z^2 = (2p)^2 = 4p^2$ which is equal to 0 mod 4 .

So $x^2 + y^2$ can never be equal to $z^2$ . Therefore, $x^2 + y^2$ is not a perfect square.

For any $n$ ,

$n^2 \equiv 0,1 \quad \left( \bmod \text{ 4} \right)$

and for any odd $k$ ,

$k^2 \equiv 1 \quad \left( \bmod \text{ 4} \right)$

For odd $x$ and $y$ ,

$x^2 + y^2 \equiv 2 \quad \left( \bmod \text{ 4} \right)$

However there cannot exist a square which is congruent to 2 modulo 4.

We are talking of Pythagorean Triple. One leg of any primary triple is odd the other even. So any derived triple by multiplying by an integer will always have one leg as even. Two legs can be even in a derived triple if multiplied by an even integer. For primary triple, the hypotenuse is also always odd since the two legs square added is addition of odd and even.