Proof that the sum of the even side and the hypotenuse of a coprime (and positive) Pythagorean triple is a square number

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For every Pythagorean triple $(x,y,z)$, with $d=\gcd(x,y,z)$;
there are coprime integers $a,b$ with different pairity, such that:
$$x=d(a^2-b^2) \ \text{and} \ y=d(2ab) \ \text{and} \ z=d(a^2+b^2); $$


Now notice that the coprime condition implies $d=1$;
so we have the following:

$$(2ab)+(a^2+b^2)=(a+b)^2.$$

Note that $(m^2+n^2)^2=(2mn)^2+(m^2-n^2)$. Since $2mn$ is always even, we have $m^2+n^2+2mn=(m+n)^2$.

As @shuri pointed out, in $(3,4,5)$, $4+5=9$ which is a perfect square. But when you do not include your coprime condition, we can multiply all the elements of triplet by an even number say $2,4$ etc to get $(x,y,z)$ of same parity and then one pair give us a perfect square $m^2+n^2+2mn$ while other one give us twice of a perfect square $(2m^2)$.