TITLE: Proof that the sum of the even side and the hypotenuse of a coprime (and positive) Pythagorean triple is a square number
QUESTION [5 upvotes]: I wish to prove that for any coprime (and positive) Pythagorean triple $(x,y,z)$, where $x$ is even, $x+z$ is always a square number.
By inspection, this appears true; $(3,4,5)$ gives $9$,  $(5,12,13)$ gives $25$, $(8,15,17)$ gives $25$, $(7,24,25)$ gives $49$ et cetera. Yet I am stuck on the proof.
I do not know the original source of the problem.

REPLY [3 votes]: 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 $(2m^2)$.