Given $\sqrt {m^2+(n+o)^2}$ is int, is it possible that atleast one of $\sqrt {o^2+(n+m)^2}$ or $\sqrt {n^2+(o+m)^2}$ is also integer?

Question

Given $m,n,o,\sqrt {m^2+(n+o)^2}\in\mathbb N$ and $o\le n\le m$, is it a guarantee that both of $\sqrt {o^2+(n+m)^2},\sqrt {n^2+(o+m)^2}$ are irrational?

What I tried: Firstly, ${m^2+(n+o)^2}\le n^2+(o+m)^2\le o^2+(n+m)^2$
So, to show $n^2+(o+m)^2$ is not int, its is enough to show $$2o(m-n)\le2\sqrt{m^2+(n+o)^2}+1$$ But this is didnt show anything useful. Any hints?

Edit: I realized this is same as asking can there exist right angled triangles with unqual hypotnuese such that sum of other two sides is equal

Answer 1

Why no?

Take $(m,n,o)=(3,3,1).$

We obtain: $$\sqrt{n^2+(o+m)^2}=\sqrt{3^2+4^2}=5.$$