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Find all of positive integers $k, m, n$ such that $$T := \frac{2kmn(m^2 - n^2)}{m^2 +n^2} \in \mathbb{Z} .$$

For example, if $(m^2 +n^2) \mid k$ then $T \in \mathbb{Z}$ but I can't find a formula of $k, m, n$. The question is that "is there a generating form of $k, m, n$?"

Thank you very much for your ideas.

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    When $k=1$ the top is the product of two legs of a Pythagorean triangle while the bottom is the hypotenuse. – 2017-01-07
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    Actually the general Pythagorean triangle is $(2mnk,(m^2-n^2)k,(m^2+n^2)k).$ This means your question is the same as asking for which Pythagorean triangles the hypotenuse divides the product of the legs. [The extra $k$ factors cancel.] – 2017-01-07
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    A strange arbitrary choice: Pick any $m,n$ at all and then just define $k$ so as to make the fraction an integer. [So maybe one wants to restrict the $k$ in some way.] – 2017-01-07
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    Yes, in fact, I want to restrict the domain of valeurs of $k$. But I can't do that. – 2017-01-07
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    mathJuan I've put up an answer which I think covers when T is an integer. – 2017-01-08

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Let $\gcd(m,n)=d$ and put $m=dx,\ n=dy$ so that $x,y$ are now coprime. [We will assume that $x>y.$]

Replacing $m,n$ by $dx,dy$ in $T$ gives $$T=\frac{d^2k\cdot 2xy\cdot(x^2-y^2)}{x^2+y^2}.\tag{1}$$ Now if $x,y$ are of opposite parity, the triple $(2xy, x^2-y^2,x^2+y^2)$ is a primitive Pythagorean triple. This means that the denominator of (1) is coprime with the second and third factors in the numerator of (1), and thus $T$ is an integer iff $x^2+y^2$ divides $d^2k.$

The remaining case is when $x,y$ are each odd. Here looking mod 4 one sees that $(x^2+y^2)/2$ is odd, and it is easily shown that if $T$ is now written as $$T=\frac{d^2k\cdot xy\cdot(x^2-y^2)}{(x^2+y^2)/2}.\tag{2}$$ then the denominator is coprime with the second and third factors in the numerator, so that $T$ is an integer in this case iff $(x^2+y^2)/2$ divides $d^2k.$

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    Can 2 divide (2xy, x^2-y^2,x^2+y^2) ? – 2017-01-08
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    @mathJuan Yes, I went too fast before. I have re-written the answer so now it should be correct. What I had assumed before was that $x,y$ could be assumed to have opposite parity, but that is not so. For example, $x=3,y=1$ gives $(x^2+y^2)/2=5,$ and it is $5$ which must divide $d^2k$ for $T$ to be an integer. – 2017-01-09
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    Yes, it is perfect for me. Thank you so much for your comments. – 2017-01-09