Find the maximum possible perimeter of a right triangle 
The ratio between the perimeter of a right triangle and its area is 2:3. The sides of the triangle are integers. Find the maximum possible perimeter of the triangle. 

If the sides of the triangle are $A$, $B$ and $C$ (the hypotenuse), I have deduced that:
$$A+B+\sqrt{A^2+B^2}=\frac{AB}{3}$$
I am stuck here. Any hints? From the above I know that $A^2+B^2$ should be a square and that $AB$ should be divisible by 3.

There are naturals $m$ and $n$ such that $m>n$, $a=m^2-n^2$, $b=2mn$ and $c=m^2+n^2$.
See here: https://en.wikipedia.org/wiki/Pythagorean_triple
Thus, $$\frac{m^2-n^2+2mn+m^2+n^2}{mn(m^2-n^2)}=\frac{2}{3}.$$
Can you end it now?
I got $56$ as the answer. 
