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A Pythagorean Right Triangle (PRT) is a right triangle all of whose sides have positive integer lengths. Find all PRTs having a hypotenuse of $65$.

We need $m^2+n^2 = 65^2$, but we know that since $65 = 5 \cdot 13$ any $m,n$ satisfying $m^2+n^2 = 5^2$ or $m^2+n^2 = 13^2$ will also satisfy $m^2+n^2 = 65^2$. How can we continue?

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    see here http://mathworld.wolfram.com/PythagoreanTriple.html and the solution is $$63^2+16^2=65^2$$ – 2017-01-01
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    @Xammm, it's $m^2 + n^2 = 65^2$ not $65$. – 2017-01-01
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    The full set is $16^2 + 63^2 = 65^2$ , $25^2 + 60^2 = 65^2$, $33^2 + 56^2 = 65^2$, $39^2 + 52^2 = 65^2$ and perhaps $0^2 + 65^2 = 65^2$. Of these $16^2 + 63^2 = 65^2$ and $33^2 + 56^2 = 65^2$ are primitive while $25^2 + 60^2 = 65^2$ and $39^2 + 52^2 = 65^2$ involve multiples of $5$ or of $13$ – 2017-01-01
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    @Guru oh you're righ. Maybe I'm still drunk xD. – 2017-01-01

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