How many right angled triangles are possible with the perpendicular side equal to 36 units. I took the side $x$ and $y$ and using Pythagoras theorem you have $(x+y)(x-y) = 1296$ and $1296$ has $25$ factors. What next?

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    $\begingroup$ I think we wants the ones with integer side lengths $\endgroup$ – Jorge Fernández Hidalgo Mar 23 '15 at 2:48
  • $\begingroup$ Do you want the sides to have integer lengths? It's not clear from your question, so I'm voting to close as unclear what is being asked. $\endgroup$ – Joel Reyes Noche Mar 27 '15 at 4:11
  • $\begingroup$ I think what you are asking for are the triples where one or the other leg is $36$ units long. Those are $f(10,8)=(36,160,164)$ and $f(9,2)=(77,36,85)$ $\endgroup$ – poetasis Jul 2 '19 at 15:35
  • $\begingroup$ If you want to match any sides this answer will show you how. $\endgroup$ – poetasis Jul 2 '19 at 15:39

We will count the integer-sided triangles. I would rather call the hypotenuse $z$ and the other leg $y$, so we want $(z-y)(z+y)=36^2$. But $z-y$ and $z+y$ must both be even, else $z$ and $y$ would not be integers. So $z+y=2\alpha$ and $z-y=2\beta$, where $\alpha$ and $\beta$ are integers and $\alpha\gt\beta$ and $\alpha\beta=18^2$.

Now $18^2=2^2\cdot 3^4$, so $18^2$ has $15$ factors $\alpha$. One of them, $18$, gives $\beta=18$, so must be discarded. Exactly half of the remaining factors give $\alpha\lt\beta$. So there are $7$ Pythagorean triangles with one leg equal to $36$.

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