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?
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 $36$.