How many integer-sided right triangles are there whose sides are combinations?

Question

How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?

Attempt:

This seems like a hard question, since I can't even think of one example to this. Mathematically we have,

$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$

where we have to find all positive integer solutions $(x,y,z)$.

I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,

$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$

then we are done.

Answer 1

Hint: Start the other way around, by the formula to generate all Pythagorean triples.

Answer 2

Solving $(1)$ for $z$, we have,

$$z = \frac{1\pm\sqrt{1\pm4w}}{2}\tag3$$

where,

$$w^2 = (x^2-x)^2+(y^2-y)^2\tag4$$

It can be shown that $(4)$ has infinitely many integer solutions. (Update: Also proven by Sierpinski in 1961. See link given by MXYMXY, Pythagorean Triples and Triangular Numbers by Ballew and Weger, 1979.)

However, the problem is you still have to solve $(3)$. I found with a computer search that with $x<y<1000$, the only integers are $x,y,z = 133,\,144,\,165$, so,

$$\left(\dfrac{133(133-1)}{2} \right)^2+\left (\dfrac{144(144-1)}{2} \right)^2 = \left(\dfrac{165(165-1)}{2} \right)^2$$

P.S. If you're curious about rational solutions, then your $(1)$ and $(2)$ have infinitely many.

Answer 3

For your equation, solutions include $$(1,1,1)\quad (1,2,2)\quad (1,3,3)\quad (1,3,3)\quad (1,4,4)\quad (1,5,5)\quad \cdots $$ Any triple with zero or one for one of $\space x,y\space $ will work. All others would probably require a brute force search.