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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.

Tito Piezas III
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user19405892
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  • Well if $x = y$ then you're looking at $45$-$45$-$90$ triangles, and certainly those don't have integer side lengths – pjs36 Feb 14 '16 at 21:17
  • @pjs36 Good point. Still trying to find solutions to this is going to be hard I think. – user19405892 Feb 14 '16 at 21:19
  • Your approach seems cool - however, the only solution for $y^2(y+1)^2+y^2(y-1)^2=z^2(z-1)^2$ with $y<1,000,000$ is $(y,z)=(1,2)$ (checked with Wolfram Mathematica 10.0). Not too promising, unfortunately... –  Feb 14 '16 at 21:42
  • For a similar equation to write the solution turned out to be. http://www.artofproblemsolving.com/community/c3046h1054874_triangular_numbers_for_pythagorean_triples And this is complicated and does not work. – individ Feb 15 '16 at 06:11
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    I might mention that this is an unsolved problem: https://www.google.co.kr/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwjos5-50JDLAhXCkJQKHQWaBckQFgggMAA&url=http%3A%2F%2Fwww.fq.math.ca%2FScanned%2F17-2%2Fballew.pdf&usg=AFQjCNEJJl_aJkyH3zZYBACg_G6J6Unyug&sig2=hjJZTOFgjmcD3-GD98RTPQ&bvm=bv.114733917,d.dGo&cad=rjt – S.C.B. Feb 24 '16 at 14:40

3 Answers3

1

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

vonbrand
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1

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.

Tito Piezas III
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  • For $(4)$, by observing the results of a computer search, if $w,x,y = 2uv,,u,,u+1$, then integer solutions can be found using the Pell equation $u^2-2v^2=-1$. – Tito Piezas III Feb 24 '16 at 23:01
0

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.

poetasis
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