# How many Pythagorean triples are there in which every member is triangular?

How many Pythagorean triples $(a,b,c)$ are there such that $a, b$ and $c$ are triangular?

Any two solutions with only $a$ and $b$ interchanged are considered equivalent.

The question of existence was posed by Zarankiewicz and answered by Sierpinski in Sur les nombres triangulaires carrés (1961):

Such a triangle is known whose sides are $t_{132} = 8778, t_{143} = 10296$ and $t_{164} = 13530$, but we do not know if there are others and if their number is finite.

A partial result: Since $t_n^2 = 1^3 + \ldots + n^3$, the equation $t_x^2 + t_y^2 = t_z^2$ is equivalent to $t_x^2$ being a sum of $k = z - y$ consecutive cubes. This leads one to consider elliptic curves. In Pythagorean triples and triangular numbers (1979) by D. W. Ballew and R. C. Weger it is shown by applying a theorem of Siegel that for a given $k$ there are only finitely many solutions. Actually finding such solutions has been done by R. J. Stroeker: On the sum of consecutive cubes being a perfect square (1995), without the restriction that the square is some $t_x^2$.

• Maybe someone should finally create an "Integral points on $K3$ surfaces" tag. Nov 13, 2014 at 18:07
• @Jeremy, there is a k3-surfaces tag but it does not seem to say anything about integer points. Nov 13, 2014 at 18:53
• @JeremyRouse, anyway, I can make a tag if you like, your best fairly short phrasing. Nov 13, 2014 at 19:31
• @JeremyRouse I don't want a tag as much as I want someone to ask the question: What is known about algorithms for finding/bounding integral points on K3 surfaces? Nov 17, 2014 at 15:46

It was proved in http://arxiv.org/abs/0811.2477 (On certain diophantine equations related to triangular and tetrahedral numbers, by Maciej Ulas) that the equation $t_x^2+t_y^2=t_z^2$ has inﬁnitely many solutions in rational numbers. It seems the number of integer solutions of this equation is still an open problem. It is also an open problem whether there exists a right triangle all sides of which are tetrahedral numbers of the form $T_n=\frac{1}{6}n(n+1)(n+2)$ (a tetrahedral version of the question asked).

I tried the following which hasn't yet returned a solution:

Suppose we want to find a Pythagorean triple $(a,b,c)$ such that $b-a=1$. Then we can parameterize such solutions by,

$\bigl (\frac{x-1}{2} \bigr )^2 + \bigl (\frac{x+1}{2} \bigr )^2 = y^2$

where (x,y) are solutions to the Pell equation $x^2 - 2y^2 = -1$. Multiplying both sides of this equation by $x^2$ we obtain,

$\bigl (\frac{x(x-1)}{2}\bigr )^2 + \bigl (\frac{x(x+1)}{2} \bigr )^2 = (xy)^2$

The first two numbers are clearly triangular, so the question now becomes:

Are there solutions to the Pell equation $x^2 - 2y^2 = -1$ such that their product is triangular?

Let $(x_n, y_n)$ be the nth positive solution to our Pell equation, i.e., $(x_1, y_1) = (1,1); (x_2, y_2) = (7,5); (x_3, y_3) = (41, 29); . . .$

Then, from known formulas for solutions to our Pell equation, we can derive a function that returns their product (steps intentionally missing), namely,

$f(n) = x_n y_n = \frac{(1 + \sqrt 2 )^{4n-2} -\ (1 - \sqrt 2 )^{4n-2}}{4 \sqrt 2}$ for $n \in N$

Then $f(1) = 1, f(2) = 35, f(3) = 1189, . . .$

Using the power of Wolfram Alpha, I found that $f(2)$ through $f(66)$ are not triangular numbers.

Perhaps there is a triangular number further in the sequence or perhaps there is some reason that this product can never be triangular?

• This is not an answer to the question. Nov 17, 2014 at 2:38
• True. But even finding one more example of such a triple would be something. So I thought I share a possible method to discover another one, and my computing are limited. Nov 17, 2014 at 2:42
• So it would be a reasonable comment. Nov 17, 2014 at 2:46