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

Question

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 $t_x^2$.

Answer 1

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).

Answer 2

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?