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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 $f(50)$ 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?

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?

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 $f(1)$ through $f(50)$ 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?

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 $f(50)$ 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?

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,

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

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,

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

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 $f(1)$ through $f(50)$ 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?

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 $f(1)$ through $f(50)$ 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?