# Square Triangles

A triangle number is one of the form $$\sum_{k = 1}^n k = \frac{n(n + 1)}{2}$$ for positive any integer n. So my question is how many triangle numbers are also square numbers; that is how many ordered pairs of positive integers $$(n,m)$$ are there such that $$f(n,m) = \frac{n(n + 1)}{2} - m^{2} = 0$$. $$(8,6),(49,35)$$ are two solutions of $$f(n,m) = 0$$..

Note by Samuel Queen
4 years, 10 months ago

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See problem 5 here for another proof that there are infinitely many triangular numbers that are perfect squares.

- 4 years, 10 months ago

\begin{align*} \frac{n(n+1)}{2} &= m^2 \\ n^2 + n &= 2m^2 \\ 4n^2 + n + 1 = 8m^2 + 1 \end{align*}

Then substitute $$x=2n+1$$ and $$y=2m$$:

\begin{align*} x^2 &= 2y^2 + 1 \\ x^2 - 2y^2 &= 1 \end{align*}

This is a well-known Pell's equation, and has infinite solutions.

- 4 years, 10 months ago

The 3rd equation must be $$4n^{2}+4n+1=8m^{2}+1.$$

- 4 years, 10 months ago

Ok so $$g(x,y) = x^{2} - 2y^{2} - 1 = 0$$ has infinitely many integral solutions. But each $$(x,y)$$ that solves $$g(x,y) = 0$$ gives a $$(n,m)$$ that solves $$f(n,m) = 4n^{2} + 4n - 8m^{2} = 0$$ only if $$x = 2k + 1$$ and $$x \ge 3$$.

- 4 years, 10 months ago

Yeah, and $$x^2-2y^2=1$$ also has infinite solutions for $$x$$ being an odd integer. Should have added that.

- 4 years, 10 months ago

Isn't $$x$$ automatically odd? (That is, if $$x,y$$ are both integers.)

- 4 years, 10 months ago