Integer solutions to $ 2 (r^2 - r) = t^2 - t $

Question

I'm trying to find integer positive solutions to the equation: $$ 2 (r^2 - r) = t^2 - t $$ So far I've been giving "test" values to t, say $t = 20$, and then solving the quadratic equation with substituted $t$.

If the resulting value of $r$ is a natural number, then I have a solution. If not, I just try with a different number.

I have been able to get some solutions with this method (such as $t = 21, r =15$ and $t = 120, r =85$) but it's very repetitive and tedious for larger values.

Is there any smarter way to get integer solutions for this equation?

Answer 1

Let $Y=2t-1$ and $X=2r-1$.

Then $Y^2-2X^2=-1$.

This is the negative Pell equation $y^2-nx^2=-1$ with $n=2$.

Solutions are $Y=$$1, 7, 41, 239,...$

and $X=$$1,5,29,169,...$,

so $t=0,4,21,120,...$

and $r=0,3,15,85,....$

[Click on the numbers to see more solutions from The On-Line Encyclopedia of Integer Sequences.]

Answer 2

If you consider the equation $$\begin{aligned} 2(r^2-r)&=t^2-t \\ \implies 2r^2-2r-(t^2-t)&=0 \end{aligned}$$ as a quadratic in $r$, then it's roots are $$r=\dfrac{1\pm \sqrt{2t^2-2t+1}}2$$ so, for $r$ to be a natural number, we take the "$+$" sign in the form for the roots, and note the term under the root-sign has to be a perfect square (actually for $r$ to be integer, it has to be an odd perfect square, but the term $2t^2-2t+1$ is already odd for $t\in \Bbb{Z}$, so it's enough to look for perfect squares) from where we get $$\begin{aligned} 2t^2-2t+1&=z^2 \\ \implies 4t^2-4t+2&=2z^2 \\ \implies (2t-1)^2-2z^2&=-1\end{aligned}$$

and as mentioned by J.W.Tanner in the comment, you have to resort to Pell's equation, because the last line is in the very form of the Pell's equation.