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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?

YuiTo Cheng
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fabrizzio_gz
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2 Answers2

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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.]

J. W. Tanner
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  • Thank you very much. I was able to solve it with Pell's equation. To get the sequence of all possible solutions, this was also helpful: https://math.stackexchange.com/questions/531833/generating-all-solutions-for-a-negative-pell-equation Would you know why the coefficients to calculate $x_k, y_k$ were $3$ and $2$? I'm guessing they come from the fundamental solutions of the positive Pell equation $x² -2y²=1$ ($x=3, y=2$) – fabrizzio_gz Sep 04 '20 at 00:08
  • note that $(1+\sqrt2)^2=3+2\sqrt2$ – J. W. Tanner Sep 04 '20 at 00:15
  • I can see that's a solution to the $x_n+y_n\sqrt2=(1+\sqrt2)^(2n−1)$ equation, but I don't understand why we take those coefficients instead of those of the fundamental solution ($x=1,y=1$). I also don't understand why we take the square. – fabrizzio_gz Sep 04 '20 at 00:22
  • Note that the values for $Y$ obey $Y_{n+2} = 6 Y_{n+1} - Y_n$ beginning with (1, 7, ...) while we also have $X$ obey $X_{n+2} = 6 X_{n+1} - X_n$ beginning with (1, 5,..) – Will Jagy Sep 04 '20 at 00:26
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    @fabrizzio_gz: To get from one $(1+\sqrt2)^{2n-1}$ term to the next, we multiply by $(1+\sqrt2)^2$. If we multiplied only by $1+\sqrt2$ we would get solutions to $x^2-2y^2=1$, not just $x^2-2y^2=-1$ – J. W. Tanner Sep 04 '20 at 00:30
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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.

Fawkes4494d3
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