# Diophantine solution set for $\frac{n(n-1)}2 = b(b-1)$

By Diophantine solution set I mean solutions where n and b are integers.

I have one solution I found by trial and error but would like to find out how to generate them.

One solution is n=21, b=15.

Additionally I am only interested in solution where n > b and n and b are both positive.

Can be refactored = 1/2 = (b^2-b)/(n^2-n)

• Note that it is just $(2n-1)^2=2\cdot(2b-1)^2-1$, a sort of Pell's equation. – punctured dusk Oct 30 '14 at 21:00

The equation reduces to $(2n-1)^2-2(2b-1)^2 = -1$ which is a generalized pell-equation. If you plug $x= 2n-1$ and $y = 2b-1$, the equation is equivalent to $$x^2-2y^2 = -1$$
Notice that $(x_0,y_0) = (7,5)$ is the minimal solution and the other solutions are obtained by the formula $$(x_n , y_n) = (3x_{n-1}+4y_{n-1} , 2x_{n-1} +3y_{n-1})$$ Now just notice that by induction, all these solutions are odd, so we can plug $n$ and $b$ back in. By the theory op Pell's equations, these are the only solutions.