Square Triangular Numbers

The number $N$ is square are triangular provided that $N = \tfrac12n(n+1) = m^2$ for some positive integers $n,m$ . Thus we require $(2n+1)^2 - 2(2m)^2 \; = \; 1$ The positive integer solutions $x,y$ of Pell's equation $x^2 - 2y^2 = 1$ are given by the formula $x_k + y_k\sqrt{2} \; =\; (1 + \sqrt{2})^{2k} \; = \; (3 + 2\sqrt{2})^k \hspace{2cm} k \in \mathbb{N}$ Since $x_1 = 3$ is odd, $y_1 = 2$ is even, and since $x_{k+1} \; = \; 3x_k + 4y_k \hspace{2cm} y_{k+1} \; = \; 2x_k + 3y_k$ we deduce by induction that $x_k$ is always odd and $y_k$ always even. Now $(3 + 2\sqrt{2})^k \; =\; x_k + y_k\sqrt{2} \hspace{2cm} (3 - 2\sqrt{2})^k \; = \; x_k - y_k\sqrt{2}$ so that $y_k \; = \; \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{2\sqrt{2}}$ and hence $N_k \; = \; \tfrac14y_k^2 \; = \; \left( \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{4\sqrt{2}} \right)^2$ is a square and triangular number for any positive integer $k$ .

$N_{k}=\left({\frac {(3+2{\sqrt {2}})^{k}-(3-2{\sqrt {2}})^{k}}{4{\sqrt {2}}}}\right)^{2}.$ This formula gives each square triangular number and as it can take any $k$ , there must be an infinite amount