, 2011

Suppose that $n^2 = T_m = \frac{m(m+1)}{2}$. Multiplying by $8$,
$$8n^2 = 4m^2 + 4m = (2m + 1)^2 - 1 \Leftrightarrow (2m + 1)^2 - 8n^2 = 1.$$
Letting $A = 2m+1$ and $B = 2n$, this gives Pell's equation, $A^2 - 2B^2 = 1$. The first solution is $A = 3, B = 2$, and the general solution is given by $a_k + \sqrt{2}b_k = (3 + 2\sqrt{2})^k$. The solution is
$$a_k = \frac{(3 + 2\sqrt{2})^k + (3 - 2\sqrt{2})^k}{2} \quad \text{and} \quad b_k = \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{2\sqrt{2}}.$$
The squares themselves are $x^2 = \frac{1}{32}((3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k)^2$ for $k \in \mathbb{N}$.