262. Find a triangular number whose square is also a triangular number.
262. From the equation $\left[\frac{x(x+1)}{2}\right]^{2}=\frac{y(y+1)}{2}$, we find:

$$
y=\frac{-1+\sqrt{2 x^{2}(x+1)^{2}+1}}{2}
$$

Let $x(x+1)=z$, then $2 z^{2}+1=t^{2}, t^{2}-2 z^{2}=1$.

Solving this Pell's equation, we get an infinite set of values $z=2,12,70,408 \ldots$. From the equation $x(x+1)=2$, we have $x=1$, the corresponding triangular number is 1. From the equation $x(x+1)=12$, we find $x=3$, the triangular number will be 6. The values $z=70$ and $z=408$ do not lead to triangular numbers.

60

It has been proven that, apart from 1 and 6, there are no triangular numbers that satisfy the requirement of the problem.