Hi Luke,

Note that triangular numbers are of the form $\frac{n(n-1)}{2}$ (so actually depending on the parity of $n, \frac{n}{2} \times (n-1)$ or $n \times \frac{(n-1)}{2}).$ The second thing to note is that $n$ and $n-1$ are relatively prime, i.e., they have no common divisors beyond 1. So if a number is a perfect square and a triangular number what does that say about each of the factors $\frac{n}{2}$ and $(n-1)$ (or $n$ and $\frac{n-1}{2}$)? Once you see this you will narrow your search down and find the next one quickly.

Penny