XI OBM

Suppose $\frac{n(n+1)}{3}=m^2$, so $n(n+1)=3m^2$. $n$ and $n+1$ are coprime so either

(1) $n=3a^2$ and $n+1=b^2$, or

(2) $n=a^2$ and $n+1=3b^2$.

But (2) implies that $3b^2-1$ is a square, whereas squares must be $0$ or $1$ mod $3$. So we must have (1).