The sum of all natural numbers $$n$$ for which $$n(n+1)$$ is a perfect square is

Since $$n$$ and $$n+1$$ are coprime, their product can be a square only when both are perfect squares. The only consecutive perfect squares are $$0$$ and $$1$$, which would give $$n=0$$. As the paper uses natural numbers beginning with $$1$$, there is no valid value of $$n$$, so the required sum is $$0$$.