Find all triplets of natural numbers $n, p$ and $q$ such that $n^{p}+n^{q}=n^{2010}$.
Let $(n, p, q)$ be a triplet that satisfies the equation. First, we check that $n$ cannot take the values 0 and 1. Therefore, we assume $n \geq 2$. Necessarily, $p$ and $q$ are less than 2009, and thus $n^{p} \leq \frac{n^{2010}}{2}$, the same for $n^{q}$, and the equation is only satisfied in the case of equality, which is when $n=2$, and $p=q=2009$.