Find all positive integers $n$ such that $n+1$ divides $2n^2 + 5n$. Be sure to verify that the $n$ found are solutions.
- We factorize: $n+1$ divides $2n^2 + 5n = n(2n + 5)$. Since $\text{GCD}(n+1, n) = \text{GCD}(n+1-n, n) = \text{GCD}(1, n) = 1$, by Gauss's lemma, since $n+1$ and $n$ are coprime, $n+1$ divides $2n + 5 = 2(n+1) + 3$, so $n+1$ divides 3. Therefore, $n+1 = 1$ or 3, so $n = 0$ or 2.

Conversely, $n=0$ works because 1 divides 0, and $n=2$ works because 3 divides $2 \times 4 + 5 \times 2 = 18$.

- We use the fact that $n+1$ divides $(n+1)(n-1) = n^2 - 1$. In particular, $n+1$ divides $2n^2 + 5n = 2(n^2 - 1) + 2 + 5n$, so $n+1$ divides $5n + 2$. Thus, $n+1$ divides $5(n+1) - 5 + 2$, so $n+1$ divides -3, so $n+1$ divides 3. Therefore, $n+1 = 1$ or 3, so $n = 0$ or 2.

Conversely, $n=0$ works because 1 divides 0, and $n=2$ works because 3 divides $2 \times 4 + 5 \times 2 = 18$.