$ \cos(\pi \sqrt {n^2 + n}) \geq 0$ $ \frac {5\pi}{2} \geq \pi \sqrt {n^2 + n} \geq \frac {3\pi}{2}$ $ \frac {5}{2} \geq \sqrt {n^2 + n} \geq \frac {3}{2}$ $ \frac {25}{4} \geq n^2 + n \geq \frac {9}{4}$ $ n^2 + n \geq \frac {9}{4}$ $ (n - \frac { - 1 \pm \sqrt {10}}{2}) \geq 0$ $ n \geq \frac { - 1 \pm \sqrt {10}}{2}$ $ \frac {25}{4} \geq n^2 + n$ $ 0 \geq (n - \left(\frac { - 1 \pm \sqrt {26}}{2}\right))$ $ n \leq \frac { - 1 \pm \sqrt {26}}{2}$ $ \frac{-1 \pm \sqrt{26}}{2} \geq n \geq \frac{-1 \pm \sqrt{10}}{2}$

Note that $ n < \sqrt{n^2+n} < n+\dfrac{1}{2}$. For all $ x \in \left( n , n+\frac{1}{2} \right)$, we have: $ \cos(\pi x) > 0$ iff $ n \in \mathbb{N}$ is even. $ \cos(\pi x) < 0$ iff $ n \in \mathbb{N}$ is odd. Therefore, the only positive integers such that $ \cos (\pi\sqrt{n^{2}+n})\ge 0$ are the even integers.