Romanian Mathematical Olympiad

a) Since $m^2 - m + 1 = (m - 1)^2 + (m - 1) + 1$ and $m - 1 \ge 0$, it follows that $m - 1 \in \mathbb{N}$ and consequently $m^2 - m + 1 \in \{n^2 + n + 1 \mid n \in \mathbb{N}\}$.

b) Write $p = k^2$, where $k$ is an integer. Since $p > 1$, we have $k \ge 2$. Now $p^2 + p + 1 = k^4 + k^2 + 1 = (k^2 + 1)^2 - k^2 = (k^2 - k + 1)(k^2 + k + 1)$. Numbers $r = k$ and $q = k - 1$, both positive integers, satisfy the claim.