Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Prove it Romania

a) Show that $m^2 - m + 1$ is an element of the set $\{n^2 + n + 1 \mid n \in \mathbb{N}\}$ , for any positive integer $m$ .

b) Let $p$ be a perfect square, $p > 1$ . Prove that there exist positive integers $r$ and $q$ such that $p^2 + p + 1 = (r^2 + r + 1)(q^2 + q + 1)$ .

Solution

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.

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