Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

Let $p(m, n)$ be the polynomial $2 m^{2} - 6 m n + 5 n^{2}$ . The range of $p$ is the set of all integers $k$ such that $k = p(m, n)$ for some integers $m$ , $n$ . Find which members of $\{1,2, \ldots, 100\}$ are in the range of $p$ . Show that if $h$ and $k$ are in the range of $p$ , then so is $h k$ .

Solution

Solution:

We have $p(m, n) = (m - 2 n)^{2} + (m - n)^{2}$ , so $p(2 a - b, a - b) = a^{2} + b^{2}$ . Hence the range of $p$ is just the sums of two squares.

$\left(a^{2} + b^{2}\right)\left(c^{2} + d^{2}\right) = (a c - b d)^{2} + (a d + b c)^{2}$ , which establishes that if $h$ and $k$ are in the range, then so is $h k$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.