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Hong Kong competition problems, 2021

Algebra Difficulty 4.8 AIME Prove it Hong Kong

Let $n$ be a positive integer. Is it possible to express $n^2 + 3n + 3$ into the form $ab$ with $a$ and $b$ being positive integers, and such that the difference between $a$ and $b$ is smaller than $2\sqrt{n} + 1$ ?

Solution

No, it is not possible. Indeed, if $n^2 + 3n + 3 = ab$ , then
$(a+b)^2 = (a-b)^2 + 4ab \ge 4ab = 4n^2 + 12n + 12 > 4n^2 + 12n + 9 = (2n+3)^2.$
Since both sides are squares, this yields $(a+b)^2 \ge (2n+4)^2$ . It follows that
$(a-b)^2 = (a+b)^2 - 4ab \ge (2n+4)^2 - (4n^2 + 12n + 12) = 4n + 4 = 4(n+1).$
Therefore, $|a-b| \ge 2\sqrt{n+1}$ .

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