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Algebra Difficulty 4.8 AIME Prove it Hong Kong

Let nn be a positive integer. Is it possible to express n2+3n+3n^2 + 3n + 3 into the form abab with aa and bb being positive integers, and such that the difference between aa and bb is smaller than 2n+12\sqrt{n} + 1?

Solution

No, it is not possible. Indeed, if n2+3n+3=abn^2 + 3n + 3 = ab, then
(a+b)2=(a−b)2+4ab≥4ab=4n2+12n+12>4n2+12n+9=(2n+3)2. (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≥(2n+4)2(a+b)^2 \ge (2n+4)^2. It follows that
(a−b)2=(a+b)2−4ab≥(2n+4)2−(4n2+12n+12)=4n+4=4(n+1). (a-b)^2 = (a+b)^2 - 4ab \ge (2n+4)^2 - (4n^2 + 12n + 12) = 4n + 4 = 4(n+1).
Therefore, ∣a−b∣≥2n+1|a-b| \ge 2\sqrt{n+1}.

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