Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.1 National Olympiad Prove it Ukraine

Determine all positive integers $n$ such that number $N$ is also integer, where
$N = \frac{2 \cdot 1}{\sqrt{1^2 + 1 + 4 + \sqrt{1^2 - 1 + 4}}} + \frac{2 \cdot 2}{\sqrt{2^2 + 2 + 4 + \sqrt{2^2 - 2 + 4}}} + \dots + \frac{2n}{\sqrt{n^2 + n + 4 + \sqrt{n^2 - n + 4}}}$
(Anikushyn A., Rublyov B.)

Solution

Let us note that
$\frac{2k}{\sqrt{k^2 + k + 4} + \sqrt{k^2 - k + 4}} = \frac{(\sqrt{k^2 + k + 4} - \sqrt{k^2 - k + 4})2k}{(k^2 + k + 4) - (k^2 - k + 4)} = \\ = \sqrt{k^2 + k + 4} - \sqrt{k^2 - k + 4} = \sqrt{k^2 + k + 4} - \sqrt{(k-1)^2 + (k-1) + 4}.$
And thus
$N = (\sqrt{6} - \sqrt{4}) + (\sqrt{10} - \sqrt{6}) + (\sqrt{16} - \sqrt{10}) + \dots + (\sqrt{n^2 + n + 4} - \sqrt{(n-1)^2 + (n-1) + 4}) = \\ = \sqrt{n^2 + n + 4} - \sqrt{4} = \sqrt{n^2 + n + 4} - 2.$
It means that $N$ is integer if and only if $n^2 + n + 4$ is a perfect square.
Since $n^2 < n^2 + n + 4 < n^2 + 4n + 4 = (n + 2)^2$ , then for every positive integer $n$ we obtain $n^2 + n + 4 = (n + 1)^2 = n^2 + 2n + 1$ . And thus $n = 3$ .

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