Competition problem — Number theory

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Number theory Difficulty 5.0 AIME Prove it Thailand

Let $n$ be a positive integer for which $5n + 1$ is a perfect square. Show that $n + 1$ is a sum of 5 perfect squares.

Solution

Let $5n + 1 = m^2 \equiv 1 \pmod{5}$ . Thus $m = 5k \pm 1$ for some integer $k$ . We have
$n + 1 = \frac{(5k \pm 1)^2 + 4}{5} = 5k^2 \pm 2k + 1 = 4k^2 + (k \pm 1)^2$
which can be written as a sum of 5 perfect squares as desired.

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