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Problem 1098

AIME late

Number theory Difficulty 5.0 Prove it Bulgarian Mathematical Olympiad · Bulgaria

The positive integers $l, m, n$ are such that $m-n$ is a prime number and $8\left(l^{2}-m n\right)=2\left(m^{2}+n^{2}\right)+5(m+n) l$ . Prove that $11 l+3$ is a perfect square.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Setting $p = m - n$ and $q = m + n$ gives
$m n = \frac{1}{4}\left(q^{2} - p^{2}\right) \quad \text{and} \quad m^{2} + n^{2} = \frac{1}{2}\left(q^{2} + p^{2}\right)$
Hence the given conditions can be written as
$8 l^{2} - 2 q^{2} + 2 p^{2} = q^{2} + p^{2} + 5 q l$
i.e.
$p^{2} = (3 q + 8 l)(q - l)$
Since $p$ is a prime number and $3 q + 8 l > q - l$ we obtain $p^{2} = 3 q + 8 l$ and $1 = q - l$ . Hence $11 l + 3 = p^{2}$ .

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