Competition problem — Number theory

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Number theory Difficulty 4.8 AIME Prove it Soviet Union

Problem:

Find unequal integers $m$ , $n$ such that $mn + n$ and $mn + m$ are both squares. Can you find such integers between $988$ and $1991$ ?

Solution

Solution:

Answer no

For example, $49 = 7^2$ , $50 = 2 \cdot 5^2$ , $8 = 2 \cdot 2^2$ , $9 = 3^2$ , so $49 \cdot 8 + 8 = 20^2$ , $49 \cdot 8 + 49 = 21^2$ .

WLOG $m < n$ . Then $mn + m = (m + h)^2$ , $mn + n = (m + k)^2$ , with $k > h$ . So $n - m = (m + k)^2 - (m + h)^2 = (k - h)(2m + k + h) > 2m$ , so $n > 3m$ . Hence we cannot have $m$ and $n$ between $988$ and $1991$ .

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