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All-Soviet-Union Mathematical Olympiad

Number theory Difficulty 4.8 AIME Prove it Soviet Union

Problem:

Find unequal integers mm, nn such that mn+nmn + n and mn+mmn + m are both squares. Can you find such integers between 988988 and 19911991?

Solution

Solution:

Answer no

For example, 49=7249 = 7^2, 50=2⋅5250 = 2 \cdot 5^2, 8=2⋅228 = 2 \cdot 2^2, 9=329 = 3^2, so 49⋅8+8=20249 \cdot 8 + 8 = 20^2, 49⋅8+49=21249 \cdot 8 + 49 = 21^2.

WLOG m<nm < n. Then mn+m=(m+h)2mn + m = (m + h)^2, mn+n=(m+k)2mn + n = (m + k)^2, with k>hk > h. So n−m=(m+k)2−(m+h)2=(k−h)(2m+k+h)>2mn - m = (m + k)^2 - (m + h)^2 = (k - h)(2m + k + h) > 2m, so n>3mn > 3m. Hence we cannot have mm and nn between 988988 and 19911991.

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