Competition problem — Number theory

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Number theory Difficulty 4.8 AIME Prove it Soviet Union

Problem:
Can both $x^2 + y$ and $x + y^2$ be squares for $x$ and $y$ natural numbers?

Solution

Solution:
No. The smallest square greater than $x^2$ is $(x + 1)^2$ , so we must have $y > 2x$ . Similarly $x > 2y$ . Contradiction.

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