Competition problem — Number theory

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Number theory Difficulty 4.8 AIME Prove it United States

Problem:
Find all ordered triples $(a, b, c)$ of positive integers with $a^{2}+b^{2}=4c+3$ .

Solution

Solution:
There are no such ordered triples. Since the right side is odd, one of $a$ and $b$ is odd and the other even. But the square of any even number is a multiple of $4$ , and the square of any odd number has a remainder of $1$ when divided by $4$ . But the right side leaves a remainder of $3$ when divided by $4$ , a contradiction. Thus, this is impossible.

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