Competition problem — Number theory

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Number theory Difficulty 5.0 AIME Prove it United States

Problem:

Are there natural numbers $a$ and $b$ that make the equation
$2 a^{2} + 1 = 4 b^{2}$
true? (The natural numbers are the counting numbers, $1, 2, 3, \ldots$ )

Solution

Solution:

The answer is no. Suppose that $a$ and $b$ are any natural numbers. Then $2 a^{2}$ is even, and so is $4 b^{2}$ . But an even number plus $1$ will be odd, not even, so the equation can never be made to work, i.e. there are no natural numbers that satisfy the equation.

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