Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Ukraine

Find all pairs of prime numbers $(a, b)$ , such that $a^b = b^a + 1$ is prime.

Solution

Answer: $(2,3)$ , $(3,2)$ , $(2,2)$ .

Clearly, either $a$ or $b$ is even. WLOG, $a = 2$ . It is easy to see that $b = 2$ and $b = 3$ satisfy the condition. Suppose that $b > 3$ . We have: $2^b = b^2 + 1$ , $b > 3$ . It is easy to see that the last expression is divisible by $3$ and cannot be prime.

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