Competition problem — Number theory

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Number theory Difficulty 4.9 AIME Prove it Slovenia

Find all prime numbers $p$ such that $p^2 + 7^3$ is a perfect cube.

Solution

Let $p^2 + 7^3 = n^3$ . Then $p^2 = n^3 - 7^3 = (n-7)(n^2 + 7n + 49)$ . Obviously, $n > 7$ and $n - 7 < n^2 + 7n + 49$ . So there is only one possible case: $n - 7 = 1$ and $n^2 + 7n + 49 = p^2$ . We conclude that $n = 8$ and $p^2 = 169$ , so $p = 13$ . The only prime number with the required property is $p = 13$ .

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