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Hong Kong competition problems

Number theory Difficulty 5.4 AIME, harder Prove it Hong Kong

Show that the equation $y^{37} \equiv x^3 + 11 \pmod{p}$ is solvable for every prime $p$ , where $p \le 100$ .

Solution

Since $p \le 100$ , we have $(p - 1, 37) = 1$ (by checking $1, 38, 75$ are not primes). Thus, there exists $k \in \mathbb{Z}^+$ such that $37k \equiv 1 \pmod{p-1}$ . Thus, by the Fermat little theorem, we have
$(11^k)^{37} = 11^{37k} \equiv 11 \pmod{p}.$
Therefore, $(x, y) = (0, 11^k)$ is a solution to the given system.

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