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📅 March 14, 2026 👤 scraface

Mar 14, 2026

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Try small values modulo 5 first:
Mod 5: \(n^3 \equiv 13 \equiv 3 \pmod{5}\)
Check cubes mod 5:
\(0^3 = 0\)
\(1^3 = 1\)
\(2^3 = 8 \equiv 3\)
\(3^3 = 27 \equiv 2\)
\(4^3 = 64 \equiv 4\)
So only \(n \equiv 2 \pmod{5}\) works.
Now lift to mod 25: Let \(n = 5k + 2\). Plug into \(n^3 \equiv 13 \pmo...

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