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So all even \(n\) satisfy \(n^3 \equiv 0 \pmod{8}\). So condition is \(n \equiv 0 \pmod{2}\), but weâll refine later.
Step 2: Solve modulo 125
n^3 \equiv 888 \pmod{125}
Note \(888 \mod 125 = 888 - 7\cdot125 = 888 - 875 = 13\), so:
n^3 \equiv 13 \pmod{125}
We search for a solution to \(n^3 \equiv 13 \pmod{125}\). Try lifting via Henselâs Lemma or trial.
Try small values modulo 5 first:
Mod 5: \(n^3 \equiv 13 \equiv 3 \pmod{5}\)
Check cubes mod 5:
\(0^3 = 0\)