Solution to puzzle 61 (One cube?)

We have 18(n² + 3) = (n + 3)³ − (n − 3)³.

By Fermat's Last Theorem, x^m + y^m = z^m has no non-zero integer solutions for x, y and z, when m > 2.  Here we need the result only for the case m = 3, which was first proved by Euler, with a gap filled by Legendre.

Hence the only values of n for which 18(n² + 3) is a perfect cube, are where n + 3 = 0 or n − 3 = 0; that is, n = ±3.

Source: Original

Back to top