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Solution to puzzle 61 (One cube?)

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

By Fermat's Last Theorem, xm + ym = zm 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

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