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Solution to puzzle 78: Perfect square

Find all integer solutions of y2 = x3 − 432.


Note that x3 = y2 + 432 is a perfect cube if and only if 63(y2 + 432) = 216(y2 + 432) is a perfect cube.
But 216(y2 + 432) = (y + 36)3 − (y − 36)3.
Hence (6x)3 + (y − 36)3 = (y + 36)3(1)

By Fermat's Last Theorem, an + bn = cn has no non-zero integer solutions for a, b and c, when n > 2.  Here we need the result only for the case n = 3, which was first proved by Euler, with a gap filled by Legendre.

However, x > 0.
Hence (1) can hold only when y − 36 = 0 or y + 36 = 0; that is, y = ±36, in which case 6x = 72.

Therefore the only solutions are x = 12, y = ±36.


Further reading

  1. Mordell Curve
  2. Louis Joel Mordell
  3. The Mathematics of Fermat's Last Theorem
  4. Diophantine Equation--3rd Powers

Source: Original

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