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Find all integer solutions of y^2 = x^3 − 432. Find all integer solutions of y^2 = x^3 − 432. @Mathematics
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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.
yay!! ur right :D
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