Find all integer solutions of y² = x³ - 432.
(In reply to Answer
by K Sengupta)
y^2 = x^3 - 432
or, y^2 + 432 = x^3
or, (y/6 + 6)^3 = (y/6 -6)^3 + x^3
Since y is an integer, (y/6 +/- 6) is always a rational number irrespective of whether y is divisible by 6 or not.
But, by Fermat's last Theorem, the sum of cubes of two nonzero rational numbers cannot be equal to the cube of a nonzero rational number.
Since each of x and (y/6 +/- 6) is a rational number, the equality in (*) is only possible, only when:
either, y/6 -6 = 0, giving y = 36, so that x= 12
or, y/6 + 6 = 0, giving y = -36, so that x= 12
or, x = 0, giving: y^2 = - 432 from the original equation, so that y is imaginary, which is a contradiction.
Consequently, (x, y) = (12, 36) and (12, -36) are the only possible solutions to the given problem.
Edited on June 10, 2008, 6:25 am