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dark_magician_92
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By solve I mean, post the answer. cool.gif I know that won't gain me any reputation, but this is so easy for me I don't want to spoil it for anyone else. That's what I get for being a programmer lol.

Edited by Darth Legion
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We remember from 7th grade algebra that a^2 –b^2 = (a+b) (a-b) and

a^3 –b^3 = (a^2+b^2+ab) (a-b)

Looking at the equation, we seek a number n to subtract from both sides whereby

Y^2 – n = x^3 – (432 + n) makes the left side a difference of squares, and the right side a difference of cubes.

By experimentation, starting with the first cube greater than 432 (8^3 = 512), we observe the following:

8^3 = 512; 512-432 = 80. 80 is not a perfect square, so we try 9 and so forth…

12^3 = 1728. 1728 – 432 = 1296, which is 36^2. So we have

y^2 - 36^2 = (x^3 - 12^3)

and

(y + 36) (y – 36) = (x^2 +12 x + 144) (x-12)

Obvious integers which make this work are at the zeroes: y = +-36 and x = 12. I don’t think there are others, but I haven’t proved it yet.

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