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How can I find integer solutions for $x^3 - y^2 = 0 $ ?
In case that there are infinite number of solutions .How can we prove that ? and how to generate first few solutions ?

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2 Answers

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Edited to match the corrected question.

HINT: If $x^3=y^2$, and $x$ is an integer, then $x$ must be a perfect square. (Why?) Thus, $x^3$ must be a sixth power. Conversely, if $x^3$ is a sixth power, it’s a perfect square, and you get a solution.

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oh sorry I meant $x^3 - y^2 = 0 $ , I edited the post – Loers Antario Sep 18 '12 at 22:48
@Loers: I suspected that you did and was in the process of addressing that possibility when you made the edit. – Brian M. Scott Sep 18 '12 at 22:50
aha , thanks , but what if y^2 was a quadratic equation in its standard form $ay^2+by+c$ , what can we do then ? – Loers Antario Sep 18 '12 at 23:02
@Loers: Cry. :-) I don’t offhand know of a general technique for such equations, but it’s not my field at all, so there may be one. – Brian M. Scott Sep 18 '12 at 23:07
up vote 3 down vote

The highest power of any prime dividing $x^3$ must be divisible by 3 and the highest power of any prime dividing $y^2$ must be divisible by 2. So the highest power of any prime dividing $x^3=y^2$ must be divisible by 6. This implies that $x^3=y^2$ can be written as $u^6$ for some integer $u$. It then immediately follows that $x=u^2$ and $y=u^3$ and conversely each such pair $x,y$ is indeed a solution to that equation, so that is exactly the solution set.

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