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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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3 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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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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$$(u^2)^3 = u^6$$

$$(u^3)^2 = u^6$$

for any integer $u$:

set $x = u^3$

and $y = u^2$

$$x^2 = (u^3)^2 = u^6 = (u^2)^3 = y^3$$

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