Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
add comment

2 Answers

up vote 6 down vote accepted

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.