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Are there any odd positive numbers that satisfy the equation:

$a^2 - b^3 = 4$ ?

I am certain that there are none but can't prove it. How would you prove that?

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If this is homework, please consider adding the homework tag. – Aryabhata Oct 16 '10 at 15:00
It's an elliptic curve :-) – Robin Chapman Oct 16 '10 at 15:09
@Robin: Are you saying it cannot be homework? :-) I believe there is an easy elementary solution to this. – Aryabhata Oct 16 '10 at 15:19
Robin just means that there are general techniques which can be brought to bear here, not that they are necessary. – Qiaochu Yuan Oct 16 '10 at 15:43
@Qioacho: Of course there are. I suppose that is what he meant and wasn't referring to my earlier comment about possible homework. Just wanted to clarify... – Aryabhata Oct 16 '10 at 16:01

1 Answer 1

Start by rewriting it as $a^2-4=b^3$, and do what comes naturally.

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So I have then $(a-2)(a+2) = b^3$. I still don't see what's next... – mgamer Oct 17 '10 at 15:39
@mgamer: What is the gcd of $a-2$ and $a+2$? – Aryabhata Oct 18 '10 at 13:31
The GCD of $a+2$ and $a-2$ is? Now each of them has to be a cube individually. – Ross Millikan Oct 18 '10 at 13:32
Can we conclude anything about gcd of $a-2$ and $a+2$? How come? – mgamer Oct 18 '10 at 13:56
Well, the gcd will also divide their difference. – Derek Jennings Oct 18 '10 at 14:07

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