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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?

1 Answers 1

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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...2010-10-17
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    @mgamer: What is the gcd of $a-2$ and $a+2$?2010-10-18
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    The GCD of $a+2$ and $a-2$ is? Now each of them has to be a cube individually.2010-10-18
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    Can we conclude anything about gcd of $a-2$ and $a+2$? How come?2010-10-18
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    Well, the gcd will also divide their difference.2010-10-18
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    So gcd of $a-2$ and $a+2$ divides 4. So we've got something like this: $4xy = b^3$. Am I correct?2010-10-18
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    @mgamer: gcd divides 4, does not mean it is 4. It could be any of 1,2 or 4. Can it be 2?2010-10-18
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    Let's take a = 25 for instance. Then a-2 equals to 23 and a+2 equals to 27. So we've got gcd(23,27) = 1. When on the other hand we take even a then a-2 and a+2 are also even. So (a-2)*(a+2) is divisible by 4. I must frankly tell you that I still don't know what I'm missing here...2010-10-18
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    @mgamer: The title of the question says that a is odd...2010-10-18
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    So the gcd((a-2),(a+2)) cannot be 2 nor 4.2010-10-18
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    I've come up with this: (a-2)(a+2) has at least two distinct prime factors while b^3 has exactly one distinct prime factor. This proves the equation is always false when a is odd. Is that correct?2010-10-18
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    @mgamer: No, it is not right. Why can't b^3 have more than one prime factors? For instance 15^3. Read Ross' comment of 15 hours ago.2010-10-19
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    @Moron, You're right. So (a-2) and (a+2) cannot be both perfect cubes because there are no perfect cubes which difference is equal to 4. It's still tricky for me to explain why (a-2) and (a+2) have to be perfect cubes individually. Is is that product of two coprimes is a perfect cube if each of that coprimes is a perfect cube?2010-10-19
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    @mgamer: If a prime power divides the product xy and x and y are co-prime. It has to divide either x or y. So if xy are co-prime and xy is a perfect cube _then_ (which is what you have to show. Not _if_ as you wrote, but that is true too) each x and y have to be perfect cubes.2010-10-19