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When I noticed that $3^3+4^3+5^3=6^3$, I wondered if there are any other times where $(a-1)^3+a^3+(a+1)^3$ equals another cube. That expression simplifies to $3a(a^2+2)$ and I'm still trying to find another value of $a$ that satisfies the condition (the only one found is $a=4$)

Is this impossible? (It doesn't happen for $3 \leq a \leq 10000$) Is it possible to prove?

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    I think you mean $a=4$, not $a=2$: $1^3 + 2^3 + 3^3 = 36$ is not a cube, but $3^3 + 4^3 + 5^3 = 6^3$2012-03-14
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    @RobertIsrael Yup, I changed that :)2012-03-14
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    Have you read [this](http://mathworld.wolfram.com/CubicNumber.html) article? It seems that that is the only three consecutive integers whose cubes sum to a cube.2012-03-14
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    It is also known as a Plato's number http://en.wikipedia.org/wiki/Plato%27s_number2012-03-14
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    @Jack is it difficult to prove? I think I've read that but still tried to find one.2012-03-14
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    If you notice, $\gcd(a,a^2+2)=\gcd(a,2)$, so the only way to get $3a(a^2+2)=b^3$ is for $a=2^m\cdot3^n$ and $a^2+2=2^h\cdot3^k\cdot c^3$ with $2,3\not\mid c$ and $m+h\equiv0\equiv1+n+k\pmod3$. Perhaps you can rule out some cases (like e.g. m=0).2012-03-14
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    @bgins Why can't $a$ itself also have other cube factors?2012-03-14
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    @Jack: I saw that article before but it doesn't seem to cite the reference.2012-03-15
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    @SivaramAmbikasaran: Indeed, I didn't find the reference either. I only found it in [this](http://books.google.com/books?id=aFDWuZZslUUC&pg=PA616&lpg=PA616&dq=%22the+only+three+consecutive+integers+whose+cubes+sum+to+a+cube%22&source=bl&ots=4pbPuvEggM&sig=aWEERAG9l_mhrhk6Uyl-i5UouqQ&hl=en&sa=X&ei=tk1hT5qrMJHmgge2t4yfAg&ved=0CC4Q6AEwAQ#v=onepage&q=%22the%20only%20three%20consecutive%20integers%20whose%20cubes%20sum%20to%20a%20cube%22&f=false) book.2012-03-15
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    @Jack Thanks. Quite interesting. I am not sure if the proof is too trivial that there is no citation.2012-03-15
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    @Arthur: You're right, it can. However, I think if it does, we can "without loss of generality" throw them away and look at the (smaller) case presented.2012-03-15
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    @bgins No, I don't think we can, since introducing cube factors to $a$ means $(a^2 + 2)$ can obtain a lot more values.2012-03-15
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    @Arthur, thanks. I addressed that in my partial answer below.2012-03-15

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