I tried brute force and got $n=3$ as a possible solution. But have been unable to find a method to get a generalized answer, if one exists. Please point me towards the right direction.
The number of positive integers $n$ for which $n^{3}+(n+1)^{3}+(n+2)^{3}=(n+3)^{3}$
2
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elementary-number-theory
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1If you expand things out and collect terms, you'll have a cubic in $n$. One of its roots is $n=3$, which means the cubic has $n-3$ as a factor. Factoring this out you'll be left with a quadratic in $n$. If there are any other integer solutions, the quadratic will factor. Happy hunting! – 2017-02-28
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1Why can't you just expand all of the terms and collect them all together to get $n^3-6n-9=0$? Then, knowing that one of the roots is 3, you can divide by $n-3$ and get $n^2+3n+3$ which doesn't have integer solutions. – 2017-02-28
2 Answers
6
If you collect terms, you get $$ 2 (x - 3) (x^2 + 3 x + 3)=0. $$ The only positive root is $3$, as the quadratic part will always be greater than zero when $x>0$.
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0Phew. Thanks. Frankly, I wasn't too keen on expanding it and fumbling around. Didn't know the expansion would turn out to be reducible. Thanks! – 2017-02-28
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0[I wasn't keen either :)](http://www.wolframalpha.com/input/?i=simplify+(x%2B3)%5E3-x%5E3-(x%2B1)%5E3-(x%2B2)%5E3) – 2017-02-28
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If you expand this equation, you have a polynomial equation of degree $3$. You have already identified one solution, so you can factor out $(n-3)$ and use the quadratic formula to identify the other two roots. If they are positive integers, you have more solutions. If not, then the only solution is the one you found already.