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Find all the positive integers $n$ such that $n^3-n$ is a perfect cube.

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closed as off-topic by T. Bongers, USER91500, Cookie, Claude Leibovici, Ted Jun 29 '14 at 7:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, USER91500, Cookie, Claude Leibovici, Ted
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If $n > 1$, then $3n^2-4n+1 = (3n-1)(n-1) > 0$.

Hence, $(n-1)^3 = n^3-3n^2+3n-1 < n^3-n < n^3$.

Since $n^3-n$ is strictly between two consecutive perfect cubes, $n^3-n$ is not a perfect cube.

The only remaining positive integer is $n = 1$, which yields $n^3-n = 0 = 0^3$.

Therefore, the only positive integer $n$ such that $n^3-n$ is a perfect cube is $n = 1$.

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Here's another way to do it. Let $n^3-n=k^3$. Then we have $$n=n^3-k^3=(n-k)(n^2+nk+k^2).$$ However, $n-k\ge1$ and $n^2>n$ whenever $n>1$. Hence $(n-k)(n^2+nk+k^2)>n$ for $n>1$, which contradicts the above equality. Finally, checking that $n=1$ works gives us our only solution.

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Or think of the fact that $$n^3-n=n(n^2-1)=(n-1)n(n+1)$$, is the product of three consecutive numbers. But any two consecutive numbers are relatively-prime, so the center number must be a perfect cube itself, say $n=m^3$. Substituting back in the top expression, we have a product $$(m^3-1)m^3(m^3+1)=m^3(m^6-1)=m^3((m^2)^3-1) $$ .

But $m^3((m^2)^3-1)$ cannot be a perfect cube, because $m^3$ is a perfect cube, but $(m^2)^3-1$ is not.

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