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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, le gâteau au fromage, Claude Leibovici, Ted Jun 29 at 7:28

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3 Answers 3

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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