Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$
$n^3-1=(n-1)(n^2+n+1)$. If $n^3-1$ is prime then $n-1=1$ or $n=2$ so $n^3-1=7$. But I need a concrete proof of it, not from intuition.
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user1942348
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7How is this not concrete? – Tobias Kildetoft Apr 29 '15 at 07:41
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3You could put in a bit more detail, as in if $n-1\gt 1$, then since $n^2+n+1\gt 1$, their product $n^3-1$ cannot be prime., – André Nicolas Apr 29 '15 at 07:46
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@AndréNicolas Thanks – user1942348 Apr 29 '15 at 07:51
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$$n^3-1=(n-1)(n^2+n+1)$$ Suppose that $n^2+n+1\le n-1\implies n^2\le -2$ which leads to contradiction.
Therefore $n^2+n+1\gt n-1$ and since $n^3-1$ is prime only possibility is $$n-1=1\implies \color{Green}{n=2}\,\,\,\,\,\text{and} \,\,\,\,\,\,\color{Red}{n^3-1=7}$$
Bumblebee
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