Question

Find the prime of the form $n^3-1$ for natural number 'n'.

Answer 1

$n^3-1=(n-1)(n^2+n+1)$

$n-1$ be equal to $1$ or $n$

For $n-1$ equal to $1$, $n=2$ and prime is $2^3-1=7$

For $(n^2+n+1)$ be $1$ or $n$

$(n^2+n+1)=1$

$n=0$ or $-1$.

And $(n^2+n+1)=n$

$n=$ No solution.

So only solution or prime of form $n^3-1$ is $7$.