$n^3 - 1 = (n-1)(n^2 + n + 1)$ $n^2 + n + 1 > 1$ for any postive integer $n$ Thus $(n-1) = 1 \Rightarrow n = 2$ Thus only solution is $n = 2$

So the prime would be 7, right?

Is thua really an Olympiad question?!

I have seen a harder AIME #4. As a generalization, if the question asked for $x^n-1$ being a prime, then the only possible solution would be $x=2$. THis may not be a solution, but there are no other solutions possible.

You mean $x=2$?

mathwiz0803 wrote: I have seen a harder AIME #4. As a generalization, if the question asked for $x^n-1$ being a prime, then the only possible solution would be $x=2$. THis may not be a solution, but there are no other solutions possible. Obviously, since that expression is divisible by $x-1$ hence it must be $1$

Pandasareamazing. wrote: So the prime would be 7, right? Yes the prime is $7$.

$n^3-1=(n-1)(n^2+n+1)$ $n^2+n+1>1$ for all natural n Implying that $n-1=1 \Rightarrow n=2$ Which is the only solution $\blacksquare$