One Less Than A Perfect Cube

The answer is true, $7$ is the only prime number that is one less than a perfect cube. This is because, if we take $n^3$ to be the general formula for a perfect cube then as $a^3-b^3=(a-b)(a^2+b^2+ab)$ ;

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

Hence, we can see that it does not form a prime number. But, when we substitute $n=2$ ; $(2^3-1^3)=(2-1)(2^2+1+2)=\boxed 7$

Which is the only exception, because the other factor becomes $1$ .

Let $p$ be a prime which is one less than a perfect cube.Then $p=n^3-1$ for some $n\in \mathbb{Z^+}$ .Factorizing,we get: $p=(n-1)(n^2+n+1)$ Now, as $p$ is prime, one of the factors must equal $1$ .It is easily seen that as $n\geq 1$ , so $n^2+n+1>1$ .Hence the only choice is that $n-1=1\implies n=2$ .This gives $p=7$ .So yes, 7 is the only prime one less than a cube.