How many prime?

$\frac {n^3-1}{5}$ is an integer $\implies n^3 \equiv 1 \pmod{5} \implies n \equiv 1 \pmod{5}$

Hence $n=5k+1$ which $\implies \frac {n^3-1}{5}=\frac {125k^3+75k^2+15k}{5}=k(25k^2+15k+3)$

So must $k=1 \implies n=6$ since $\frac {n^3-1}{5}$ is a prime.

Verifying if $n=6$ is a possible value: $\frac {6^3-1}{5}=43$ which is a prime

so there is only one solution hence the answer is 1

$n^3-1=(n-1)(n^2+n+1)=5p$ . By Euclidean algorithm:

$\gcd(n-1,n^2+n+1)=\gcd(n-1,(n-1)(n+2)+3)\in\{1,3\}$

If $\gcd=3$ , then $9\mid 5p$ , impossible. So $\gcd=1$ and four cases:

• $n-1=1, n^2+n+1=5p$ , impossible ( $n=2$ and $n^2+n+1=7\neq 5p$ ).
• $n-1=5, n^2+n+1=p$ gives $(n,p)=(6,43)$ .
• $n-1=p, n^2+n+1=5$ , impossible, since $n^2+n+1\neq 5$ .
• $n-1=5p, n^2+n+1=1$ , impossible, since $n^2+n+1>1$ .

We can't have them both negative, e.g. $n-1=-1, n^2+n+1=-5p$ , since $n^2+n+1>0$ .