Find all primes $p$ for which $19 p - 1$ is a perfect cube

Question

I set the equation to $19 p - 1 = n^3$, then got $19 p =(n+1)(n^2-n+1)$. I don't know what to do now.

Answer 1

$$ 19p = (n+1)(n^2 - n + 1)$$

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

Hence we have the following cases:

1. If $gcd(n+1, n^2 - n + 1) = 3$, there are no solutions since we have $3^2$ on the RHS

2. If $gcd(n+1, n^2 - n + 1) = 1$,

   2a.$19 = n+1 \implies p = n^2-n+1 = 307$

   2b. $19 = n^2 - n + 1 \implies n^2-n-18 = 0 \implies n=\frac{1\pm\sqrt{73}}{2}$

Hence $\boxed{p = 307}$ (case 2a) is the only solution. $19*307 - 1 = 18^3$

Answer 2

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

$\implies(n+1,n^2-n+1)$ must divide $3$

If $3$ divides $n+1,3^2$ will divide $19p$ which is impossible $\implies p\ne19$

Now $n+1$ must divide $19p$ so, is one of $1,p,19,19p$

Check all four cases

Answer 3

Because $19$ is a prime, as long as $n+1,n^2-n+1 \gt 1,$ which is any time $n \gt 1,$ you need one of the factors on the right to be $19$ and one to be $p$. That is not many possibilities.