Prime numbers $p$ such that $7p+1$ is a cube

Question

I am stuck on this assignment. I have to find every prime number $p$ such that $7p+1$ is a cube number. After exploring enough I must say there is no prime $p$ satisfying this condition. I have tried solving it by using mod $3$ but I can not go on. Hope someone might give me a hint.

Answer 1

Since $7 p + 1$ is a cubic, we may write $7 p + 1 = x^3$, then reduce each side by $7$ we have $x^3 \equiv 1 \pmod{7}$, which solves to $x \equiv 1, 2 \text{, or } 4 \pmod{7}$.

We can write $x = 7 k + (1, 2 \text{, or } 4)$. And $p = \dfrac{\bigl(7 k + (1, 2 \text{, or } 4)\bigr)^3 - 1}{7}$ should be a prime.

Which is impossible (see below).

---

Upd: Sorry I forgot to check the cases when $k = 0, 1$, because the factorization may be $1 \cdot p$ a prime. Thanks @IvanKaznacheyeu and @JiazeZhang !

When $k = 0$ and $x = 7 \cdot 0 + 2 = 2$, we have $\dfrac{x^3 - 1}{7} = 1$ not a prime.

When $k = 1$ and $x = 7 \cdot 1 + 1 = 8$, we have $\dfrac{x^3 - 1}{7} = 73$ which is a prime.

Therefore, the only solution is $p = 73$ and $7 \cdot 73 + 1 = 512 = 8^3$.

Answer 2

Modular arithmetic (as used by Pink Rabbit) is the long way to the answer. The comment by Ivan Kazanacheyeu shows the way to go.

We can rewrite $7p+1=a^3$ as $7p=a^3-1=(a-1)(a^2+a+1)$

We can show that $p\ne 2$ because $7\cdot 2+1$ is not a cube. We can show that $p\ne 7$ because $7\cdot 7+1$ is not a cube. Hence $p$ is an odd prime other than $7$.

The number $7p$ has two distinct factors, and the number $(a-1)(a^2+a+1)$ has two distinct factors. Hence either $7=a-1$ and $p=a^2+a+1$, or $p=a-1$ and $7=a^2+a+1$. There are no other possibilities.

The second case yields $a=2 \Rightarrow p=1$ which is not a prime. The first case yields $a=8 \Rightarrow p=73$, which is the solution found by Pink Rabbit.