Find all primes p and q such that $343p^3-q^3$ is also a prime

Question

Find all primes $p$ and $q$ such that $343p^3-q^3$ is also a prime.

I have no idea where to start. Does $343=7^3$ help?

Answer 1

Hint: Write $343p^3-q^3=(7p)^3-q^3$. Now use $a^3-b^3=(a-b)(a^2+ab+b^2)$.

Answer 2

I find a solution and prove it is the only one. Because of the "also" in the title we assume $p$ and $q$ primes.We consider only positive primes. $$343p^3-q^3=\text{ prime }\\(7p)^3-q^3=(7p-q)(49p^2+7pq+q^2)=\text{ prime }$$ This is possible only if $$(7p-q)=1\text{ and } 49p^2+7pq+q^2=\text{ prime }\\(7p-q)=\text{ prime }\text{ and } 49p^2+7pq+q^2=1$$

The second case is easily seen to be impossible hence we have the system $$\begin{cases}7p-q=1\\49p^2+7pq+q^2=\text{ prime }\end{cases}$$ Because of the particular solution $(p,q)=(1,6)$ the general solution of the first equation is $$\begin{cases}p=t+1\\q=7t+6\end{cases}$$ It follows we have to get prime solutions of the diophantine equation $$147t^2+273t+127=r\text{ prime }$$ with the restriction $t+1$ and $7t+6$ primes.

A solution is given for $t=1$. We have $$147+273+127=547$$ thus the solution $$\color{red}{(p,q,r)=(2,13,547)}$$ On the other hand, if $t$ is even then $q$ is even greater than $2$ and if $t$ is odd and greater than $1$ then $p$ is even greater than $2$. In both cases $p$ and $q$ cannot be both prime as required. Thus the given solution is the only one.