If $p>5$ is prime, $2p+1$ is a prime, $\frac{4p+1}{3}$ is prime, $8p+1$ is prime, Then $p \equiv 29 (mod \; 30)$ Assume that $p>5$ is prime, $2p+1$ is a prime, $\frac{4p+1}{3}$ is prime, $8p+1$ is prime. Then I want to prove that $p  \equiv 29 (mod \; 30).$
First of all I have to show that $4p+1$ is a multiple of 3 if $2p+1$ is prime and $p > 5.$
Any help will be appreciated.
Thank you so much.

We have $p\equiv \pm 1\pmod{3}$. But if $p\equiv 1\pmod{3}$ then $2p+1$ is divisible by $3$, so $p\equiv -1\pmod{3}$, and therefore $p\equiv 5\pmod{6}$.
Also, $p$ is congruent to one of $1,2,3,4$ modulo $5$. Since $2p+1$ is prime, we cannot have $p\equiv 2\pmod{5}$. Since $\frac{4p+1}{3}$ is prime, $p$ cannot be congruent to $1$ modulo $5$. And since $8p+1$ is prime, $p$ cannot be congruent to $3$ modulo $5$. So $p\equiv 4\pmod{5}$.
Since $p\equiv 5\pmod{6}$ and $p\equiv 4\pmod{5}$, we have $p\equiv 29\pmod{30}$.
