# Infinite primes of the form $2kp+1$, $p$ is an odd prime and $k\in\mathbb{N}$

How do I prove that there are infinitely many primes of the form $$2kp+1$$ such that $$p$$ is an odd prime and $$k\in\mathbb{N}$$. The hint in the book I am using suggests considering the number $$(2q_1q_2\cdots q_r)^p-1$$

I assumed a finite number of primes of the form $$2k+1,$$ namely $$q_1,q_2,\cdots q_r$$. We can see that the number $$n={\underbrace{(2q_1q_2\cdots q_r)}_Q}^p-1\equiv1(\mod p)$$.

Hence, $$n$$ is of the form $$2kp+1$$ such that $$q_1,q_2\cdots q_r\nmid n$$. $$n$$ can be factorized as $$(Q-1)(1+Q+Q^2+Q^3\cdots+Q^{p-1})$$. Now, both $$(Q-1), (1+Q+Q^2+\cdots Q^{p-1})$$ are of the form $$2kp+1$$. But how do I prove that one of them is either prime or has a new factor of the form $$2kp+1$$.

A similar question is asked here: Prove that there are infinitely many primes of form 2kp+1 where p is an odd prime. I could not find any useful solution out of it since it was too short.

• All prime factors of $$\frac{Q^p-1}{Q-1}$$ are of the form $2kp+1$. – Daniel Fischer Jan 26 '20 at 18:54
• @DanielFischer How? I might be missing something very elementary here so please feel free to point out. My understanding is this $\frac{Q^p-1}{Q-1}=1+Q+Q^2\cdots Q^{p-1}\equiv1(\mod p)$. Here's why I am getting a little confused? Could there be two factors of the above expression of the form $2kp-1$ so that eventually the remainder is $1\mod p$. Please comment. – PythonSage Jan 26 '20 at 18:58
• Let $r$ be a prime factor of $\frac{Q^p-1}{Q-1}$. What is the order of $Q$ modulo $r$? – Daniel Fischer Jan 26 '20 at 19:00
• @DanielFischer I guess $p$ Since $r\mid 1+Q+Q^2\cdots Q^{p-1},\therefore Q(1+Q+Q^2\cdots Q^{p-2})\equiv -1(\mod r)\implies Q(1+Q+Q^2\cdots+Q^{p-1})-Q^p\equiv -1\mod r$. – PythonSage Jan 26 '20 at 19:10
• Yes, the order is $p$. Since $Q^p \equiv 1 \pmod{r}$ it can only be either $1$ or $p$. Thus it remains to see that it isn't $1$, i.e. $Q \not\equiv 1 \pmod{r}$. Now $\gcd\bigl(Q-1,\frac{Q^p-1}{Q-1}\bigr) = \gcd(Q-1,p)$, and I've overlooked that you don't have $p$ in the list of factors of $Q$, so it might just happen that $p$ also divides $\frac{Q^p-1}{Q-1}$. Well. Then one would have to show that that's not a power of $p$. It's simpler to add $p$ to the factors of $Q$. – Daniel Fischer Jan 26 '20 at 19:20

$$1,2p+1,2(2p)+1,3(2p)+1,\cdots$$
Since $$1$$ is relatively prime to $$2p$$, by Dirichlet's theorem on arithmetic progressions we know this sequence has infinitely many prime numbers in it.