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.

  • $\begingroup$ All prime factors of $$\frac{Q^p-1}{Q-1}$$ are of the form $2kp+1$. $\endgroup$ – Daniel Fischer Jan 26 '20 at 18:54
  • $\begingroup$ @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. $\endgroup$ – PythonSage Jan 26 '20 at 18:58
  • $\begingroup$ Let $r$ be a prime factor of $\frac{Q^p-1}{Q-1}$. What is the order of $Q$ modulo $r$? $\endgroup$ – Daniel Fischer Jan 26 '20 at 19:00
  • $\begingroup$ @DanielFischer I guess $p$ Since $r\mid 1+Q+Q^2\cdots Q^{p-1},\therefore Q(1+Q+Q^2\cdots Q^{p-2})\equiv -1(\mod r)\implies Q(1+Q+Q^2\cdots+Q^{p-1})-Q^p\equiv -1\mod r$. $\endgroup$ – PythonSage Jan 26 '20 at 19:10
  • $\begingroup$ 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 $Q$. $\endgroup$ – Daniel Fischer Jan 26 '20 at 19:20

Consider the arithmetic progressions


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.

  • $\begingroup$ Thanks. I forgot to mention I used Drichlett to solve it. But I am looking for a proof along the lines of what I have outlined in the question. $\endgroup$ – PythonSage Jan 26 '20 at 18:51

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