Prove that there are infinitely many primes of form $2kp+1$ where $p$ is an odd prime

Suppose there are only finitely many primes of form $2kp+1$ : $$p_1,p_2,\cdots, p_r$$

I am trying to mimic euclid's infinte prime proof. But not getting a suitable number to consider. Any help ?

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    $\begingroup$ What type of numbers has only prime factors of the form $2kp+1$? $\endgroup$ – Daniel Fischer Sep 11 '14 at 20:02
  • $\begingroup$ I know odd prime factors of $2^p-1$ are of form $2kp+1$ $\endgroup$ – rrr Sep 11 '14 at 20:04

Let $f(x)=1+x+x^2+\cdots +x^{p-1}$. let $n=2p_1p_2\cdots p_r$. Then $f(n)$ must have a prime factor, $q$. Show that $q\equiv 1\pmod p$. Show that $q\neq p_i$ for any $i$.

  • $\begingroup$ Thanks you :) how is this different from considering the number $N = 2p_1p_2\cdots p_r + 1$ ? I see both numbers leave a remainder $1$ when divided by $p_i$; So I think $q \equiv 1 \pmod p$ follows immediately $\endgroup$ – rrr Sep 11 '14 at 20:08
  • $\begingroup$ It is not true for your $N$ that any factor $q$ has $q\equiv 1\pmod p$. The above essentially makes sure that is true. $\endgroup$ – Thomas Andrews Sep 11 '14 at 20:13
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    $\begingroup$ Just because $N\equiv 1\pmod {p_i}$ doesn't mean that if $q\mid N$ then $q\equiv 1\pmod p$. For example, if $r=1$, $p=5$ and $p_1=11$ then $2\cdot 11+1=23$ and so the only $q$ is $q=23$ and $23\not\equiv 1\pmod 5$. $\endgroup$ – Thomas Andrews Sep 11 '14 at 20:16

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