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in Euclid for infinite primes we define $p_0= 2, p_n = p_0p_1...p_{n-1}+1$ that way we get a series so that for every $n \in \mathbb{N}$ we get that $p_n$ is divided by a prime not one of $ p_0,...,p_{n-1} $.

my question is: does for every $n \in \mathbb{N}$ we get that $ p_n $ is a prime itself ?

i will be grateful if someone will give me a counterexample or a proof for this statement.

Leyla Alkan
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augustus yostefanus f' the iii
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  • Trial and error should quickly get you a counterexample. – lulu Mar 21 '18 at 18:21
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    No. See what happens at (I think) $11$. – Ethan Bolker Mar 21 '18 at 18:21
  • i dont think i can reach that level of calculation – augustus yostefanus f' the iii Mar 21 '18 at 18:24
  • To be clear: this is NOT the usual construction found in Euclid. Euclid takes the first $n$ primes and multiplies them together, adding $1$ to the product. That is not what you have written. Yours becomes composite almost immediately. Unless I botched it, your $p_5=1807=13\times 139$ . But maybe you defined $p_n$ incorrectly? – lulu Mar 21 '18 at 18:26
  • @lulu I think you mean $p_4=1807$ because the sequence starts with $p_0=2.$ – bof Mar 21 '18 at 18:28
  • @bof Yes, you are correct. Doesn't change the counterexample though. – lulu Mar 21 '18 at 18:29
  • You wrote $p_n=\dots$ where you really want $p_n$ to be the $n$th prime. – saulspatz Mar 21 '18 at 18:30
  • Your assertion "$p_n$ is divided by a prime not one of $p_0,\dots,p_{n-1}$" would also be true if you simply defined $p_n=10^{n+1}.$ I think what you wanted to say was "$p_n$ is divided by a prime which does not divide any of $p_0,\dots,p_{n-1}.$" – bof Mar 21 '18 at 18:34
  • You've added an original twist and Euclid's construction... but no matter. More importantly 2) Euclid's proof doesn't prove that $K=p_1p_2...p_n + 1$ is prime. It proves that none of $p_i$ are prime factors of $K$. And that in turn proves that as $K$ must have prime factors, there must be other primes than just $p_i$. And as that can happen for any finite group of $p_i$ there must be an infinite number of primes.
    • – fleablood Mar 21 '18 at 18:43
  • we define $p0=2,p_n=p_0p_1...p_{n−1}+1$. Actually, we never did any such thing. We often (although Euclid did not) say let $p_0 =2; p_1 = 3, p_2 = 5$....so that $p_i =$ the $i$th prime and let $N = p_0...p_n + 1$. NO-ONE ever claimed that $N$ was the $n+1$th prime. – fleablood Mar 21 '18 at 18:46