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Let $p_1,p_2,p_3,\cdots$ be all the primes sorted in an increasing order.

Is $p_1p_2p_3\cdots p_i + 1$ is always prime? Why? How can I prove that?

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    $\begingroup$ Try proving that it is not always a prime. That has a higher chance of success. $\endgroup$ – Daniel Fischer Mar 20 '14 at 22:16
  • $\begingroup$ let me try. Ty @DanielFischer $\endgroup$ – SonicFancy Mar 20 '14 at 22:17
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    $\begingroup$ $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1 = 59 \cdot 509$. $\endgroup$ – Eric Towers Mar 20 '14 at 22:18
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    $\begingroup$ +1 for actually asking this question; many people believe it without even realising... $\endgroup$ – preferred_anon Mar 20 '14 at 22:28
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    $\begingroup$ Prime numbers of the form $p_1p_2\cdot\cdot\cdot p_n+1$ are usually called Euclid primes or primorial primes (oeis.org/A014545) $\endgroup$ – Alessandro Codenotti Mar 20 '14 at 22:45
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$P_n=p_1p_2\cdots p_n+1$ can't be divisible by any of $p_1,p_2,\ldots,p_n$, and so $P_n$ must be prime if $P_n<p_{n+1}^2.$ So in particular 3 is prime since $3<3^2$, 7 is prime since $7<5^2,$ and 31 is prime since $31<7^2$. But the argument doesn't work for larger values, and in particular it fails for $13\#+1$, $17\#+1$, $19\#+1$, etc.

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