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Q.To show that the number $N=(P_1....P_n)+1$ is not always prime (where $P_1,...,P_n$)are the first n primes) , find an n for which $(P_1...P_n)+1$ is not prime

My attempt

since $P_1,...,P_n$are the first n primes so $P_1=2$

so product is even so N is odd number

$N=2k+1$

since odd number {1,3,5,7,9,....} is not always prime so N is not always prime

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  • $\begingroup$ This proof does not work. You can't get all odd numbers through this procedure. Just because you get an odd number doesn't mean that every odd number appears (you only get a subset). $\endgroup$
    – Michael Burr
    Commented Jan 10, 2017 at 15:48
  • $\begingroup$ Just find a specific example of when $1$ plus the product of the first $n$ primes isn't prime. It's true that this is "not always prime" but what you've written doesn't demonstrate this. $\endgroup$
    – pjs36
    Commented Jan 10, 2017 at 15:49
  • $\begingroup$ Just check the first few cases. You'll find it before very long. $\endgroup$
    – Michael Burr
    Commented Jan 10, 2017 at 15:50
  • $\begingroup$ Yes not all odd numbers are prime, but not all odd numbers are of your form either. Just try small $n$, you'll find a good example. $\endgroup$
    – lulu
    Commented Jan 10, 2017 at 15:50
  • 1
    $\begingroup$ A014545 $\endgroup$
    – Amin235
    Commented Jan 10, 2017 at 16:16

2 Answers 2

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$2\cdot3\cdot5\cdot7\cdot11\cdot13+1=30,031=59\cdot509$

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The number $N=p_1\cdots p_n+1$ is often prime, for the first $n$ primes $p_i$, but not always. The first example is $$ N=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13+1=30031=59\cdot 509. $$ The next one is $$ N=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17+1=19 \cdot 97\cdot 277. $$

Edit: We will not obtain all odd numbers by this (as remarked by Michael, and user190080 in his answer, unfortunately deleted now).

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  • $\begingroup$ I just deleted mine and upvoted yours, maybe you could add the information, why the attempt of the OP doesn't work... $\endgroup$
    – user190080
    Commented Jan 10, 2017 at 16:04

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