# if we have: $[(p_1)(p_2)(p_3)…(p_n)]+1$ & $p_n$ is the nth prime number, & $p_1,p_2…$ are all the prime numbers below $p_n$, is it always prime

if we have the following: $[(p_1)(p_2)(p_3)......(p_n)]+1$ where $p_n$ is the nth prime number, and $p_1,p_2...$ are all the prime numbers below $p_n$, is $[(p_1)(p_2)(p_3)......(p_n)]+1$ a prime number?

When i plugged in random numbers into the thing I thought that it was true but then obviously there are prime number generators on computers so it seemed as if $[(p_1)(p_2)(p_3)......(p_n)]+1$ shouldn't always be prime for a number $n$

How do i prove or demonstrate that $[(p_1)(p_2)(p_3)......(p_n)]+1$ isn't always prime?

• What does your notation, $[(p_1)(p_2)]$, mean? – infinitylord Jun 2 '17 at 19:54
• I think it is just multiplication, the $[]$ are probably redundant, maybe it seemed clearer to OP – Gregory Jun 2 '17 at 19:55
• Counterexample is a perfectly valid method. Try $n=6$. – Rocket Man Jun 2 '17 at 19:55
• $(2)(3)(5)(7)(11)(13) + 1 = 30031$. What are the factors of that? – Gregory Jun 2 '17 at 19:57
• I am sure Gregory knew $30031= 59 \times 509$ @infinitylord – Donald Splutterwit Jun 2 '17 at 20:18