# help me to prove this conjecture

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• Jul 16th 2007, 08:04 AM
Discrete
Quote:

Originally Posted by CaptainBlack
As every prime p not equal to 2 is odd, p+1 is even and so not a prime.

But 2 is prime and 2+1=3 which is also prime.

RonL

[I must say that I am lost, what is the conjecture - or whatever - that is under discussion here?]

I just realized that what you said is right.. so then how would I correct my proof to proof or disproof this conjecture??
• Jul 17th 2007, 04:43 AM
topsquark
Quote:

Originally Posted by Discrete
I just realized that what you said is right.. so then how would I correct my proof to proof or disproof this conjecture??

Simply restrict your proof to a product of primes that has at least two primes in it.

-Dan
• Jul 17th 2007, 07:07 AM
Discrete
Quote:

Originally Posted by topsquark
Simply restrict your proof to a product of primes that has at least two primes in it.

-Dan

but then it won't satisfy for all condition if I restrict it

BTW can I use this counterexample

say that the number is 3 x 5 x 7 x 11 x 13

and the first 3 prime number is 3,5,and 7... if we use multiply these three number we get 105 but then we add 1 to this number it becomes 106 which is not a prime number.. therefore the conjecture is false
• Jul 17th 2007, 08:32 AM
topsquark
Quote:

Originally Posted by Discrete
but then it won't satisfy for all condition if I restrict it

I don't see that to be a disagreeable issue. For instance there are many formulas that are only applicable to certain circumstances. And there are numerous proofs I've seen that use Mathematical Induction where the theorem isn't true for the first few values of n but are true for, say, n = 3 and above.

-Dan
• Jul 17th 2007, 09:05 AM
Discrete
Quote:

Originally Posted by topsquark
I don't see that to be a disagreeable issue. For instance there are many formulas that are only applicable to certain circumstances. And there are numerous proofs I've seen that use Mathematical Induction where the theorem isn't true for the first few values of n but are true for, say, n = 3 and above.

-Dan

is my counter example right?
• Jul 17th 2007, 09:24 AM
topsquark
Maybe you'd better restate what it is you are trying to prove.

I had thought you were trying to show that
$\displaystyle 2 \cdot 3 \cdot 5 \cdot ~ ... ~ \cdot p + 1$
is not divisible by any prime p in the multiplicative sequence.

-Dan
• Jul 17th 2007, 09:53 AM
Discrete
no that's not the case, I am trying to prove that p1 x p2 x p3 x pn + 1 is a prime number for every positive integer n. where p1 x p2 x pn is the first n prime number.. I am not trying to proof any divisible here
• Jul 17th 2007, 10:47 AM
ThePerfectHacker
Quote:

Originally Posted by Discrete
no that's not the case, I am trying to prove that p1 x p2 x p3 x pn + 1 is a prime number for every positive integer n. where p1 x p2 x pn is the first n prime number.. I am not trying to proof any divisible here

That is false.
$\displaystyle 2+1 = \mbox{ prime}$
$\displaystyle 2\cdot 3 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11+1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509$
• Jul 17th 2007, 10:50 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
That is false.
$\displaystyle 2+1 = \mbox{ prime}$
$\displaystyle 2\cdot 3 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7 + 1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11+1 = \mbox{ prime}$
$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509$

Gosh I could have said that:cool:

RonL
• Jul 17th 2007, 11:11 AM
Discrete
I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:

$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509$

can you explain
• Jul 17th 2007, 12:24 PM
CaptainBlack
Quote:

Originally Posted by Discrete
I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:

$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509$

can you explain

BECAUSE IT IS A COUNTER EXAMPLE TO YOUR CONJECTURE AND ONE COUNTER EXAMPLE DISPROVES A CONJECTURE.

2, 3, 5, 7, 11, 13 are the first six primes and one more than their product is composite.

RonL
• Jul 17th 2007, 12:27 PM
topsquark
Quote:

Originally Posted by Discrete
I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:

$\displaystyle 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509$

can you explain

Because you are trying to show that for any sequence of primes, their product plus one is a prime. This is not true because
$\displaystyle 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$
which your proposition states should be a prime number is not a prime number.

-Dan
• Jul 17th 2007, 12:36 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Because you are trying to show that for any sequence of primes, their product plus one is a prime. This is not true because
$\displaystyle 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$
which your proposition states should be a prime number is not a prime number.

-Dan

In fact:

$\displaystyle 1+\prod_{i=1}^{N}p_i \ \ \ \mbox{ is composite for }N=6, 7, 8,9,10,12$

RonL
• Jul 17th 2007, 01:50 PM
Discrete
I got it now, thanks!
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