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

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- Jul 16th 2007, 08:04 AMDiscrete
- Jul 17th 2007, 04:43 AMtopsquark
- Jul 17th 2007, 07:07 AMDiscrete
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 AMtopsquark
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 AMDiscrete
- Jul 17th 2007, 09:24 AMtopsquark
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 AMDiscrete
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 AMThePerfectHacker
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 AMCaptainBlack
- Jul 17th 2007, 11:11 AMDiscrete
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 PMCaptainBlack
- Jul 17th 2007, 12:27 PMtopsquark
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 PMCaptainBlack
- Jul 17th 2007, 01:50 PMDiscrete
I got it now, thanks!