# help me to prove this conjecture

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• Jul 15th 2007, 09:06 AM
Discrete
help me to prove this conjecture
Conjecture: If p1 * p2 * p3 * ... pn + 1 is prime for every positive integer n, where p1,p2,...,pn are the n smallest prime number

So far, I have tried this

Proof (direct) :

Let pj be a prime number which is greater than 1

(p1*p2*p3*p4*...pn) % pj = 0
1 is divisible by pj

Therefore this conjecture is true?
• Jul 15th 2007, 09:11 AM
CaptainBlack
Quote:

Originally Posted by Discrete
Conjecture: If p1 * p2 * p3 * ... pn + 1 is prime for every positive integer n, where p1,p2,...,pn are the n smallest prime number

So far, I have tried this

Proof (direct) :

Let pj be a prime number which is greater than 1

(p1*p2*p3*p4*...pn) % pj = 0
1 is divisible by pj

Therefore this conjecture is true?

2*3*5*7*11*13 + 1= 30031 = 59*509

RonL
• Jul 15th 2007, 09:20 AM
Discrete
Quote:

Originally Posted by CaptainBlack
2*3*5*7*11*13 = 30030 = 59*509

RonL

what do you mean
• Jul 15th 2007, 09:26 AM
red_dog
Quote:

Originally Posted by CaptainBlack
2*3*5*7*11*13 = 30030 = 59*509

RonL

I think Discrete mean the following:
$p_1\cdot p_2\cdot \ldots \cdot p_n+1$ is prime, where $p_1,p_2,\ldots , p_n$ are the first $n$ primes.
• Jul 15th 2007, 09:56 AM
Discrete
Quote:

Originally Posted by red_dog
I think Discrete mean the following:
$p_1\cdot p_2\cdot \ldots \cdot p_n+1$ is prime, where $p_1,p_2,\ldots , p_n$ are the first $n$ primes.

that is right red dog, that's the question and is my proof right?
• Jul 15th 2007, 10:03 AM
ThePerfectHacker
Quote:

Originally Posted by Discrete
Therefore this conjecture is true?

No, it just shows that the number is divisible by a different prime.
• Jul 15th 2007, 10:33 AM
CaptainBlack
Quote:

Originally Posted by red_dog
I think Discrete mean the following:
$p_1\cdot p_2\cdot \ldots \cdot p_n+1$ is prime, where $p_1,p_2,\ldots , p_n$ are the first $n$ primes.

There was a typo I missied the +1.

2*3*5*7*11*13 + 1 = 30031 = 59*509

I make that one more than the product of the first 6 primes is composite.
One counter example is sufficent to refute a conjecture.

RonL
• Jul 15th 2007, 11:05 AM
Discrete
but how do I want to proof this by using direct, contraposition or contradiction proof?
• Jul 15th 2007, 11:08 AM
CaptainBlack
Quote:

Originally Posted by Discrete
but how do I want to proof this by using direct, contraposition or contradiction proof?

Suppose the conjecture true, show the counter example, which gives
a contradiction, hence the supposition is false.

RonL
• Jul 15th 2007, 11:20 AM
Discrete
Quote:

Originally Posted by CaptainBlack

Suppose the conjecture true, show the counter example, which gives
a contradiction, hence the supposition is false.

RonL

yes I know this is the way of how to proof it by contradiction but then for the case in this question, how do I deal with it?
• Jul 15th 2007, 12:14 PM
CaptainBlack
Quote:

Originally Posted by Discrete
yes I know this is the way of how to proof it by contradiction but then for the case in this question, how do I deal with it?

RonL
• Jul 15th 2007, 09:06 PM
Discrete
Is it possible to use the proof that I have mentioned in my first post?? 1 is not divisible by any prime numbers, so therefore any prime number which is added by 1 becomes no prime number, right??
• Jul 16th 2007, 04:25 AM
topsquark
Quote:

Originally Posted by Discrete
Is it possible to use the proof that I have mentioned in my first post?? 1 is not divisible by any prime numbers, so therefore any prime number which is added by 1 becomes no prime number, right??

Not by any prime in your list used in constructing the number, yes. If that's what you are trying to prove then the proof is correct. As in CaptainBlack's example
$2 \cdot 3\cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1 = 30031 = 59 \cdot 509$

Note that 30031 is not divisible by any of 2, 3, 5, 7, 11, or 13, but it is divisible by the primes 59 and 509.

-Dan
• Jul 16th 2007, 06:05 AM
CaptainBlack
Quote:

Originally Posted by Discrete
Is it possible to use the proof that I have mentioned in my first post?? 1 is not divisible by any prime numbers, so therefore any prime number which is added by 1 becomes no prime number, right??

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?]
• Jul 16th 2007, 06:19 AM
ThePerfectHacker
This is how I understand this user.

Let $p_1,...,p_n$ be the first $n$ primes.

Then, $m=p_1p_2...p_n+1$ is not divisible by any of these $p_i$. (Remember Euclid's Proof).

So, (and that is what he claims) it must mean that $m$ is prime itself because it is not divisible by any of those primes.

That is his conjecture.

But that is false because we only showed that $m$ is divisible by another prime. Either itself, if is it, or a completely different prime if it is not.
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