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Whether I am correct or wrong I don't know. If there are any corrections, please let me know.

Let $p_n$ = product of all primes. (of course we can go still beyond as we know $p_n$ is infinite). Now consider $M = p_n + 1$ and $N = p_n - 1$. As we know that, by Euclid theorem, no prime of $p_n$ will divide $M$ as it leaves remainder $1$. So, $M$ is prime. At this same time, no prime of $p_n$ can divide $N$ as it leaves the remainder $1$. So, $N$ must be prime. Now, $M - N = (p_n +1 ) - (p_n -1) = 2$, i.e. $(M, N)$ is prime pair with difference $2$. So, if we extend primes still we can see the difference $2$. So twin primes are not finite.

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    Actually a slight variation of this argument, where it is repeated for all multiples of the product $p_n$, correctly proves that if the number of primes is finite, then the number of twin primes is infinite. I find this quite amusing.2013-04-19

2 Answers 2

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(I will use the notation $p_n=$ the $n$th prime number, not $p_n=$ product of the first $n$ prime numbers as the OP is using.)

Your argument is incorrect, since neither $M=(p_1\cdots p_n)+1$ nor $N=(p_1\cdots p_n)-1$ have to be prime. All you know is that $M$ and $N$ are not divisible by the primes $p_1,\ldots,p_n$.

For some specific examples:

$(2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13)+1=30031=59\cdot 509$ $(2\cdot 3\cdot 5\cdot 7)-1=209=11\cdot 19$

As you can see, $209$ is not divisible by any of $2,3,5,7$, and $30031$ is not divisible by any of $2,3,5,7,11,13$, but this has not prevented them from being divisible by other primes, and hence composite.

See also the discussions here and here.

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    Thank you so much for your comment on my proof. I understand that, where I am wrong.2011-12-18
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The flaw in your proof lies with your assumption $M=p_n+1$ is a prime, which is not. The proof of actually only uses the fact that there is a prime dividing this product.

Or other wise you might have been talking about Primorial primes .

Thank you