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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    $\begingroup$ Good idea. To be precise, let $Q_n$ be the product of all primes that are less than or equal to the $n$-th prime $p_n$. Then $Q_n+1$ is not divisible by any prime $\le p_n$. That does not mean that $Q_n+1$ is prime. The same consideration holds for $Q_n-1$. $\endgroup$ Dec 18, 2011 at 6:28
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    $\begingroup$ I hope you don't take offense, but one can tell without even reading it that this argument is wrong. That's fine! This is a very hard problem. I think you mean for $p_n$ to be the product of the first $n$ primes. Then, as in Euclid's proof of the infinitude of primes, you form $p_n + 1$. This is not divisible by the first $n$ primes for the reason you give, but why do you think it must be prime? For example, $p_6 + 1 = 59\cdot 509$. $\endgroup$ Dec 18, 2011 at 6:33
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    $\begingroup$ A minor nitpick (which doesn't affect your argument): when you divide $N$ by one of the first $n$ primes, you get a remainder of $-1$. $\endgroup$ Dec 18, 2011 at 6:42
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    $\begingroup$ Welcome to MSE. By the way, this might interest: you en.wikipedia.org/wiki/Brun%27s_constant that is $\sum_\text{twinprimes} 1/p$ converges which is not the case for $\sum_\text{primes} 1/p$. $\endgroup$ Dec 18, 2011 at 6:45
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    $\begingroup$ 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. $\endgroup$ Apr 19, 2013 at 16:02

2 Answers 2


(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.

  • $\begingroup$ Thank you so much for your comment on my proof. I understand that, where I am wrong. $\endgroup$
    – mahi
    Dec 18, 2011 at 15:23

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


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