1
$\begingroup$

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.

  • 2
    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 $Q_n-1$.2011-12-18
  • 2
    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, $p_6 + 1 = 59\cdot 509$.2011-12-18
  • 1
    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 $-1$.2011-12-18
  • 1
    Welcome to MSE. By the way, this might interest: you http://en.wikipedia.org/wiki/Brun%27s_constant that is $\sum_\text{twinprimes} 1/p$ converges which is not the case for $\sum_\text{primes} 1/p$.2011-12-18
  • 0
    Thank you very much for all members, who showed my errors. Thanks again.2011-12-18
  • 0
    This is one of the most widely published false proofs in the usenet newsgroup sci.math. Search on Archimedes Plutonium twin primes sci.math.2011-12-23
  • 2
    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