13
$\begingroup$

I am absolutely sure this is wrong but I can't find why.

For every integer $n$ there exist a finite number of primes less than $n$. Take the set containing those primes and multiply them together to get $x$. Aren't $x+1$ and $x-1$ prime, implying there is an infinite number of twin primes?

Follow up question is there guaranteed to be a prime between n and $x^{.5}$? What about for large n? this prime wouldn't have to devide x just exist in the given range

$\endgroup$
10
  • 17
    $\begingroup$ Why can we deduce that x+1 and x-1 are prime? Some primes between n and sqrt(x) might divide x+1 or x-1. $\endgroup$
    – Did
    Commented May 16, 2016 at 14:51
  • 1
    $\begingroup$ Yes, x+1 and x-1 being prime would imply that there are an infinite number of twin primes... Unfortunately, neither x+1 nor x-1 are necessarily prime. $\endgroup$ Commented May 16, 2016 at 14:51
  • 1
    $\begingroup$ if it were more than root x it would have to be multiplied by something less then root x. $\endgroup$ Commented May 16, 2016 at 17:18
  • 2
    $\begingroup$ @N.S.JOHN Yes, necessarily less than sqrt(x). No nonprime number N has only factors greater than sqrt(N). $\endgroup$
    – Did
    Commented May 16, 2016 at 17:19
  • 1
    $\begingroup$ @N.S.JOHN: But $11 \leq \sqrt{209}$, so there's no point in looking at anything past $\sqrt{209}$. $\endgroup$
    – jwodder
    Commented May 16, 2016 at 18:11

3 Answers 3

30
$\begingroup$

Let $n = 8$. Then all primes less than $8$ are $7, 5, 3, 2$. The product of these is $x = 210$.

$x + 1 = 211$ which is prime, $x - 1 = 209 = 11\times19.$

$\endgroup$
4
  • 8
    $\begingroup$ This isn't necessarily a reason why your method doesn't work, but it certainly is a counter-example, within the first 10 positive integers. $\endgroup$ Commented May 16, 2016 at 14:51
  • 4
    $\begingroup$ @Ed_4434 It's actually in the first 8! $\endgroup$
    – bjb568
    Commented May 16, 2016 at 21:51
  • 12
    $\begingroup$ @bjb568 My god you're right. My numerical intuition has gone out the window. becomes Engineer $\endgroup$ Commented May 16, 2016 at 21:53
  • 11
    $\begingroup$ @bjb568 8! > 10 $\endgroup$
    – Steve Cox
    Commented May 16, 2016 at 23:17
7
$\begingroup$

Your proof most likely stems off of a proof that there are an infinite number of primes. It says that if $\mathbb{P}$ is the set of all primes, and $|\mathbb{P}|<\aleph_0$, then intuitively $$\left(\pm1+p_{|\mathbb{P}|}\#\right)\notin\mathbb{P} \:\text{and}\:\forall p\in\mathbb{P},p\nmid\left(\pm1+p_{|\mathbb{P}|}\#\right)$$ (where $p_n\#$ is the $n$th primorial) implying that $\left(\pm1+p_{|\mathbb{P}|}\#\right)$ is prime, proving that there are an infinite number of primes by reductio ad absurdum. However, since $|\mathbb{P}|=\aleph_0$, $\pm1+p_n\#$ is prime is not necessarily true for arbitrary $n$.

The type of primes that are in the form of $\pm1+p_n\#$ are called primorial primes and you can read more about them here.

$\endgroup$
1
  • 7
    $\begingroup$ That primoral is not always prime. The idea is that it's coprime to the current list of primes, so more primes must exist. $\endgroup$ Commented May 16, 2016 at 23:05
1
$\begingroup$

I had the same question once I heard about the twin prime conjecture. The confusion stems from the following.

Take the first n primes: $p_1, p_2, .... p_n$, and define:

$$A_n = p_1 p_2 .... p_n +1$$ $$B_n = p_1 p_2 .... p_n -1$$

Now we know that neither $A_n$ nor $B_n$ is divisible by any of the primes $p_1, p_2, .... p_n$.

However this does not rule out the possibility that there may exist other prime numbers greater than $p_n$ that divide $A_n$ or $B_n$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .