Twin prime conjecture proof error

Question

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

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. — Did, May 16 '16 at 14:51
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. — barak manos, May 16 '16 at 14:51
Your question is a kind of duplicate of this question (which is a duplicate of this question). — barak manos, May 16 '16 at 15:47
if it were more than root x it would have to be multiplied by something less then root x. — shai horowitz, May 16 '16 at 17:18
@N.S.JOHN Yes, necessarily less than sqrt(x). No nonprime number N has only factors greater than sqrt(N). — Did, May 16 '16 at 17:19
@N.S.JOHN: But $11 \leq \sqrt{209}$, so there's no point in looking at anything past $\sqrt{209}$. — jwodder, May 16 '16 at 18:11

score 30 · Accepted Answer · answered May 16 '16 at 14:48

30

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, $x - 1 = 209 = 11\times19.$

answered May 16 '16 at 14:48

Edward Evans

4,590

8

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. – Edward Evans May 16 '16 at 14:51
4

@Ed_4434 It's actually in the first 8! – bjb568 May 16 '16 at 21:51
12

@bjb568 My god you're right. My numerical intuition has gone out the window. becomes Engineer – Edward Evans May 16 '16 at 21:53
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@bjb568 8! > 10 – Steve Cox May 16 '16 at 23:17

score 7 · Answer 2 · answered May 16 '16 at 16:53

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.

That primoral is not always prime. The idea is that it's coprime to the current list of primes, so more primes must exist. — Michael Klein, May 16 '16 at 23:05

score 1 · Answer 3 · answered Oct 23 '17 at 04:13

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 $B_n$.

Twin prime conjecture proof error

3 Answers3