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My professor introduced the Goldbach Conjecture to me a couple months ago, and I've been intrigued by it ever since. The Goldbach Conjecture states that every even number greater than 4 can be written as the sum of two primes.

If we look at the number $2m$, where we let $m$ be the number containing every prime greater than $2$, then $2m$ = $2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdots$

Then every number less than $2m -1$ will be contained in $2m$, so no matter what number $x$ you subtract from $2m$, you get

$2m - x = x\cdot\left(\frac{2m}{x}-1\right)$ which is composite, so this even number can't be written as sum of two primes.

I'm assuming this is not a valid counter-example, but why not?

Ash
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    Lol, nice try.${}{}{}$ – Asinomás Jul 26 '16 at 18:27
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    Can you clarify about the construction of $m$? It's not clear if an infinite number of primes is involved or not. In the former case, $m$ is not a real number because the product diverges, as there are infinitely many primes (the first proof of this fact, credited to Euclid, is really slick). – rubik Jul 26 '16 at 18:29
  • The bracketed expression (2m/x – 1) won't always be an integer, e.g. when x = 5^2 (as will also be the case for any other x comprising multiple powers of the prime factors in 2m which still satisfies x < 2m - 1). If x is always an integer but (2m/x – 1) isn't, then one cannot say that x.(2m/x – 1) is always a composite - have I missed something here? – Jimbo Nov 25 '22 at 11:49

3 Answers3

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$m$ doesn't exist, since there are infinitely many primes.

In a little more detail: the product of infinitely many natural numbers is not, in general, a natural number. Similarly, the sum of infinitely many natural numbers is not, in general, a natural number: for example, $$1+1+1+1+...$$ is not a natural number. The Goldbach conjecture is a statement about natural numbers, so looking to infinite products for a counterexample isn't going to work.

There are contexts where we can make sense of an infinite expression like the above (see e.g. Why does $1+2+3+\cdots = -\frac{1}{12}$?), but it's something that takes serious work to make precise - and often results in surprising properties, or the failure of properties we usually take for granted.

Addressing rubik's comments: what if we only use finitely many primes in the construction of $m$?

Well, then the whole shebang breaks down! Suppose $$m=3\cdot 5\cdot . . . \cdot p_k$$ is the product of the first $(k-1)$-many primes bigger than $2$. Then their may be a prime which is $<2m-1$, but which is not a factor of $2m$ - namely, $p_{k+1}$! So we can't conclude that $2m$ is a counterexample to Goldbach.

J. W. Tanner
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Noah Schweber
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  • $m=0{}{}{}{}{}$ – Asinomás Jul 26 '16 at 18:27
  • @CarryonSmiling Um, what? The OP has defined $m$ as $3\cdot 5\cdot 7\cdot . . .$ - how on earth is that $0$? – Noah Schweber Jul 26 '16 at 18:28
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    well, maybe with the OP's first definition $m=0$ works.(an integer containing all primes $>2$) – Asinomás Jul 26 '16 at 18:29
  • @rubik I disagree, I don't see that in the question at all. Their argument crucially relies on the assumption that $m$ is divisible by every prime; until they weigh in otherwise, I think my answer addresses their question. – Noah Schweber Jul 26 '16 at 18:35
  • @NoahSchweber Thank you for your answer, it was very clear and I understand now why it's not valid. – Ash Jul 26 '16 at 18:40
  • your definition of the goldbach conjecture is incorrect. every even number >2 is prime. you forgot 4=2+2 – OEIS Feb 25 '22 at 23:56
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There are infinitely many primes, so $m$ doesn't exist.

Pretend like $m$ could exist i.e. there are finitely many primes. Then clearly $m+1$ has a prime factor, say $q$. Since $m$ is the product of all primes, $q$ must be a prime factor of $m$ as well. Thus, $q$ must be a prime factor of $m+1-m=1$ as well, yet $1$ has no prime factors. Therefore, $m$ cannot exist, so there must be infinitely many primes.

JasonM
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Apart from the concern about the nature of $ m $ as mentioned by others, your claim that every number less than $2m - 1$ is contained in $2m $ is not true. For example consider number 4; then for $ x=4$ you do not conclude anything.

J S
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