# Goldbach Conjecture Counterexample

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?

• Lol, nice try.${}{}{}$ – Jorge Fernández Hidalgo Jul 26 '16 at 18:27
• 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

$$m$$ 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.

• $m=0{}{}{}{}{}$ – Jorge Fernández Hidalgo 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
• well, maybe with the OP's first definition $m=0$ works.(an integer containing all primes $>2$) – Jorge Fernández Hidalgo 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

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.

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.