Goldbach Conjecture Counterexample

Question

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?

Answer 1

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

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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.

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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.

Answer 2

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.

Answer 3

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.