Although there is not algorithm that can solve the halting problem, it is still meaningful to talk about an imaginary computer that has some procedure that decides whether a given program halts or not, called an oracle for the halting problem.
If we had an oracle for the halting problem, then we could immediately verify the truth of the Goldbach conjecture (as well as many others). Heres how:
Consider the following program:
- Let $n=2$.
- Increase $n$ by $2$
- Check to see if $n$ can be written as a sum of two primes (we can do this just by checking all of the (finitely many) primes less than n.
- If we could write $n$ as a sum of two primes, go back to step $2$. Otherwise, stop.
The only way for this program to halt is if, at some time during step $4$, we cannot write $n$ as a sum of two primes. Since $n$ is always even and at least $4$, then, the program halts only if the Goldbach conjecture is false. Since, furthermore, $n$ will run through all even integers greater than or equal to $4$ if the program does not halt, the program also always halts if the Goldbach conjecture is false.
Now, we can take this program, and run our oracle for the halting problem on it.
If the Goldbach conjecture is true, the oracle will say that the program does not halt, and if the Goldbach conjecture is false, the oracle will say that the program will halt.
Therefore, if the halting problem were solvable, the Goldbach conjecture (and many other problems) would also be easily solvable.
$m$ 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.
Best Answer
Your argument correctly shows that the Goldbach conjecture cannot be false and undecidable. However, it can be true and undecidable.
This is equivalent to saying that it can be true in the standard model of, say, Peano arithmetic, but false in some nonstandard model: in a nonstandard model there may be a nonstandard counterexample to Goldbach's conjecture which is not an ordinary integer, and so which cannot be written down in the usual sense.
The reason it can be true and yet assuming it's false doesn't lead to a contradiction is that if it's undecidable, Peano arithmetic doesn't know it's true, so Peano arithmetic can't use the fact that it's true in a proof.