If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the conjecture proved? If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the Goldbach conjecture proved?
If ZFC+CH implies Goldbach, and if the Goldbach turn out to be false, then it would mean that ZFC+CH is not consistent, but we know that ZFC+CH is consistent assuming that ZFC is consistent...
What do you think?
 A: Because the Goldbach conjecture is an arithmetic statement, it is absolute between any two models which agree on the natural numbers.
Now, given any model of $\sf ZFC$, $M$, there is a forcing extension $M[G]$ with the same ordinals (and in particular, the same natural numbers, which are the just the finite ordinals), in which $\sf CH$ holds. Or, better yet, simply  consider $L^M$, which is an inner model with the same ordinals (and, again, the same natural numbers), in which $\sf CH$ holds.
Therefore, if you can prove Goldbach, Riemann, or the ABC Conjecture, assuming $\sf CH$, you may as well have proved it. Using $L$ will also tell you that using the Axiom of Choice was redundant, so in fact the proof is in $\sf ZF$ and not $\sf ZFC$.
So, to sum this up, if you prove that $\sf ZFC+CH$ implies Goldbach's conjecture, and then you prove that Goldbach's conjecture is false, you've proved that $\sf ZF$ is inconsistent. Which, to my taste, is a far bigger result than Goldbach's conjecture (although others may disagree).
