I know just as much as the next guy that this will be a far way off the mark, which is why I have phrased the question as "why is this wrong", not "is this wrong" – so let's do what mathematicians do best and poke holes in this (admittedly short) "proof" of the old-Gold conjecture 🙂
The conjecture goes like this, we all know:
$\forall n \in \mathbb{N}$, there exist two primes $p_a$ and $p_b$ such that $p_a+p_b=2n$. (statement 1)
For a prime $n$ (let's call it $p_N$ for simplicity), the solution to (1) is trivial, namely $p_N+p_N=2n$.
For a composite $n>3$, without loss of generality, let us say that $2\lt p_a \lt n \lt p_b \lt 2n-2$, and start by assuming the negation, namely:
$\exists n$ such that no $p_a,p_b$ satisfy (1). (statement 2)
Taking (1) mod $p_a$ and mod $p_b$,
$p_a\equiv2n\not\equiv0$ (mod $p_b$), and $p_b\equiv2n\not\equiv0$ (mod $p_a$). The non-equivalence to zero is guranteed by the bounds on a composite $n$, because neither $p_a$ nor $p_b$ can be equal to an even prime nor to $n$ itself. If, say, $p_a \mid n$, then $n=p_a +m\cdot p_a$, which would make $p_b=p_a+2m\cdot p_a$, which is clearly false by the definition of a prime. An equivalent, symmetrical argument can be made for if $p_b\mid n$.
In accordance with (2), we can see that gcd$(n,p_a) = 1$, and similarly that gcd$(n,p_b) = 1$, for all $p_a,p_b\in(2,2n-2)$. However, if gcd$(x,p_k)=1 ,\forall p_k\le \sqrt{x}$, that implies $x$ is prime. It is apparent that $\sqrt{n}\in(2,2n-2)$, and thus implies that n is prime. This contradicts with our assumption that n is composite, namely "If a composite number does not satisfy Goldbach's conjecture, it is a prime number, which satisfies Goldbach's conjecture".
QED?
P.S. Thank you to the MSE community at large for contuing to entertain "proofs" on long-standing questions. It is thanks to people like you that maniacs like me who believe they can solve these questions can avoid embarassing themselves to their PhD committee!
Best Answer
The mistake is "for all $p_a, p_b \in (2,2n-2)$", and is a case of obfuscated context.
While it is true that if $\gcd (x,p_k) = 1, \forall p_k \le \sqrt x$ then $x$ is prime, not all primes in between $2$ and $2n-2$ is coprime to $n$. As shown in the first long paragragh, only primes satisfying statement 1 is coprime to $n$. If a prime $p < \sqrt n$ does not satisfy statement 1, i.e. $2n- p$ is composite, we cannot draw any conclusion on $\gcd(p,n)$. Any odd prime divisor of $n$ provides a counterexample.