# Why is this proof of Goldbach's Conjecture flawed?

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!

• First off, do not use $n$ and $p_n$ interchangeably. Pick one. For instance, if $n=2$, then $p_2=3$. Notice $2$ and $p_2$ are two different numbers. Secondly, how exactly are you getting the congruence $p_a\equiv 2n$ mod $p_b$? Where is that coming from? Feb 12, 2021 at 21:16
• "In accordance with (2), we can see that gcd..." - could you explain why gcd(n,p)=1? I'm failing to see that. Feb 12, 2021 at 21:17
• @runway44 Changed $p_n$ to $p_N$ to clarify use of variables, good catch. I am arriving at the congruence just by taking (1) modulo $p_a$ or $p_b$ Feb 12, 2021 at 21:21
• But you just said you started by assuming the negation, so if there are no $p_a$ and $p_b$ for which $p_a+p_b=2n$ then how are you taking the equation mod stuff? Feb 12, 2021 at 21:24
• @lisyarus The gcd(n,p) is either 1 or p, corresponding to a non-zero or zero residue of n mod p, respectively. By the modular equivalences, n has a non-zero residue over all p_a,p_b on (2,2n-2) Feb 12, 2021 at 21:32

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.

• I'm having trouble following. Could you provide a concrete counter example to help illustrate your main point? Feb 12, 2021 at 21:58
• Take $2n = 18$. We know that $5+13 = 7+11 = 18$, so each of $5,7,11,13$ is coprime to $9$. That part of your argument is correct. However, $2,3 \le \sqrt {9}$ are primes not included in the above list, so we cannot say $\gcd (p, 9) = 1$ for all $p \le \sqrt 9$, and we hence cannot conclude that $9$ is prime. Feb 13, 2021 at 2:53
• I see it now! Sneaky. Thank you for your insight! Feb 13, 2021 at 18:01

I also think there is a problem here:

If, say, $$p_a∣n$$, then $$n=p_a+m\cdot p_a$$, which would make $$p_b=p_a+2m\cdot p_a$$

Since we are assuming that there are no $$p_a, p_b$$ satisfying $$p_a+p_b=2n$$:

if $$p_b$$ is another prime, then $$p_a+p_b\neq 2n$$, so in the quoted line it actually follows that $$p_b\neq p_a+2m\cdot p_a$$ (but as you noted, this is already impossible as $$p_b$$ should be prime and we haven't gained anything)

• I think it's okay here. I definitely fudged up the order in which the facts were presented, but I wasn't relying on the negation to formulate this. It is only required here that $n$ be composite. Thank you for your insight! Feb 12, 2021 at 21:26
• Well, you are requiring that $n$ is composite and that there is some $p_b$ such that $p_a+p_b=2n$, and only then it follows that $p_a$ does not divide $n$. Feb 12, 2021 at 21:31