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".


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!

  • 1
    $\begingroup$ 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? $\endgroup$
    – runway44
    Feb 12, 2021 at 21:16
  • $\begingroup$ "In accordance with (2), we can see that gcd..." - could you explain why gcd(n,p)=1? I'm failing to see that. $\endgroup$
    – lisyarus
    Feb 12, 2021 at 21:17
  • $\begingroup$ @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 $p_b$ $\endgroup$ Feb 12, 2021 at 21:21
  • 1
    $\begingroup$ 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? $\endgroup$
    – runway44
    Feb 12, 2021 at 21:24
  • $\begingroup$ @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) $\endgroup$ Feb 12, 2021 at 21:32

2 Answers 2


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.

  • $\begingroup$ I'm having trouble following. Could you provide a concrete counter example to help illustrate your main point? $\endgroup$ Feb 12, 2021 at 21:58
  • $\begingroup$ 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. $\endgroup$
    – player3236
    Feb 13, 2021 at 2:53
  • $\begingroup$ I see it now! Sneaky. Thank you for your insight! $\endgroup$ 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)

  • $\begingroup$ 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! $\endgroup$ Feb 12, 2021 at 21:26
  • 1
    $\begingroup$ 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 $n$. $\endgroup$
    – hgmath
    Feb 12, 2021 at 21:31

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