Can somebody please tell me what is wrong with the following "proof" of the strong Goldbach Conjecture?

For every even number $n$, there are $\frac{n}{4}$ pairs of odd numbers $[a, b]$ such that $a+b=n$ and $a\le b$. If either $a$ or $b$ is not prime, it must be divisible by some prime $p\le\sqrt{n}$. Since either $a$ or $b$ can be divisible by $p$, each prime $p$ can eliminate a fraction of $\frac{p-2}{p}$ pairs. Finally, $[1, n-1]$ is not a prime pair. We can construct a sieve such that the number $k$ of prime pairs $[a, b]$ is

$$k = \frac{n}{4} \cdot \frac{3-2}{3} \cdot \frac{5-2}{5} \cdot \frac{7-2}{7} \cdot \frac{11-2}{11} \dotsm \frac{p_{max}-2}{p_{max}} - 1$$

Since $p_{max}\le\sqrt{n}$

$$k \ge \left \lfloor{\frac{n}{4} \cdot \frac{1}{\sqrt{n}} - 1}\right \rfloor $$

($\left\lfloor{\cdot}\right\rfloor$ is the floor function)

In fact, this sieve eliminates too many pairs $[a, b]$, which doesn't matter here.

For all even $n\ge64$, we get $k\ge1$, so there always is at least one pair of primes $[a, b]$. For $n\ge4$ and $n<64$, such pairs are known already, so for all even $n\ge4$ there are primes $[a, b]$ such that $n=a+b$.

Again, what's wrong with this? Thanks.

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    $\begingroup$ Welcome to MSE. Your formatting is very bad, please consider learning a little bit of MathJax meta.math.stackexchange.com/questions/107/… $\endgroup$ – Ewan Delanoy Oct 3 '14 at 16:27
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    $\begingroup$ You seem to be overly optimistic at the ability for a sieve to arrive at a lower bound as good as $n/4 \prod_p (1-2/p)$, which is only a heuristic. Without any error estimation, what makes you think this is a proof? $\endgroup$ – Erick Wong Oct 3 '14 at 16:36
  • $\begingroup$ I pretty much know this must be wrong, but I don't see how. As far as I can see, every prime can eliminate only so many pairs, and only primes p<=sqrt(n) are relevant. Again, I'd be happy if somebody can explain why this doesn't work. $\endgroup$ – tomerius Oct 3 '14 at 16:44
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    $\begingroup$ You seem to be trying to reinvent sieve theory. What you wrote is a useful heuristic method (googling things like "Cramer random model", "Hardy-Littlewood k-tuples conjecture" should give you some idea of just how useful). The problem is precisely that you make some small error for each prime, and these errors add up. In fact, they add up to more than you can afford (unless you find some way of controlling the error, which noone has done for this problem so far). $\endgroup$ – Jakub Konieczny Oct 3 '14 at 17:33
  • $\begingroup$ I'm pretty sure I can't reinvent sieve theory. I thought the approximation is on the safe side because (1-2/p) not only eliminates multiples of p but also p itself which should more than offset for any fractions left. $\endgroup$ – tomerius Oct 3 '14 at 17:45

You are stating that for any prime $p\le\sqrt{n}$, the number $k$ of pairs is EXACTLY reduced by the factor $\frac{p-1}{p}$. For your proof to be correct, you need to proof/refer to a valid proof that shows that this approximation (it is an approximation, just try it with e.g. $n=2^l$ (with $l\in\mathbb{N}$ and $l\ge4$) and you will find it does not evaluate to a natural number, though $k$ definitely is one) is a lower bound for $k$.

  • $\begingroup$ I'm probably not getting away with "it's obvious that ...", no? Thanks. $\endgroup$ – tomerius Oct 3 '14 at 17:14
  • $\begingroup$ Yes, it's an approximation so the first equation should have been k >= n/4 * (3-2)/3 * (5-2)/5 * (7-2)/7 * (11-2)/11 * ... * (pmax-2)/pmax - 1 $\endgroup$ – tomerius Oct 3 '14 at 17:31
  • $\begingroup$ Yes. But you still need to prove, that this holds for any $n\ge64$... $\endgroup$ – marzipankaiser Oct 3 '14 at 17:34

You assume that the elimination by sieving is independent. This has to be proved - there may be many cases where different sieves eliminate the same value.

  • $\begingroup$ Yes. If a number can be divided by primes p and q, it's eliminated twice. But this is why we presumably can multiply (p-2)/p*(q-2)/q ... $\endgroup$ – tomerius Oct 3 '14 at 16:58

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