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Goldbach's conjecture is:

Every even integer greater than $2$ can be expressed as the sum of two primes.

Extension of Goldbach's conjecture is:

Every number from $p\mathbb{Z}$ greater than $p$ can be expressed as the sum of $p$ primes

where $p\mathbb{Z}=\lbrace 0, \pm p, \pm 2p, \pm3p,\dots\rbrace$.

You can find a counterexamples?

For example if $p=5$:

  • $10=2+2+2+2+2$
  • $15=3+3+3+3+3$
  • $20=2+3+3+5+7$
  • ...
  • $100=47+2+17+17+17$

Who can make this programming?

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First of all we will assume that the Goldbach's conjecture is true, so given an integer $n=kp$ with $k\geq 4$ ($k=3,2$ are easy), we will prove that $n$ is the sum of $p$ primes.

If $p=2,$ it's the Goldbach's conjecture and it's true, so we can assume that $p>2$.

Case 1 If $n$ is even, let's write $$n=\underbrace{2+\cdots+2}_{p-2}+(n-2(p-2)),$$ and because $(n-2(p-2))=(k-2)p+4\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.

Case 2 If $n$ is odd, let's write $$n=3+\underbrace{2+\cdots+2}_{p-3}+(n-2(p-3)-3),$$ and because $(n-2(p-3)-3)=(k-2)p+3\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.

So we proved the following statement: $$\text{Goldbach's conjecture} \implies \text{Extension of Goldbach's conjecture}$$

But Goldbach's conjecture is a particular case of the extension of Goldbach's conjecture (when $p=2$), hence: $$\text{Goldbach's conjecture} \iff \text{Extension of Goldbach's conjecture}$$

Conclusion The extension of Goldbach's conjecture as you define it is equivalent to the Goldbach's conjecture itself, and hence no one will ever find a counterexample or prove it unless (s)he solves Goldbach's conjecture.

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  • $\begingroup$ I think that this can be modified to : every integer larger than $2p$ is the sum of $p$ primes $\endgroup$ – Elaqqad Jul 9 '15 at 11:30
  • $\begingroup$ Always it surprised me that nobody expressed Goldbach's conjecture as follows: "Every integer >3 is equidistant from two primes". It appears to me so nice. (and the 3 could be incorporated with the convention of 1 be prime). $\endgroup$ – Piquito Jul 9 '15 at 12:59
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    $\begingroup$ A remark: in order to prove this equivalence you don't need to assume Golbach's conjecture is true as you did. You have just a logic equivalence. $\endgroup$ – Piquito Jul 9 '15 at 13:07
  • $\begingroup$ @Ataulfo I did not assume it for proving the equivalence, it's just the first thing to do in order to prove an implication, I assume Goldbach's conjecture is true and then I proved the other result. This is not general assumption in the proof. $\endgroup$ – Elaqqad Jul 10 '15 at 11:50
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Suppose Goldbach is true. Let n be your candidate number. Then either n is even or odd. If even, then it is the sum of 2 primes. If odd then n - 3 is even so n is the sum of 3 primes. In no case would you need more than 3 primes.

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  • $\begingroup$ You may not need more than 3 primes, but can you use $p$ primes? $\endgroup$ – jwodder Jul 9 '15 at 12:33
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    $\begingroup$ Sure. Inductively suppose you have settled the question for L - 1 and all numbers less than n. Then just look at n - 2. $\endgroup$ – lulu Jul 9 '15 at 12:38

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