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Can anyone clarify why induction method fails for this conjecture?

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    Have you tried to apply induction? If you try you should immediately detect the difficulties.2012-06-07
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    You can try it. If you fail you answered your own question. If you don't fail you will be pretty famous soon. Sounds like a win-win.2012-06-07
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    ok. Will try to apply. :)2012-06-07
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    If the conjecture is ever proved, it seems very likely that some sort of induction will be used somewhere.2012-06-07
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    ***Boring Comment***: if $p$ is prime, then $p+2$ is *not prime in general* (even if $p$ is odd). You should be able to see this is the case if you list the first few prime numbers. (*Hint*: "Few" $\geq 4$ if you restrict to odd prime numbers.) If $p$ and $p+2$ are prime numbers, then we say that the pair $(p,p+2)$ is a **twin prime**. An open problem is to determine whether or not there are infinitely many twin primes.2012-06-07
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    What makes you think it fails? ${}{}{}$2012-06-07
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    One should think it fails because Goldbach's Conjecture is unsolved.2012-06-07
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    Not so. It might work, but nobody has yet been clever enough to see how to carry out the induction step.2012-06-07
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    Leaked from the $\epsilon_0$ edition of *Proofs from the Book:* "I have discovered a truly marvelous proof of Goldbach's conjecture. But, alas, induction up to $\epsilon_0$ is too large to fit in Peano arithmetic."2012-06-07
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    I have that edition, it is superb. Much better than the edition preceding it!2012-06-07
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    It might also be possible to prove Goldbach's conjecture using modular representation theory of finite groups, but then again it might not. Why *should* one think a proof by induction is possible?2012-06-07
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    As far as I know, there is no proof that induction fails.2012-06-07
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    One does not need a proof that certain approaches fail to come to the conclusion that certain approaches may well be a waste of time.2012-06-07
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    @Number Do you want to say that a proof via transfinite induction exists ? Or do I misunderstand your comment ? Or do you refer to Fermat with his last theorem ?2018-08-09

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Let's prove the conjecture by induction.

Claim: For every even number $n≥4$, there exist primes $p$ and $q$ such that $p+q=n$.

Base case: $n=4$. Let $p=q=2$.

Induction step: Say that we know that the claim is true for every even number $k ≤ n$. We would like to prove that it is true for $n+2$ as well.

We have available for each even number $k≤n$ two primes, $p(k)$ and $q(k)$, with $p(k)+q(k) = k$. We need to find prime numbers $p$ and $q$ with $p+q = n+2$.

At this point I do not know how to proceed. Please help me out. How can I construct the desired $p$ and $q$ here?

Perhaps it can be done. But as far as I know nobody has yet thought of a way to do it.

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    Why can't we say that we know that n+2 is even, so the induction hypothesis applies, so the conjecture being true for n (n is even) implies that it's true for n+2 since n+2 is also even?2018-11-05
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    The induction hypothesis in my proof is that the conjecture is true for every even number **up to** $n$. You cannot immediately conclude from this that it is true for $n+2$ also.2018-11-07