Take the 2-minute tour ×

Can anyone clarify why induction method fails for this conjecture?

share|improve this question
One should think it fails because Goldbach's Conjecture is unsolved. –  rschwieb Jun 7 '12 at 11:37
Have you tried to apply induction? If you try you should immediately detect the difficulties. –  rschwieb Jun 7 '12 at 11:37
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. –  Simon Markett Jun 7 '12 at 11:40
ok. Will try to apply. :) –  john Jun 7 '12 at 11:42
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. –  Amitesh Datta Jun 7 '12 at 12:06
show 8 more comments

1 Answer

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.

share|improve this answer
add comment

Your Answer


