# Regarding the Goldbach conjucture

My question is regarding the Goldbach conjecture. The conjecture states that all even numbers can be expressed as a sum of two primes.

So in order to prove the conjecture, is it enough to show that all prime numbers are odd numbers, i.e., that all numbers which cannot be factorized are always odd numbers, since all even numbers can expressed as a sum of two odd numbers?

Alright, I get the logic, but can anyone prove that all prime numbers are odd numbers.

• You are making logical mistake. The fact that all primes except 2 are odd does not imply that all even numbers can be written as a sum of some primes. – Wojciech Karwacki Jan 11 '16 at 6:42
• If it were enough, it wouldn't be unsolved. – fosho Jan 11 '16 at 6:43
• The integers of form $2^k+1$ for $k=1,2,3...$ are all odd, and start out $3,5,9,17....$ yet even though $16$ is even it is not a sum of two of these particular odd numbers. – coffeemath Jan 11 '16 at 6:58
• Think about it for a second -- if a number is not odd, then it is even, so it is divisible by two and thus it cannot be a prime number (the number 2 being the only exception here of course). So clearly all prime numbers (except 2) must be odd. – ASKASK Jan 11 '16 at 7:25

• "All primes are odd numbers" $\Rightarrow$ "Only even numbers are the sum of two primes"
• "All primes are odd numbers" $\not\Rightarrow$ "All even numbers are the sum of two primes"

As a side-note, keep in mind that $2$ is also prime, hence not all primes are odd numbers.

Just because every sum of pairs of primes (other than 2) is even, does not mean that every even number is the sum of a pair of primes (other than 2).

Analogously, every square is a rectangle - but not every rectangle is a square.

For a less silly analogy, consider the "Square Goldbach Conjecture" - that every even number is the sum of the squares of two primes. The sum of the squares of two primes (other than 2) is still even, so your argument would apply - but it's easy to check that e.g. 6 is a counterexample.

"Alright, I get the logic, but can anyone prove that all prime numbers are odd numbers."

Counter example, 2 is prime.

Now excluding two, suppose you have some prime $p$ not equal to 2 that is even. Because $p$ is even it can be written in the form $2k$. But then 2 divides 2k implies 2 divides $p$ which is a contradiction to $p$ being prime. So all primes not equal to 2 must be odd.

Define the set P' = Primes / { 3 }, that is all primes other than 3. Your argument would apply exactly the same to P'. But 6 is not the sum of two elements of P'. On the other hand, if we added a few even numbers then the result wouldn't change.

At the very least you need not just that the set of primes contains all odd numbers (except p = 2), but a lot of odd numbers.

No, the set $A=\{3\}$ also have those properties, but not all even numbers can be expressed as a sum of two from that set.

That is the problem here is that while all primes (except 2) are odd it doesn't mean that all odd numbers are prime - which your argument would require.

Since the conjucture is not to my knowledge not proven, yet it has been around for a couple of hundred years. It's probably due to that it's not that easy to prove (or disprove).