Goldbach Conjecture

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paras007

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Can anyone tell whether my proof is correct?
Let x and y be two non-prime odd numbers and z be any even number.

Now,

every even number can be expressed as the sum of two odd numbers,

Therefore,

x+y=z

Let k be the common factor of x and y.

k is odd _(since x and y are odd)

Therefore,

k(x'+y')=z

=> x'+ y'= m _( since k is a factor of z)

m is even _(since even÷odd=even)

Now x' and y' are prime numbers and m is even.

Hence, every even number can be expressed as the sum of two prime numbers.
 
I see two holes in your proof:
  1. Generally, it is not true that x' and y' are prime. You can me them co-prime if you choose the greatest common factor for 'k'.
  2. While 'z' is any even number your 'm' is not arbitrary but specific to your construction. I.e., you have to prove that any even 'm' can end up in the middle of your proof.
 
1)x' and y' are always prime, that's not my point. But for even number on the right hand side, you can find a pair of prime numbers using this procedure.
2) m is z/k, if z is arbitrary, then z/k is also arbitrary.
 
paras007 said:
1)x' and y' are always prime, that's not my point. But for even number on the right hand side, you can find a pair of prime numbers using this procedure.
2) m is z/k, if z is arbitrary, then z/k is also arbitrary.
Click to expand...
I'll add an illustration to my previous point: z = 38, and since you proof assumes arbitrary choice of x,y we'll set x = 3, y = 35. You end up since k = 1 you end up with x' = x = 3, y'=y = 35, but y is not prime.
 
Take x=21, y=93, z=114
Then
x'=7,y'=31,k=3,z=38
It will work for every z.
 
In math examples cannot be used to prove a statement, but counterexamples can be used to disprove it.
 
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