# Help! What is wrong with this proof?

1. Oct 3, 2011

### mathew3

An Arithmetic Solution to the Goldbach Conjecture
Prove that any and every even integer >4 may be expressed as the sum of at least some 2 prime integers.

Proof D.
1. We may regard prime integer multiplication as being equivalent to prime integer summation, i.e. 2x3=2+2+2
2. Therefore, given the Fundamental Theorem of Arithmetic, any and every even integer >4 may be expressed as the summation of a series of prime integers, i.e.,
I = Pa+…+Pb+…+Pc+…
where I is any integer >1 and P is some prime integer
3. Any and every even integer must equal the summation of some two odd integers, therefore
E = Oa +Ob
where E is any and every even number and O is some odd integer.
4. Given the FTOA it must also be the case that for any even integer >4

E = Pa+…+Pb+…+Pc+…

5. Therefore for any and every even integer >4

Pa+…+Pb+…+Pc+…= E = Oa +Ob [4]
6. Therefore the sum Oa + Ob must equal a summation of a series of primes.
7. Since there are at least two addends comprising the Oa +Ob summation then each addend is allowed to be a prime number.
8. E, in this case, must meet two conditions:
a. E must be composed of 2 and only 2 odd integers.
b. E must be a summation of primes.
9. In order to satisfy both conditions a and b then it must be the case that the two odd integers, Oa and Ob must sum as primes where

Pa+Pb = E= Oa +Ob [5]
10. Therefore any and every even integer >4 may be expressed as the sum of at least some 2 prime integers.

2. Oct 3, 2011

### pwsnafu

9 does not follow from 8.

Eg. Lets do 12. 12 = 22 times 3 = 22 + 22 + 22 = 2 + 2 + 2 + 2 + 2 + 2.

Now write 12 as a sum of two odd integers. I can choose anything. 12 = 9+3.
So
2 + 2 + 2 + 2 + 2 + 2 = 9 + 3

If 9 is to be true we get 2 = 9 = 3, which is nonsense.

3. Oct 3, 2011

### mathew3

Your reply is of course true on its face. However, it seems that this implies that 8 is false or at least the two conditions cannot exits simultaneously. Otherwise 9 has to follow, or so it seems to me. So why is 8 wrong?

4. Oct 3, 2011

### pwsnafu

Here is 8a.
12 = 3 + 3 + 3 + 3

Edit: Further, take a closer look at what I previously wrote:
12 = 2 + 2 + 2 + 2 + 2 + 2. This satisfies 8b, but not 8a.
12 = 9+3. This satisfies 8a, but not 8b.

Last edited: Oct 3, 2011
5. Oct 3, 2011

### mathew3

Again you are correct but your cited case doesn't apply in this case. Only the 2 integer set is under consideration. eg.
1. Fact: I have 1 dollars worth of change in my pocket. That could be 100 pennies or 10 dimes... up to or 1 silver dollar. (E)
2. Fact: I only have two coins in my pocket. (Oa +Ob)
3. Conclusion: Each coin is a half dollar (Pa+Pb=Oa +Ob)

6. Oct 3, 2011

### pwsnafu

Please quote where precisely you require the two integer set as a requirement.

Edit: look, Steps 1, 2 and 4 talk about one possible way to write out the sum, in this case every term is prime. Step 3 describes a different sum, in this case only two odd numbers. At no point in Step 3 do you require primality, and at no point in 1 do you require the number of terms to be two. You can't just say at Step 9 "oh, they were the same sum all along".

7. Oct 3, 2011

step3

8. Oct 3, 2011

### pwsnafu

Here is Step 3 again

9. Oct 3, 2011

### mathew3

I find myself agreeing with most of your statements. I wholeheartedly agree with you up to your last sentence. I'm pretty sure equation [4] is pretty much irrefutable. The question is is there anything that requires that the the number of prime addends must be greater than 2. It seems to me equation [4] indicates it is true, covering a range of 2 to 300,000 (Schnirelmann?) prime addends. The conjecture doesn't say 5 primes or ten primes or a thousand primes but 2 primes. Equation [4] certainly allows this. Equation [5] states this.
Again, [5] doesn't demand that the two odds must always and exclusively be primes.(Neither does the conjecture) But it does demand that there must always be a summation of primes that equals the summation of the two odds. Here I bring up the loose change analogy once again.

10. Oct 4, 2011

### pwsnafu

*facepalm*

From http://en.wikipedia.org/wiki/Goldbach%27s_conjecture" [Broken]:
Now that I look back at your first post
I should have picked that up on the first read.

Last edited by a moderator: May 5, 2017
11. Oct 4, 2011

### mathew3

Oh, by the way. The greater than 2 holds if we consider 1 as a prime, which was the case when the conjecture was first presented. The greater than 4 holds if 1 is considered as non prime which is the case in modern number theory.

12. Oct 4, 2011

### ramsey2879

I don't get what you are saying. The Goldbach conjecture required that every even number greater than 4 be the sum of two odd primes. You haven't proven that by statements 8 and 9 since a string of more than 2 primes does not equate to a string of 2 primes.

13. Oct 4, 2011

### mathew3

Fact: every even number>4 must be the sum of some 2 odd integers
Fact: every even number >4 must be the sum of at least 2 primes

Before I go further are we agreed on these 2 facts?

14. Oct 4, 2011

### ramsey2879

yes but 1) what do you mean by "some 2 odd integers?" and 2) how do you get to prove the Goldbach conjecture from that?

Last edited: Oct 4, 2011
15. Oct 4, 2011

### mathew3

Great. We are agreed on those two, I think. It is critical to the proof that we establish than any and every even number greater >4 is the sum of some 2 odd numbers. I don't know the theorem offhand but this has to be true.

The proof proceeds in 4 critical steps
a. establish that any even integer>4 can be composed of at least 2 prime addends; thus step 2.
b. establish that any even integer>4 can be composed of 2 odd integers; thus step 3.
c. equate a and b. Thus step 5, equation [4]
d. By considering the special case where our number of odd integer addends is restricted to 2 and only 2 addends then this imposes steps 9, equation [5]. Again I refer you to the previous loose change analogy.

Thus any even integer>4 can be expressed as the sum of two primes.

The proof is remarkably simple. The problem seems to be is it logically legitimate to advance from equation [4] to equation [5]. I think it is but I would of course like to hear reasons to the contrary.

16. Oct 4, 2011

### ramsey2879

You have Pa + Pb + ... Pe = Oa + Ob , How do you get from that to Pa + Pb = Oa + Ob??? Step 5 does not do that.

17. Oct 4, 2011

### mathew3

By restricting our numbers of addends to be 2 and only 2 addends then the number of primes are necessarily restricted to 2 primes thus Pa + Pb = E= Oa + Ob.

18. Oct 4, 2011

### ramsey2879

You can't do that without first establishing there corresponds at least one set of 2 prime addends that add up to each possible value of Oa + Ob. Thus you have no proof of the conjecture.

19. Oct 4, 2011

### pwsnafu

Except that it's not an analogy. Or at least an analogy relevant to the problem. And anytime someone resorts to analogies rather than straight logic to prove mathematics, it's time to walk away.

The "loose change analogy" relevant to this thread is:
1. Fact: I have 1 dollars worth of change in a pocket. That could be 100 pennies or 10 dimes... up to or 1 silver dollar. (E)
2. Fact: I only have two coins in a pocket. (Oa +Ob)

At no point do you say the change is in the same pocket. See post #6.

20. Oct 4, 2011

### mathew3

Step 3 and the two equal signs allows me to place this restriction.