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Partial Proof of the Goldbach Conjecture
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Research
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By Goldbach
Posted Mon Oct 31, 2005 at 09:18:53 AM PDT
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The Goldbach Conjecture can be stated as follows: for every natural number m there exist primes p and q such that T(p)+T(q)=2m, where T(N) is the totient function (the number of relative primes to N less than N; relative primes are numbers that do not share any common divisors, i.e., 9 and 4.) By definition, the totient value of a prime p is p-1, i.e., T(p)=p-1. For instance, let m=1; then T(p)+T(q)=2, (p-1)+(q-1)=2, p+q=4, and since both p and q are prime, p=q=2. Now let us examine the case p=2 and q is odd. We see that T(p)=T(2)=2-1=1, so the Goldbach Conjecture (GC) becomes T(q)=2m-1. Since q is odd, T(q)=q-1 is even; but 2m-1 must be odd (definition of odd.) Therefore the GC does not hold in this case. In other words, it has been shown that the Goldbach Conjecture can be stated as follows: all even numbers greater than 4 are the sum of 2 odd primes. I will proceed by showing all the possible values of m that I have been able to prove.
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| The Goldbach Conjecture can be stated as follows: for every natural number m there exist primes p and q such that T(p)+T(q)=2m, where T(N) is the totient function (the number of relative primes to N less than N; relative primes are numbers that do not share any common divisors, i.e., 9 and 4.) By definition, the totient value of a prime p is p-1, i.e., T(p)=p-1. For instance, let m=1; then T(p)+T(q)=2, (p-1)+(q-1)=2, p+q=4, and since both p and q are prime, p=q=2. Now let us examine the case p=2 and q is odd. We see that T(p)=T(2)=2-1=1, so the Goldbach Conjecture (GC) becomes T(q)=2m-1. Since q is odd, T(q)=q-1 is even; but 2m-1 must be odd (definition of odd.) Therefore the GC does not hold in this case. In other words, it has been shown that the Goldbach Conjecture can be stated as follows: all even numbers greater than 4 are the sum of 2 odd primes. I will proceed by showing all the possible values of m that I have been able to prove.
The case p=q is a conditional statement, since this implies T(p)+T(q)=2m, 2T(p)=2m, T(p)=p-1=m. It is obvious that p-1 is a natural number for all p; because of this, there exist p=q for all m of the form T(p)=m such that T(p)+T(q)=2m.
The case m=p is rather interesting since it brings in the concept of twin primes (two numbers are called twin primes, or simply twins, if both p and p+2 are primes.) So we make the following deductions: T(p)+T(q)=2m, (p-1)+T(q)=2p, q-1=p+1,q=p+2. Again, this is a conditional statement that is satisfied by certain p=m,q but not all p=m,q. So the GC is true for m=min{p,q} and p,q are twins; in other words, I have shown the GC to be true if m is the smaller prime in a set of twin primes or if m=T(p).
Of course, there are an infinite number of cases that could be proven (for example, try to prove that the GC holds for all m=p-q.) The question is which cases need to be proven to show that the GC is true for all natural numbers. Any suggestions as to this would be greatly appreciated. |
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