PROOF OF THE GOLDBACH CONJECTURE
By
Kerry M. Evans
Dedicated to the dear memory of my Father
Dr. Roy L. Evans
Two-hundred, fifty years have lapsed since Christian Goldbach proposed (in 1742) that every positive even Integer, not 2, can be expressed as the sum of two primes. This manuscript attempts to show that the proposition is correct as a consequence of Wilson’s Theorem. Little effort will be required to understand the proof as it is not complicated. Many attempts have been made to solve the problem over the years, so that the author must acknowledge that it is only by a stroke of good fortune that this “discovery,” as simply and straightforwardly as it presents itself, is the result only of astute observation. The method of proof is simple once the preliminary groundwork is presented.
SECTION I.
The first section is begun with the statement of a version of Wilson’s Theorem . As the method rests nearly entirely on the development of this theorem, there must be little doubt as to the few limitations of the theorem. Wilson’s Theorem has been known for a long time and its power as a two-sided implication of primality is sorely overlooked.[Key to conventions follows proof].
Theorem I-a. Wilson’s Theorem.
A stricty positive Integer, m, is composite if and only if, [(m-1)!]4 = 0 (mod m) when 0! Is defined 0! = 1.
The real power of this theorem is a result of the fact that if m is a composite number, then and only then (m-1)! = 0 (mod m). The only limitation to this result is that the Integer, 1, may be considered to have a dual nature, i.e. 0! = 1 = 0 (mod 1). This minor obscurity will be circumvented in the succeeding proof so that it in fact presents no hindrance to the accessability to sums of primes. Wilson’s Theorem allows the general definition of the problen of prime sums as follows. Let n be a strictly positive Integer with m a non-negative Integer less than n. Then 2*n is the sum of n-m and n+m, both prime or 1, if and only if the following congruences hold simultaneously.
Congruence I-a.
i.) [(n+m-1)!]4 = 1 (mod n+m)
ii.) [(n-m-1)!]4 = 1 (mod n-m)
The first congruence here is multiplied through by (n-m) and the second multiplied through by (n+m) to obtain an equivalent form of the definition.
Congruences I-b.
i.) (n-m)*[(n+m-1)!]4 = (n-m) (mod n2- m2)
ii.) (n+m)*[(n-m-1)!]4 = (n+m) (mod n2- m2)
If one notes the equivalence of the modulii of these congruences, then it follows that they may be added in the same modulus.
Congruence I-c.
(n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 = 2*n (mod (n-m)*(n+m))
The product of multipliers not 1 or 2 is always greater than their sum. And since (n-m)*[(n+m-1)!]4 and (n+m)*[(n-m-1)!]4 are congruent to 0 modulo [(n-m)*(n+m)] for n-m and n+m when either or both of the latter are composite, 2*n <> 0 + 0 (mod (n-m)*(n+m)), 2*n <> (n-m) + 0 and 2*n <> 0 + (n+m). all taken modulo (n2-m2) for composite and mixed sums. The same facts are true for 2 and any greater positive Integer. If (n-m) is 1 and (n+m) is composite, then 2*n <> (n+m)*(0!) + 0 (mod n^2– m^2). Therefore, since all the cases of composite sums have been considered, the addition of congruences I-b. i.) and ii.), to obtain congruence I-c. holds reversibly.
Lemma I-a.
Let n and m be as defined, then 2*n is the sum of (n-m) ,and (n+m) both either prime or 1, if and only if,(n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 – 2*n = 0 (mod n2 – m2). Since (n-m)*[(n+m-1)!]4 + (n+m)[(n-m-1)!]4 - 2*n = 0 (mod n2 – m2) then (n-m)[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 – 2*n = 0 or (n2 – m2) (mod 2*(n2-m2)). It is therefore necessary to consider the consequences of placing a 2 in the modulus of congruence I-c..
CongruenceI-d.
(n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 - 2*n = 0 (mod 2*(n2 – m2)) .
Cases of Proof :
i.) n2 – m2 is even. Then n2 – m2 = 4 and congruence I-d. is equivalent to 0.
ii.)(n-m) and (n+m) are both odd primes. Then congruence Id. is even while while n2- m2 is odd. Therefore, congruence I-d. is congruent to 0.
iii.)(n-m) = (n+m) = 1. Then congruence I-d. is even while n2-m2 is odd. Therefore, congruence I-d. is congruent to 0.
iv.)(n-m) = 1 while (n+m) equals an odd prime. Then congruence I-d. is odd while n2 - m2 is also odd with 0 even. Therefore congruence I-d. is congruent to n2 – m2. These results lead to the statement of a Lemma.
Lemma I-b.
Let n and m be as defined. Further, let (n-m) = 1. Then (n+m) is an odd prime if and only if,
[(n+m-1)!]4 = (n+m+1) (mod 2*(n+m))
PROOF: According to the previous analysis,(1)*[(n+m-1)!]4 + (n+m)*(0!) - 2*n = n2 – m2 (mod 2*(n2 – m2)= 2*(n+m))
[(n+m-1)!]4 + (n+m)-(n+m+1) = (n+m) (mod(2*(n+m))
[(n+m-1)!]4 = (n+m+1) (mod 2*(n+m))
The proof of the converse follows from lemma I-a..
Lemma I-c.
Let n and m be as defined. Then 2*n is the sum of (n+m) and(n-m), both prime or both 1, if and only if,
(n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-n)!]4 – 2*n = 0 (mod 2*(n2- m2))
PROOF: Since (n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 = 0 (mod 2*(n2 – m2)) are congruent to 0 modulo 2*(n2 –m2) if (n-m) and (n+m) are composite, proof of the converse follows from lemma I-a..
SECTION II.
In this section, a congruence for which the factorials of (n+m) and (n-m) are represented generally in terms of the unknowns, x and y, will be obtained. Due to the observation that this congruence is solvable for x=y (mod 2*(n2 – m2)), x is substituted for y to yield a corresponding congruence that has a solution for x. It will be proven that a general solution for x exists and may be wholly characterized in general terms as to imply that the congruence of lemma I-c. holds for any arbitrary n.
The defining congruences (congruences I-a.) are changed to hold for unknown values of x and y.
Congruences II-a.
i.) [(n+m-1)!]4 = x (mod 2*(n+m))
ii.)[(n-m-1)!]4 = y (mod 2*(n-m))
These simultaneous congruences are multiplied and added as before modulo (2*(n2 – m2)) to obtain a solvable expression for x and y in terms of n and m.
Congruences II-b.
(n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 = (n-m)*x + (n+m)*y (mod 2*(n2 – m2)).
As a consequence of the fact that m may be 0, regardless of which n, implies that x is always evaluated as 0 or 2n, or both, (consider n=1, m=0) regardless of the value of n, which holds in the general case of solutions. Therefore, letting n be arbitrary, somewhere over the range of m it is possible to identify a global solution for x which is defined in terms of 0, 2n, or both.
The question arises as to when the following congruences are solvable and generally speaking, what the value of x must be.
Congruence II-c.
i.) 2*n*x = (n-m)*[(n+m-1)!]4 + (n+m)*[(n-m-1)!]4 (mod 2*(n2–m2))
ii.) x = [(n+m-1)!]4 (mod 2*(n+m))
iii.) x = [(n-m-1)!]4 (mod 2*(n-m))
Let m=0 and n equal 1,2, or 3. For all of these values, x takes on the 2n case of the congruence. There is, however, no conclusion that m is necessarily 0, only that the 2n solution of x corresponds to a generalization of Theorem I.-c..Now let m<>0, n=2 or n=3. For no value of m (1 or [n-1])does x assume a value of 0. Thus the solution of x in terms of two primes is identical, having excluded the x=0 solution and otherwise generalized the default result.
Again, the solutions for x for, n=1, n=2 or n=3 suffice to wholly characterize a general solution for x. These solutions are definitive, modulo 2*(n2 – m2). Specifically because x is not solved in sums to 0 for these n where all possible terms except the 2*n solution, as such, limit the extent of possible evaluations of the general case.
For all n then, x is either (n2 – m2 + 1) or 1 for some value of m. If n is 3 or greater, the solution is in odd primes, since n2 – m2 + 1 = (n+m+1) (mod (2*(n+m)).
Once again, in no case should it be taken so as to imply that m is necessarily 0. As noted earlier, while m=0 implies that an evaluation for x exists in terms of 2*n it is only as m ranges and such solution may occur for a value of (n-m), 3 or greater, where n=4 or greater such that m<>0.
Substitution of the general solution for x in congruence II-c. yields the following theorem.
Theorem II-a. Goldbach’s Theorem.
Let n and m be as defined. Then the congruence,
(n-m)*[(n+m-1)!]4 + (n+m)* [(n-m-1)!]4 = 2*n*1 =2*n*(n2 – m2 + 1) = 2*n (mod 2*(n2 – m2)) holds for all n.
The representation of Goldbach’s Theorem in terms of a congruence is not rigorously correct. (For example, |m| < n has no meaning relative to an implicitly defined congruence.) However, it is possible, by means of the function, {z}, where {z} represents the greatest Integer in z, to place the necessary relations in explicit equality.
Theorem II-b.
Let f(r,s) = r – s*{r/s}. Then the following are solved for some value of m, an Integer, for all defined values of n, a natural number, where 2*n is the sum of (n+m) and (n-m), both prime.
.
i.)f([n-f(|m|,n)]*([(n+f(|m|,n)-1]!)4) + [n+f(|m|,n)]*([n-f(|m|,n)-1] !)4 , 2*|n2-(f(|m|,n))2|) = 2*n
As before, n, natural, m=0, implies f(2*n*x ,2*|n2 – (f(|m| , n))2|) = 0 or
f(2*n*x , 2*|n2-(f(|m| , n))2| ) = 2*n, where the solution of x satisfies the following equalities:
ii.) x = f(([n+f(|m|,n) – 1]!)4 , 2*|n+f(|m|,n)|)
iii.) x = f(([n-f(|m|,n) – 1]!)4, 2*|n-f(|m|,n)|),
although for n=2 and n=3, f(2*n*x, 2*|n2– (f(|m|,n))2 |) <>0. Therefore, f(2*n*x,2*|n2 – f( |m|,n)2 |) = 2*n for all n. And consequently, the solutions of x in ii,) and iii,) are (n+m+1) and (n-m+1), respectively for (n+m) and (n-m) are prime. QED
KEY
|x| denotes the absolute value of x
* denotes multiplication
{x} denotes the greatest Integer in x when x = |x|
a<>b denotes a not equal b
^ denotes exponentiation in the absence of superscripts
Kerry M. Evans
Email: [email protected]
See: geocities.com/kerryme47714/
Also: geocities.com/kerrymerry2000/