Kerry Evans

617 Washington Ave., Apt. A

Evansville, IN 47713

812-647-4938

mrevanskme@hotmail.com

Proof of the Goldbach Conjecture

by

Kerry M. Evans

Dedicated in dear memory of my Father

Dr. Roy L. Evans

Nearly two-hundred, seventy 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 shows 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 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 its’ few limitations. 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 positive Integer, q, is
composite if and only if, [(q−1)!]^{4} ≡ 0
(mod q) when 0! is defined 0! = 1. Otherwise [(q−1)!]^{4} ≡1
(mod q).

The real power of this theorem is that it unambiguously defines composite Integers as distinct from primes. The only limitation to this result is that the Integer, 1, while definitely composite, 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. It subsequently presents no hindrance in accessibility to sums of primes. Wilson’s Theorem allows the general definition of the problem of prime sums as follows.

Let n be a strictly positive Integer with m a nonnegative Integer less than n. Then 2∙n is the sum of n−m and n+m, prime or 1, if and only if the following congruencies hold simultaneously.

Congruencies I-a.

i.)
[(n+m−1)!]^{4} ≡ 1 (mod n+m)

ii.)
[(n−m−1)!]^{4} ≡ 1 (mod n−m)

The first congruence is multiplied through by n−m and the second multiplied through by n+m to obtain an equivalent form of the definition.

Congruencies I-b.

i.)
(n−m)∙[(n+m−1)!]^{4} ≡ (n−m) (mod n^{2}− m^{2})

ii.)
(n+m)∙[(n−m−1)!]^{4} ≡ (n+m) (mod n^{2}− m^{2})

Noting the equivalence of the moduli of these congruencies, arguments are added to inherit their simultaneous properties in a single expression.

Lemma I-a.

Suppose n and m are 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 (mod [n^{2}− m^{2}].

PROOF:

The product of multipliers
not 1 or 2 is always greater than their sum. Therefore
(n−m)∙[(n+m−1)!]^{4} and (n+m)∙[(n−m−1)!]^{4} are
congruent to 0 modulo [n^{2}− m^{2}] when
n+m and n−m, respectively, are independently composite (besides
1). Thus, 2∙n ≢ 0 + 0, 2∙n ≢ (n−m) + 0 and 2∙n ≢ 0 + (n+m), all
taken modulo [n^{2}− m^{2}], for composite and partially composite terms not
1. If n−m is 2 and n+m is a greater even composite then 2∙n ≢
(n+m)∙1 + 0 (mod [n^{2}− m^{2}]). The same fact applies to 1 and any greater odd
Integer. Since all relevant combinations of composite terms have
been considered, the addition of arguments of congruencies I-b. i.) and ii.),
to obtain the congruence of Lemma I-a holds reversibly.

Because (n−m)∙[(n+m−1)]^{4} +
(n+m)∙[(n−m−1)!]^{4} − 2∙n ≡ 0 (mod [n^{2}− m^{2}])
implies (n−m)∙[(n+m−1)!]^{4} + (n+m)∙[(n−m−1)!]^{4} −
2∙n ≡ [0 or (n^{2}− m)^{2}] (mod 2∙[n^{2}− m^{2}]), the
result of multiplying 2 in the modulus of Lemma I-a. has significant consequence.

Lemma I-b.

(n−m)∙[(n+m−1)!]^{4} +
(n+m)∙[(n−m−1)!]^{4} − 2∙n ≡ 0 (mod 2∙[n^{2}− m^{2}]) if n+m
and n−m are both prime or both 1.

PROOF:

i.) [n^{2}− m^{2}] is even. Then [n^{2}− m^{2}] = 4 and
both terms (of Lemma I-b.) are equivalent to 0 (across sides).

ii.) n−m and n+m are both odd primes. Then the terms
are even, while [n^{2}− m^{2}] is odd. Therefore, both terms are equivalent to 0.

iii.) n−m = n+m = 1. Then the terms are even while [n^{2}− m^{2}] is
odd. Therefore, both terms are equivalent to 0.

iv.) n−m=1 while n+m equals an odd prime. Then the term
left of equivalence is odd as [n^{2}− m^{2}] not
0. Thus the congruence fails.

Similar considerations lead to the statement of a Lemma.

Lemma I-c.

p is an odd prime if and
only if, [(p−1)!]^{4} ≡
(p+1) (mod 2∙p) since [(p−1)!]^{4} is
even. Proof of the converse follows from Wilson’s Theorem.

Theorem I-b.

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−1)!]^{4} − 2∙n ≡ 0 (mod 2∙[n^{2}− m^{2}]).

PROOF:

Since (n−m)∙[(n+m−1)!]^{4} +
(n+m)∙[(n−m−1)!]^{4} ≠ 2∙n (mod 2∙[n^{2}− m^{2}]) if n+m
or n−m, inclusively, are composite [unless n±m=1], proof follows similar to
that of Lemma I-a. subject to Lemma I-c..

SECTION II.

In this section, a
congruence for which the factorials relative to n+m and n−m are expressed
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∙[n^{2}− m^{2}]), x is substituted for y to yield a corresponding congruence
that has a solution for x. Proof follows that a general solution for
x exists and may be wholly characterized in general terms to imply that the
congruence of Theorem I-b.. holds for an arbitrary n.

The defining congruencies (congruencies I-a.) are changed to hold for unknown values of x and y.

Congruencies II-a.

i.) [(n+m−1)!]^{4} ≡ x (mod 2∙[n+m])

ii.) [(n−m−1)!]^{4} ≡ y
(mod 2∙[n−m])

These simultaneous
congruencies are multiplied with arguments added as before modulo (2∙[n^{2}− m^{2}]) to
obtain a solvable expression for x and y.

Congruence II-b.

(n−m)∙[(n+m−1)!]^{4} +
(n+m)∙[(n−m−1)!]^{4} ≡ (n−m)∙x + (n+m)∙y (mod 2∙[n^{2}− m^{2}]).

The fact that m may be 0, regardless of which n, uniquely implies that 2∙n∙x (where x ≡ y) is always evaluated in terms of 0, 2∙n or both (i.e. n=1), such that x holds in the universal case of solutions. Therefore, letting n be arbitrary, over the range of m it is possible to identify a global solution for x defined in terms of n and m.

Congruencies II-c.

2∙n∙x ≡ (n−m)∙[(n+m−1)!]^{4} +
(n+m)∙[(n−m-1)!]^{4} (mod 2∙[n^{2}− m^{2}])

Let m=0 and n equal 1, 2, or 3. For all of these values, x is evaluated in terms of the 2∙n case (on the right) of congruence II-c-i.; however, no conclusion follows that 2∙n∙x is necessarily 0, only that the 2∙n case of 2∙n∙x may correspond to a generalization of Theorem I.-b.. So considering, let m vary from 0 with n=2 or n=3. For no value of m (0, 1 or [n−1]) does x assume a value of 0. Thus the solution of x in terms of two primes or mutual terms of 1 is general, excluding the 2∙n∙x ≡ 0 solution and otherwise satisfying the 2∙n∙x ≡ 2∙n default.

Again, the solutions for x
when n≡1, n≡2, or n≡3 suffice to wholly characterize a solution for
x. These particular n are definitive, modulo 2∙[n^{2}− m^{2}],
specifically because 2∙n∙x is not commonly solved in sums for these n in any
possible terms of m except the 2∙n solution, excluding all other possible
evaluations in the general sense.

Thus for each n, x is
either (n^{2}− m^{2 }+ 1) or 1 (among others for n±m=2) for some
corresponding value of m. If n is 3 or greater, the solution is in
odd primes, since:

(n^{2}− m^{2 }+ 1) ≡
(n±m+1) (mod (2∙[n±m]).

In no case should it be taken to imply that m is necessarily 0. As noted earlier, while m=0 implies that a general solution for 2∙n∙x exists in terms of 2∙n, it is only for some value of m as m varies. Such solution may occur for a value of n−m = 3 or greater, where n=4 or greater, such that m≠0, depending on .

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∙(n^{2}− m^{2 }+ 1) ≡
2∙n (mod 2∙[n^{2}− m^{2}]) holds for each n at some m respectively.

The representation of Goldbach’s Theorem in terms of congruence is somewhat conceptually restrictive. Also, n and m cannot be defined positive absent of the specification, n = ∣n∣ and m = ∣m∣. Accordingly, by means of the function, {z}, where {z} represents the greatest Integer in z when z ≥0, the necessary implicit relations can be placed in “virtual” equality consequent to the following explicit conditions.

Theorem II-b.

If f(r,s) =
r−(s∙{r/s}) when r/s ≥ 0, then the following is solved for the __explicit__conditions
on Integers m=∣m∣ and n=∣n∣, where n > m, such that 2∙n is the sum of n+m
and n−m, both prime or both 1, if and only if:

f(2·n·x , 2·[n^{2}‒m^{2}]) =
f([n−m]∙([n+m−1]!)^{4} + [n+m]∙([n−m−1]!)^{4 }, 2∙[n^{2}−m^{2}])
= f(2∙n , 2∙[n^{2}− m^{2}]), where
x satisfies:

i.) x = f([(n+m−1)!]^{4 },
2∙[n+m])

ii.) x = f([(n−m−1)!]^{4 },
2∙[n−m]).

PROOF:

As before, n, arbitrary,
and m variable imply 2∙n∙x = 0 or 2∙n∙x = 2∙n, (m may be 0). Thus for n=2 and
n=3, f(2∙n∙x , 2∙[n^{2}−m^{2}]) ≠ 0
for any case of m. Therefore, f(2∙n∙x , 2∙[n^{2}−m^{2}]) =
f(2∙n , 2∙[n^{2}−m^{2}]) holds for all n, at some instance of
m. Consequent to explicit conditions, n > m ≥ 0 and
2∙n = (n+m) + (n−m) , the latter are well-defined and exclude any other
implicit unequal values. The solution of x (1 or [n^{2}− m^{2 }+ 1]
among others for n±m=2) in i.) and ii.) above is x= 1
for n±m=1 and n±m= 2. Otherwise, x = (n^{2 }– m^{2} +
1) = (n±m+1) for n±m, odd primes.

Finally, despite n (n≥2), somewhere over the range of m an eligible pair of primes exist that satisfy general solution of in Theorem II-b.. 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 equivalent to b

a ≡ b denotes a congruent to b

Beal: http://beal.yolasite.com

Fermat: http://fermat.yolasite.com

Goldbach: http://goldbach.yolasite.com

ADDENDUM

Let n>m>0. Let n+m be composite. Let n-m be prime. Suppose (2m+1)! Is relatively prime to n+m. Solve the following congruences (Chinese Remainder Algorithm) for x and y.

(2m+1)!*x = 1 mod (n+m)

(2m+1)!*y = 1 mod (n-m)

Let y minus x = z. Then,

(2m+1)! = t mod (n+m)

(2m+1)! = u mod (n-m)

[t*y – 1] = r mod (n+m)

[(u*x – 1] = s mod (n-m)

Then r minus s minus z where any of r, s or z is 0 fails to be relatively prime to n+m and not a multiple of n+m when x, y, t, u, r and s are least positive and where any but not all of r, s or z may be 0 instead.

This is true except when u*y = v mod (n+m) and v minus 1 fails to be relatively prime to n+m and is not a multiple of n+m or when r minus s minus z is 0. Similarly for t*x = w mod (n-m). So that w minus 1 fails to be relatively prime to n+m but is not a multiple of it. v and w are least positive.

KEY

*denotes multiplication

=denotes congruence