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 n2− m2)
ii.) (n+m)∙[(n−m−1)!]4
≡ (n+m) (mod n2− m2)
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 [n2− m2].
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 [n2− m2] 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 [n2− m2],
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 [n2− m2]). 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 [n2− m2]) implies (n−m)∙[(n+m−1)!]4
+ (n+m)∙[(n−m−1)!]4
− 2∙n ≡ [0 or (n2− m)2]
(mod 2∙[n2− m2]), 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∙[n2− m2])
if n+m and n−m are both prime or both 1.
PROOF:
i.) [n2− m2] is
even. Then [n2− m2]
= 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 [n2− m2] is
odd. Therefore, both terms are
equivalent to 0.
iii.) n−m = n+m = 1. Then the
terms are even while [n2− m2] 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 [n2− m2]
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∙[n2− m2]).
PROOF:
Since
(n−m)∙[(n+m−1)!]4 + (n+m)∙[(n−m−1)!]4
≠ 2∙n (mod
2∙[n2− m2]) 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∙[n2− m2]),
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∙[n2− m2]) 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∙[n2− m2]).
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∙[n2− m2])
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∙[n2− m2], 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 (n2− m2 + 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:
(n2−
m2 +
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∙(n2−
m2 + 1) ≡ 2∙n (mod 2∙[n2− m2])
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·[n2‒m2]) =
f([n−m]∙([n+m−1]!)4 + [n+m]∙([n−m−1]!)4 ,
2∙[n2−m2]) =
f(2∙n , 2∙[n2− m2]), 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∙[n2−m2]) ≠ 0 for any case of
m. Therefore, f(2∙n∙x
, 2∙[n2−m2]) = f(2∙n , 2∙[n2−m2]) 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 [n2− m2 +
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
= (n2 – m2 + 1) = (n±m+1) for n±m,
odd primes.
Finally, despite n (n≥2), somewhere over the range of m
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
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