About an even as the sum or the difference of two primes
Jamel Ghannouchi
To cite this version:
Jamel Ghannouchi. About an even as the sum or the difference of two primes. Bulletin of Mathematical Sciences and Applications (www.bmsa.us), 2013, 2 (3), pp.44-52. <hal-00862954v3>
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About an even as the sum or the difference of two primes
Jamel Ghanouchi
Abstract
(MSC=11) The present algebraic development begins simply by an exposition of
the data of the problem. Our calculus is supported by a reasoning which must
lead to an impossibility. We define the primal radius : For all x an integer greater
or equal to 3, we define a primal number r for which x−r and x+r are prime numbers. We see then that Goldbach conjecture would be verified because 2x = (x +
r) + (x − r). We prove the existence of r for all x ≥ 3. We prove also the existence,
for all x′ an integer, of a primal radius r ′ for which x′ +r ′ and r ′ −x′ are prime numbers strictly greater than 2. De Polignac conjecture would be quickly verified because 2x′ = (x′ +r ′ )−(r ′ −x′ ).
Introduction
Goldbach and de Polignac conjectures seem actually impossible to be solved. Everyone has remarked the similarity between them. One stipulates that each even
is the sum of two primes when the other stipulates that it is always the difference of two primes and that there is an infinity of such couples. In fact, we have
used that similarity to solve those very old problems, we have considered in this
research the conjectures as two faces of the same problem. For this, we have defined a notion : the primal radius. It allowed finally to prove the conjectures.
The Goldbach conjecture
Goldbach conjecture, fruit of personal works and correspondences between the
mathematicians of the XVIII century (Leonard Euler, who was born exactly 300
years before 2007, was one of them), stipulates that an even number is always
equal to the sum of two prime numbers. Let us suppose there exists an integer x ;
x ≥ 3, for which, for all p1 ≥ 3, p2 ≥ 3 prime numbers b 6= 0. We can pose for p1
and p2 distinct prime numbers verifying p1 > x > p2 .
2x = p1 + p2 + 2b
b depends of p1 and p2 . Then
x=
p1 + p2
+b
2
1
And for all x, p1 , p2 , exists y whose expression is
y=
p1 − p2
+b
2
We pose

x1



x2
x


 3
x4
= p1 + 2b
= p2 − 2b
= p2 + 2b
= p1 − 2b

2
2
2
2
x = p1 +p
+ b = p1 +x
+ 2b = x1 +p
= x1 +x
+b

2
2
2
2


p1 +x3
x1 +x3
x4 +x3
x4 +x2

 = 2 = 2 − b = 2 + b = 2 + 3b
2
2
2
2
⇒
y = p1 −p
+ b = p1 −x
= x1 −p
= x1 −x
−b
2
2
2
2

p1 −x3
x1 −x3
x4 −x3

2
 = 2 + 2b = 2 + b = 2 + 3b = x4 −x
+b

2

x1 + x2 = p 1 + p 2
LEMMA 1 The following formula

2
2
2
2
x = p1 +p
+ b = p1 +x
+ 2b = x1 +p
= x1 +x
+b

2
2
2
2


p
+x
x
+x
x
+x
x
+x
1
3
1
3
4
3
4
2

=
=
−
b
=
+
b
=
+
3b

2
2
2
2
p1 −x2
x1 −p2
x1 −x2
2
y = p1 −p
+
b
=
=
=
−b
2
2
2
2

p1 −x3
x1 −x3
x4 −x3
x4 −x2


=
+
2b
=
+
b
=
+
3b
=
+b

2
2
2
2

x1 + x2 = p 1 + p 2
imply that ∃p1 , p2 prime numbers which verify
b=0
. In fact, our hypothesis (b 6= 0 for all p1 , p2 ) can not imply b = 0 without any
condition. If the calculus leads to b = 0, the hypothesis is false because the formal
calculus can not lead to b = 0.
Proof of lemma 1 If x is a prime number, 2x = x + x is the sum of two primes,
then
p1 − p2 6= 0
we will suppose firstly that
(x1 − x2 )(x1 + x3 ) 6= 0
let
(
x1 −x2
p1 −p2
p1 −p2
x1 −x2
=
=
p1 −p2 +4b
p1 −p2
x1 −x2 −4b
x1 −x2
= 1 + p14b
−p2
= 1 − x14b
−x2
we pose
(
k = p12b
−p2
k′ = − (p1 −p2b2 +4b)
if
kk′ = 0 ⇒ b = 0
2
we will suppose
kk′ 6= 0
But ∀x, y, ∃φ verifying x = φy
x + y = (φ + 1)y = x1 6= 0
x − y = (φ − 1)y = p2 6= 0
and ∀k, k ′ , ∃α verifying k = αk ′
⇒k=
−2b
2b
=α
p1 − p2
x1 − x2
⇒ x1 − x2 = −α(p1 − p2 )
⇒ x1 − x2 − p1 + p2 = 4b = −(α + 1)(p1 − p2 )
−(α + 1)
(p1 − p2 )
4
p1 + p2 α + 1
p1 + p2
+b=
−
(p1 − p2 )
⇒x=
2
2
4
p1 − p2
p1 − p2 α + 1
⇒y=
+b=
−
(p1 + p2 )
2
2
4
⇒b=
or
x=
(1 − α)p1 + (3 + α)p2
φ
=
p2
4
φ−1
and
y=
let
(
(1 − α)(p1 − p2 )
1
=
p2
4
φ−1
x1 +x3
p1 +p2
p1 +p2
x1 +x3
=
=
p1 +p2 +4b
p1 +p2
x1 +x2 −4b
x1 +x3
= 1 + p14b
+p2
= 1 − x14b
+x3
we pose
(
m = p12b
+p2
m′ = − x12b
+x3
mm′ 6= 0
and ∀m, m′ , ∃β verifying m = βm′
⇒m=
−2b
2b
=β
p1 + p2
x1 + x3
⇒ x1 + x3 = −β(p1 + p2 )
⇒ x1 + x3 − p1 − p2 = 4b = −(β + 1)(p1 + p2 )
−(β + 1)
(p1 + p2 )
4
p1 + p2 (β + 1)
p1 + p2
+b=
−
(p1 + p2 )
⇒x=
2
2
4
⇒b=
3
⇒y=
p1 − p2 (β + 1)
p1 − p2
+b=
−
(p1 + p2 )
2
2
4
thus
x=
1−β
1−β
φ
p1 +
p2 =
p2
4
4
φ−1
y=
β+3
1
1−β
p1 −
p2 =
p2
4
4
φ−1
we resume
x=
1−β
1−β
φ
(1 − α)p1 + (3 + α)p2
=
p1 +
p2 =
p2
4
4
4
φ−1
and
y=
1−β
β+3
1
(1 − α)(p1 − p2 )
=
p1 −
p2 =
p2
4
4
4
φ−1
b=
−(α + 1)
−(β + 1)
(p1 − p2 ) =
(p1 + p2 )
4
4
−2 − α − β
β−α
p1 =
p2
⇒
4
4
β−α
−2 − α − β
⇒
=
p2
p1
but
4x − 4y + 4b = 2p1 + 2p2 + 4b − 2p1 + 2p2 − 4b + 4b = 4p2 + 4b = 4P2
And
4x + 4y − 4b = 2p1 + 2p2 + 4b + 2p1 − 2p2 + 4b − 4b = 4p1 + 4b = 4P1
Thus
4P1 = 4p1 + 4b = 4p1 − (α + 1)(p1 − p2 ) = (3 − α)p1 + (1 + α)p2
= 4p1 − (β + 1)(p1 + p2 ) = (3 − β)p1 − (1 + β)p2 =
1
(6 − α − β)p1 + (α − β)p2 )
2
And
4P2 = 4p2 + 4b = 4p2 − (α + 1)(p1 − p2 ) = −(1 + α)p1 + (5 + α)p2
1
= 4p2 −(β +1)(p1 +p2 ) = −(1+β)p1 +(3−β)p2 = (−(2+α+β)p1 +(8+α−β)p2 )
2
But p1 and p2 are primes.
(β − α)p1 = (−α − β − 2)p2
α+1
)(P1 − P2 ))
4
5+α
1+α
= (β − α)((
)P1 − (
)P2 )
4
4
α+1
)(P1 − P2 ))
= (−α − β − 2)(P2 − b) = (−α − β − 2)(P2 + (
4
= (β − α)(P1 − b) = (β − α)(P1 + (
4
= (−α − β − 2)((
3−α
1+α
)P1 +
P2 )
4
4
Thus
((β − α)(5 + α) + (α + β + 2)(1 + α))P1
= ((−α − β − 2)(3 − α) + (β − α)(1 + α))P2
And
(β − α)P1 = (−α − β − 2)P2 + b(β − α + α + β + 2)
= (−α − β − 2)P2 − (β + 1)(
α+1
)(P1 − P2 )
2
Hence
(2β − 2α + (β + 1)(α + 1))P1
= (−2α − 2β − 4 + (β + 1)(α + 1))P2
Thus
((β − α)(3 + α) + (α + 1)2 )P1
= ((−α − β − 2)(1 − α) − (α + 1)2 )P2
= ((β + 1)(3 + α) + (α + 1)(α + 1 − 3 − α))P1
= (−(β + 1)(1 − α) − (α + 1)(α + 1 + 1 − α))P2
= ((β + 1)(3 + α) − 2α − 2)P1
= (−(β + 1)(1 − α) − 2α − 2)P2
Hence
((β + 1)(3 + α) − 2(α + 1))(p1 + p2 )P1
= (−(β + 1)(1 − α) − 2(α + 1))(p1 + p2 )P2
= ((α + 1)(p1 − p2 )(3 + α) − 2(α + 1)(p1 + p2 ))P1
= (−(α + 1)(p1 − p2 )(1 − α) − 2(α + 1)(p1 + p2 ))P2
= (α + 1)((−α + 1)p1 − (3α + 5)p2 )P1
= (α + 1)(−(α + 3)p1 − (1 + 3α)p2 )P2
If
((1 − α)p1 − (5 + 3α)p2 )P1 + ((α + 3)p1 + (1 + 3α)p2 )P2 = 0
= ((1 − α)p1 − (5 + 3α)p2 )(p1 + b) + ((α + 3)p1 + (1 + 3α)p2 )(p2 + b)
= (1 − α)p21 + (1 + 3α)p22 − 2(α + 1)p1 p2 + 4b(p1 − p2 ) = 0
= −(1 + α)p21 + 2p21 + 2(1 + α)p22 + (α + 1)p22 − 2p22 − 2(α + 1)p1 p2 + 4b(p1 − p2 )
= (α + 1)(−p21 + 3p22 − 2p1 p2 ) + 2(p21 − p22 ) + 4b(p1 − p2 )
= (α + 1)(p2 − p1 )(3p2 + p1 ) + 2(p21 − p22 ) + 4b(p1 − p2 )
= 4b(3p2 + p1 ) + 2(p21 − p22 ) + 4b(p1 − p2 )
= 4b(2p1 + 2p2 ) + 2p21 − 2p22 = 0
It means
4b = p2 − p1 = −(α + 1)(p1 − p2 ) = −(β + 1)(p1 + p2 )
5
⇒ α = 0,
But
β=
−2p2
p1 + p2
4k2 k′2 = (k + k′ )2 = k2 + k′2 + 2kk′
Thus
(4k′2 − 1)k2 − 2k′ k − k′2 = 0
δ = k′2 + k′2 (4k′2 − 1) = 4k′4
k′ − 2k′
−k′
−k′
=
=
4k′2 − 1
4k′2 − 1
2k′ + 1
k=
1 − 2k′ = 1 ⇒ k′ = 0 ⇒ k = 0 ⇒ 4b = p2 − p1 = 0
Impossible, we have supposed x1 − x2 6= 0. And as
((−α + 1)p1 − (3α + 5)p2 )P1 6= (−(α + 3)p1 − (1 + 3α)p2 )P2
We deduce that
α + 1 = 0 ⇒ 4b = −(α + 1)(p1 − p2 ) = 0
P1 and P2 are p1 and p2 and
2x = p1 + p2
Thus α = β = −1. b can not be different of zero. If
(x1 − x2 )(x1 + x3 ) = 0 ⇒ (x4 + x2 )(x4 − x3 ) 6= 0
let
(
x4 +x2
p1 +p2
p1 +p2
x4 +x2
=
=
p1 +p2 −4b
p1 +p2
x4 +x2 +4b
x4 +x2
= 1 − p14b
+p2
= 1 + x44b
+x2
we pose
(
let
(
2k + 1 = 1 − p14b
+p2
2k′ + 1 = 1 + x44b
+x2
x4 −x3
p1 −p2
p1 −p2
x4 −x3
=
=
p1 −p2 −4b
p1 −p2
x4 −x3 +4b
x4 −x3
= 1 − p14b
−p2
= 1 + x44b
−x3
we pose
(
2m + 1 = 1 − p14b
−p2
2m′ + 1 = 1 + x44b
−x3
if
kk′ = 0 ⇒ b = 0
we will suppose
kk′ 6= 0
But ∀x, y, ∃φ verifying x = φy
x + y = (φ + 1)y = x1 6= 0
x − y = (φ − 1)y = p2 6= 0
6
and ∀k, k ′ , ∃α verifying k = αk ′
⇒k=
2b
−2b
=α
p1 + p2
x4 + x3
⇒ x4 + x3 = −α(p1 + p2 )
⇒ x4 + x3 − p1 − p2 = 4b = −(α + 1)(p1 + p2 )
−(α + 1)
(p1 + p2 )
4
p1 + p2 α + 1
p1 + p2
+b=
−
(p1 + p2 )
⇒x=
2
2
4
p1 − p2
p1 − p2 α + 1
⇒y=
+b=
−
(p1 + p2 )
2
2
4
⇒b=
or
y=
(1 − α)p1 − (3 + α)p2
1
=
p2
4
φ−1
and
x=
let
(
(1 − α)(p1 + p2 )
φ
=
p2
4
φ−1
x4 −x3
p1 −p2
p1 −p2
x4 −x3
=
=
p1 −p2 −4b
p1 −p2
x4 −x3 +4b
x4 −x3
(
m = − p12b
−p2
m′ = x42b
−x3
= 1 − p14b
+p2
= 1 + x44b
−x3
we pose
mm′ 6= 0
and ∀m, m′ , ∃β verifying m = βm′
⇒m=−
2b
2b
=β
p1 − p2
x4 − x3
⇒ x4 − x3 = −β(p1 − p2 )
⇒ x4 − x3 − p1 + p2 = 4b = −(β + 1)(p1 − p2 )
−(β + 1)
(p1 − p2 )
4
p1 − p2 (β + 1)
p1 − p2
+b=
−
(p1 − p2 )
⇒y=
2
2
4
p1 + p2 (β + 1)
p1 + p2
+b=
−
(p1 − p2 )
⇒x=
2
2
4
⇒b=
thus
y=
1−β
1−β
1
p1 −
p2 =
p2
4
4
φ−1
x=
β+3
φ
1−β
p1 +
p2 =
p2
4
4
φ−1
7
we resume
x=
1−α
1−α
φ
(1 − β)p1 + (3 + β)p2
=
p1 +
p2 =
p2
4
4
4
φ−1
and
y=
1−α
α+3
1
(1 − β)(p1 − p2 )
=
p1 −
p2 =
p2
4
4
4
φ−1
b=
−(α + 1)
−(β + 1)
(p1 + p2 ) =
(p1 − p2 )
4
4
β −α
2+α+β
⇒
p1 =
p2
4
4
2+α+β
β−α
=
⇒
p2
p1
but
4x − 4y + 4b = 2p1 + 2p2 + 4b − 2p1 + 2p2 − 4b + 4b = 4p2 + 4b = 4P2
And
4x + 4y − 4b = 2p1 + 2p2 + 4b + 2p1 − 2p2 + 4b − 4b = 4p1 + 4b = 4P1
Thus
4P1 = 4p1 + 4b = 4p1 − (α + 1)(p1 + p2 ) = (3 − α)p1 − (1 + α)p2
= 4p1 − (β + 1)(p1 − p2 ) = (3 − β)p1 + (1 + β)p2 =
1
(6 − α − β)p1 − (α − β)p2 )
2
And
4P2 = 4p2 + 4b = 4p2 − (α + 1)(p1 + p2 ) = −(1 + α)p1 + (3 − α)p2
1
= 4p2 −(β +1)(p1 −p2 ) = −(1+β)p1 +(5+β)p2 = (−(2+α+β)p1 +(8−α+β)p2 )
2
But p1 and p2 are primes.
(β − α)p1 = (α + β + 2)p2
β+1
)(P1 − P2 ))
4
5+β
1+β
= (β − α)((
)P1 − (
)P2 )
4
4
β+1
)(P1 − P2 ))
= (α + β + 2)(P2 − b) = (α + β + 2)(P2 + (
4
1+β
3−β
= (α + β + 2)((
)P1 +
P2 )
4
4
= (β − α)(P1 − b) = (β − α)(P1 + (
Thus
((β − α)(5 + β) − (α + β + 2)(1 + β))P1
= ((α + β + 2)(3 − β) + (β − α)(1 + β))P2
8
And
(β − α)P1 = (α + β + 2)P2 + b(−β + α + α + β + 2)
α+1
)(P1 − P2 )
= (α + β + 2)P2 − (β + 1)(
2
Hence
(2β − 2α + (β + 1)(α + 1))P1
= (2α + 2β + 4 + (β + 1)(α + 1))P2
Thus
((β − α)(3 + β) − (2α + β + 3)(β + 1))P1
= ((α + β + 2)(1 − β) − (β + 1)(α + 1))P2
= ((α + 1)(−3 − β − 2 − 2β) + (β + 1)(β + 1 + 3 + β))P1
= ((β + 1)(1 − β) − (α + 1)(−β − 1 + 1 − β))P2
= ((α + 1)(−5 − 3β) + 2(β + 1)(β + 2)P1
= ((β + 1)(1 − β) + 2(α + 1)β)P2
Hence
((β + 1)(−5 − 3β) + 2(β + 1)(β + 2))(p1 − p2 )P1
= ((β + 1)(1 − β) + 2(α + 1)β)(p1 + p2 )P2
= ((β + 1)(p1 − p2 )(−5 − 3β) + 2(β + 1)(β + 2)(p1 − p2 ))P1
= ((β + 1)(p1 + p2 )(1 − β) + 2(β + 1)β(p1 − p2 ))P2
= (β + 1)((−β − 1)p1 + (β + 1)p2 )P1
= (β + 1)((β + 1)p1 + (1 − 3β)p2 )P2
If
((−1 − β)p1 − (1 + β)p2 )P1 − ((β + 1)p1 + (1 − 3β)p2 )P2 = 0
= ((−1 − β)p1 − (1 + β)p2 )(p1 + b) − ((β + 1)p1 + (1 − 3β)p2 )(p2 + b)
= (−1 − β)p21 + (−1 + 3β)p22 − 2(β + 1)p1 p2 + b((−2 − 2β)p1 − (2 + 4β)p2 ) = 0
= −(1 + β)p21 + 3(1 + β)p22 − 4p22 − 2(β + 1)p1 p2 + b((−2 − 2β)p1 − (2 + 4β)p2 )
= (β + 1)(−p21 + 3p22 − 2p1 p2 ) − 4p22 ) + 2b((−1 − β)p1 − (1 + 2β)p2 )
= (β + 1)(p2 − p1 )(3p2 + p1 ) − 4p22 + 2b(−p1 − p2 − β(p1 + 2p2 ))
= 4b(3p2 + p1 ) − 4p22 + 2b(−p1 − p2 − β(p1 + 2p2 ))
p2 − p1
=
(2p1 + 10p2 − β(p1 + 2p2 )) − 4p22 = 0
4
Impossible. And as
((−β − 1)p1 − (β + 1)p2 )P1 6= ((β + 1)p1 − (1 − 3β)p2 )P2
We deduce that
β + 1 = 0 ⇒ 4b = −(β + 1)(p1 − p2 ) = 0
P1 and P2 are p1 and p2 and
2x = p1 + p2
Thus α = β = −1. And
b=0
9
Theorem
p1 + p2
2
We did not pose any condition on p1 and p2 . As p2 < x < p1 there are not an
infinity of such primes.
∀x ≥ 3, ∃p1 ≥ 3, p2 ≥ 3, x =
THEOREM OF THE PRIMAL RADIUS There exists r a primal radius for
which x + r, x − r are prime numbers ∀x ≥ 3.
Proof of theorem of primal radius For all x ≥ 3 exists p1 ≥ 3, p2 ≥ 3 prime
numbers for which
x=
p1 + p2
2
r=
p1 − p2
2
if
then
x + r = p1
x − r = p2
Corollary Between x and 2x exists always a prime number x < p1 = x + r <
2x = p1 + p2 . If 2z + 1 is an integer stricly greater than 8, exists always a prime p3 ,
which can be 3, for which 2z + 1 = p3 + 2x = p3 + p1 + p2 .
de Polignac conjecture de Polignac conjecture stipulates that an even number
is always equal to the difference between two prime numbers and that there is an
infinity of such prime numbers. Let x an integer and p1 , p2 prime numbers strictly
greater than 2.
2x = p1 − p2 + 2bp1 ,p2 = p1 − p2 + 2b
For commodity, we have suppressed the indexes, but b depends of p1 and p2 .
p1 − p2
+b
x=
2
But for all x, p1 , p2 , b exists y whose expression is
p1 + p2
y=
+b
2
We pose

x1 = p1 + 2b



x2 = p2 − 2b
x = p2 + 2b


 3
x4 = p1 − 2b

p1 +p2
p1 +x2
2
2
y = 2 + b = 2 + 2b = x1 +p
= x1 +x
+b

2
2


p1 +x3
x1 +x3
x4 +x3
x4 +x2

 = 2 = 2 − b = 2 + b = 2 + 3b
2
2
2
2
⇒
x = p1 −p
+ b = p1 −x
= x1 −p
= x1 −x
−b
2
2
2
2

p1 −x3
x1 −x3
x4 −x3
x4 −x2


 = 2 + 2b = 2 + b = 2 + 3b = 2 + b

x1 + x2 = p 1 + p 2
10
LEMMA 2 The following formula

2
2
2
2
+ b = p1 +x
+ 2b = x1 +p
= x1 +x
+b
y = p1 +p

2
2
2
2


p1 +x3
x1 +x3
x4 +x3
x4 +x2

 = 2 = 2 − b = 2 + b = 2 + 3b
2
2
2
2
x = p1 −p
+ b = p1 −x
= x1 −p
= x1 −x
−b
2
2
2
2

p1 −x3
x1 −x3
x4 −x3

2
 = 2 + 2b = 2 + b = 2 + 3b = x4 −x
+b

2

x1 + x2 = p 1 + p 2
imply that ∃p1 , p2 an infinity of couples which verify
b=0
Proof of lemma 2 If x = 0 is a prime number, 2x = p − p an infinity of couples
of primes, and
p1 − p2 6= 0
we will suppose firstly that
(x1 − x2 )(x1 + x3 ) 6= 0
let
(
x1 −x2
p1 −p2
p1 −p2
x1 −x2
=
=
p1 −p2 +4b
p1 −p2
x1 −x2 −4b
x1 −x2
= 1 + p14b
−p2
= 1 − x14b
−x2
we pose
(
k = p12b
−p2
k′ = − (p1 −p2b2 +4b)
if
kk′ = 0 ⇒ b = 0
we will suppose
kk′ 6= 0
But ∀x, y, ∃φ verifying y = φx
x + y = (φ + 1)x = x1 6= 0
x − y = (φ − 1)x = p2 6= 0
Thus ∀k, k′ , ∃α verifying k = αk ′
⇒k=
−2b
2b
=α
p1 − p2
x1 − x2
⇒ x1 − x2 = −α(p1 − p2 )
⇒ x1 − x2 − p1 + p2 = 4b = −(α + 1)(p1 − p2 )
−(α + 1)
(p1 − p2 )
4
p1 + p2
p1 + p2 α + 1
⇒y=
+b=
−
(p1 − p2 )
2
2
4
⇒b=
11
⇒x=
p1 − p2 α + 1
p1 − p2
+b=
−
(p1 + p2 )
2
2
4
or
y=
(1 − α)p1 + (3 + α)p2
φ
=
p2
4
φ−1
and
x=
let
(
1
(1 − α)(p1 − p2 )
=
p2
4
φ−1
x1 +x3
p1 +p2
p1 +p2
x1 +x3
=
=
p1 +p2 +4b
p1 +p2
x1 +x2 −4b
x1 +x3
= 1 + p14b
+p2
= 1 − x14b
+x2
we pose
(
m = p12b
+p2
m′ = − x12b
+x2
mm′ 6= 0
and ∀m, m′ , ∃β verifying m = βm′
⇒m=
2b
−2b
=β
p1 + p2
x1 + x3
⇒ x1 + x3 = −β(p1 + p2 )
⇒ x1 + x3 − p1 − p2 = 4b = −(β + 1)(p1 + p2 )
−(β + 1)
(p1 + p2 )
4
p1 + p2 (β + 1)
p1 + p2
+b=
−
(p1 + p2 )
⇒y=
2
2
4
p1 − p2
p1 − p2 (β + 1)
⇒x=
+b=
−
(p1 + p2 )
2
2
4
⇒b=
we deduce
y=
1−β
φ
1−β
p1 +
p2 =
p2
4
4
φ−1
x=
1−β
β+3
1
p1 −
p2 =
p2
4
4
φ−1
we resume
y=
(1 − α)p1 + (3 + α)p2
1−β
1−β
φ
=
p1 +
p2 =
p2
4
4
4
φ−1
and
x=
1−β
β+3
1
(1 − α)(p1 − p2 )
=
p1 −
p2 =
p2
4
4
4
φ−1
b=
−(β + 1)
−(α + 1)
(p1 − p2 ) =
(p1 + p2 )
4
4
β−α
−2 − α − β
⇒
p1 =
p2
4
4
12
⇒
−2 − α − β
β−α
=
p2
p1
but
4y − 4x + 4b = 2p1 + 2p2 + 4b − 2p1 + 2p2 − 4b + 4b = 4p2 + 4b = 4P2
And
4x + 4y − 4b = 2p1 + 2p2 + 4b + 2p1 − 2p2 + 4b − 4b = 4p1 + 4b = 4P1
Thus
4P1 = 4p1 + 4b = 4p1 − (α + 1)(p1 − p2 ) = (3 − α)p1 + (1 + α)p2
= 4p1 − (β + 1)(p1 + p2 ) = (3 − β)p1 − (1 + β)p2 =
1
(6 − α − β)p1 + (α − β)p2 )
2
And
4P2 = 4p2 + 4b = 4p2 − (α + 1)(p1 − p2 ) = −(1 + α)p1 + (5 + α)p2
1
= 4p2 −(β +1)(p1 +p2 ) = −(1+β)p1 +(3−β)p2 = (−(2+α+β)p1 +(8+α−β)p2 )
2
But p1 and p2 are primes.
(β − α)p1 = (−α − β − 2)p2
α+1
)(P1 − P2 ))
4
1+α
5+α
)P1 − (
)P2 )
= (β − α)((
4
4
α+1
= (−α − β − 2)(P2 − b) = (−α − β − 2)(P2 + (
)(P1 − P2 ))
4
1+α
3−α
= (−α − β − 2)((
)P1 +
P2 )
4
4
= (β − α)(P1 − b) = (β − α)(P1 + (
Thus
((β − α)(5 + α) + (α + β + 2)(1 + α))P1
= ((−α − β − 2)(3 − α) + (β − α)(1 + α))P2
And
(β − α)P1 = (−α − β − 2)P2 + b(β − α + α + β + 2)
= (−α − β − 2)P2 − (β + 1)(
α+1
)(P1 − P2 )
2
Hence
(2β − 2α + (β + 1)(α + 1))P1
= (−2α − 2β − 4 + (β + 1)(α + 1))P2
Thus
((β − α)(3 + α) + (α + 1)2 )P1
= ((−α − β − 2)(1 − α) − (α + 1)2 )P2
13
= ((β + 1)(3 + α) + (α + 1)(α + 1 − 3 − α))P1
= (−(β + 1)(1 − α) − (α + 1)(α + 1 + 1 − α))P2
= ((β + 1)(3 + α) − 2α − 2)P1
= (−(β + 1)(1 − α) − 2α − 2)P2
((β + 1)(3 + α) − 2(α + 1))(p1 + p2 )P1
= (−(β + 1)(1 − α) − 2(α + 1))(p1 + p2 )P2
= ((α + 1)(p1 − p2 )(3 + α) − 2(α + 1)(p1 + p2 ))P1
= (−(α + 1)(p1 − p2 )(1 − α) − 2(α + 1)(p1 + p2 ))P2
= (α + 1)((−α + 1)p1 − (3α + 5)p2 )P1
= (α + 1)(−(α + 3)p1 − (1 + 3α)p2 )P2
And as
((−α + 1)p1 − (3α + 5)p2 )P1 6= (−(α + 3)p1 − (1 + 3α)p2 )P2
We deduce that
α + 1 = 0 ⇒ 4b = −(α + 1)(p1 − p2 ) = 0
If
(x1 − x2 )(x1 + x3 ) = 0 ⇒ (x4 + x2 )(x4 − x3 ) 6= 0
let
(
x4 +x2
p1 +p2
p1 +p2
x4 +x2
=
=
p1 +p2 −4b
p1 +p2
x4 +x2 +4b
x4 +x2
= 1 − p14b
+p2
= 1 + x44b
+x2
we pose
(
let
(
2k + 1 = 1 − p14b
+p2
2k′ + 1 = 1 + x44b
+x2
x4 −x3
p1 −p2
p1 −p2
x4 −x3
=
=
p1 −p2 −4b
p1 −p2
x4 −x3 +4b
x4 −x3
= 1 − p14b
−p2
= 1 + x44b
−x3
we pose
(
2m + 1 = 1 − p14b
−p2
2m′ + 1 = 1 + x44b
−x3
2
2
2
2
= p1 +p
then p1 = p2 = y or p1 +p
= p1 +p
⇒ p1 = p2 = 0 and it is
⇒ y = p1 +3p
4
2
2
4
impossible, thus
(x1 − x2 )(x1 + x3 ) 6= 0
and
b=0
The primal radius r here is equal to
r=
p1 + p2
2
14
with
x=
p1 − p2
2
then
x + r = p1
and
r − x = p2
Its existence is proved. p1 and p2 are an infinity of couples of primes, because we
did not specify any condition on them.
Conclusion
The conclusion is that Goldbach conjecture and de Polignac conjecture, which
seem so inaccessible, are in fact true. Because, we have defined the primal radius.
The conjecture of the primal radius seems to be a consequence of Goldbach and
de Polignac conjectures. In fact, if we had not its concept in mind, there would
not be the present proof of those conjectures.
Références
[1] Van der Corput, J. G. "Sur l’hypothese de Goldbach" Proc. Akad. Wet. Amsterdam , 41, 76-80 (1938).
[2] Estermann, T. "On Goldbach’s problem : proof that almost all even positive
integers are sums of two primes" Proc. London Math. Soc. 2 44, 307-314 (1938).
[3] Sinisalo, Matti K. "Checking the Goldbach Conjecture up to 4 1011" Mathematics of Computation 61, 931-934 (1993).
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