CONFIRMATION of the GOLDBACH BINARY CONJECTURE Mohammed Ghanim* * Ecole Nationale De Commerce Et De Gestion B.P 1255 Tanger Maroc

Home , Christian Goldbach

[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 GLOBAL JOURNAL OF ADVANCED ENGINEERING TECHNOLOGIES AND SCIENCES CONFIRMATION OF THE GOLDBACH BINARY CONJECTURE Mohammed Ghanim* * Ecole Nationale de Commerce et de Gestion B.P 1255 Tanger Maroc

DOI: 10.5281/zenodo.1250336 ABSTRACT I show that the Goldbach (1690-1764) conjecture, remained open since 1742, is true by using some elementary tools of mathematics. The proof is essentially based on (the American Mathematician Lowell) Schoenfeld (1920- 2002) inequality that I have proved in [5] and on the intermediate value theorem.

KEYWORDS: prime integer, prime-counting function, Goldbach conjecture, Schoenfeld inequality, intermediate value theorem 2010 Mathematics Subject Classification: A 11 xx (Number theory)

INTRODUCTION Definition 1: We call « the Goldbach conjecture » or « the Goldbach strong conjecture » or « the Goldbach binary conjecture » or « the Goldbach problem » (according to D.Hilbert) or « the Goldbach theorem » (according to G.H.Hardy) the following assertion: “any even integer greater than 4 is the sum of two prime integers”.

History: the Goldbach conjecture was announced by the German Mathematician Christian Goldbach (1690-1764) in a letter addressed to the Swiss Mathematician Leonhard Euler (1707-1783) on 7 June 1742 [6]. It has remained, without a rigorous poof from this date, although many attempts by the greatest mathematicians.

In 1900, the German Mathematician David Hilbert (1862-1943) said in his conference delivered before the second international congress of mathematicians hold at Paris in the 8th point about «the prime numbers problems »: « … and perhaps after an exhaustive discussion of the Riemann formula on prime numbers we will be in a position to reach the rigorous solution of the Goldbach problem i.e.: if any even integer is a sum of two positive prime integers? …» [8].

In 1940, the English Mathematician G.H.Hardy (1877-1947) writed: « it exists some theorems such ‘the Goldbach theorem” which did not be proved and which any stupid could conjecture » [7] [13]

In 1977, the American Mathematician (of Polish descent) H.A.Pogorzelski (1922-2015) [11] affirmed to prove the Goldbach conjecture, but his proof is not generally accepted.

In 2000, Faber and Faber devoted $1000000 for any one proving the Goldbach conjecture between March 2000 and March 2002, but no one could give a proof and the question remained open [3][13].

However, the Goldbach conjecture was verified for all the entire even values of the integer 푛, 4 ≤ 푛 ≤ 푚, where 푚 = 104 by Desboves in 1885, 푚 = 105 by N.Pipping in 1938, 푚 = 108 by M.L.Stein and M.L.Stein in 1965, 푚 = 2. 1010 by A.Granville and J.Van der lune and H.J.J Te Riele in 1989, 푚 = 4. 1011 by M.K.Sinisalo in 1993, 푚 = 1014 by J.M.Deshouillers and H.J.J.Te Riele and Y.Saouter en 1998, 푚 = 4. 1014 by J.Richstein in 2001, 푚 = 2. 1016 by T.Oliveira E Silva on 3/14/2003 and 푚 = 6. 1016 by T.Oliveira E Silva on 10/3/2003 (See[13] and its references)

Finally The Nice University (France) devoted online, since 1999, a site [17] giving, for higher values of chosen even natural numbers, all there Goldbach-decompositions in sums of two prime natural numbers.

The note: my purpose in the present brief note is to show that the Goldbach conjecture, remained open since 1742, is true by using some elementary tools of mathematical analysis. The proof is based essentially on the intermediate value theorem and the Schoenfeld inequality that I have proved in [5]. The main result of the present work is:

Theorem: ∀푛 ∈ ℕ, 푛 ≥ 2, ∃(푝, 푞) two prime numbers such that: 2푛 = 푝 + 푞.

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[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 The paper is organized as follows. §1 is an introduction containing the necessary definitions and some history. The §2 gives the ingredients of the proof of our main result. The §3 gives the proof of our main result. §4 is the conclusion. The §5 gives the references of the paper for further reading.

INGREDIENTS OF THE PROOF: Definition2: a natural integer p is said to be prime if its set of divisors is {1, p}.

Definition3: Define the set ℙof prime integers by: ℙ = {푝 ∈ ℕ, 푝 is prime}= {2=푝1,3=푝2,5=푝3,7=푝4,11=푝5,13=푝6,17=푝7,19=푝8,23=푝9, ...,푝푘 …}.

thd Proposition1: (Euclid (3 century before Jesus Christ)) (See [2]) ℙ is an increasing infinite sequence(푝푘)푘∈ℕ∗. ∗ ∗ i.e.: ∀푘 ∈ ℕ 푝푘+1 > 푝푘 i. e ∀푘 ∈ ℕ 푝푘+1 ≥ 푝푘 + 1.

Proposition2: (some properties of the cardinality of a finite set) (See [16]) the cardinality of a finite set A, denoted card (A), is a natural number: the number of elements in the set A. We have the following properties: (1)푐푎푟푑(∅) = 0 (∅ is the empty set) and 푐푎푟푑({푎}) = 1 (2)퐴 ⊂ 퐵 ⇒ 푐푎푟푑(퐴) ≤ 푐푎푟푑(퐵) (3)푐푎푟푑(퐴 ∪ 퐵) = 푐푎푟푑(퐴) + 푐푎푟푑(퐵) − 푐푎푟푑(퐴 ∩ 퐵) (4)푐푎푟푑(퐴\퐵) = 푐푎푟푑(퐴) − 푐푎푟푑(퐴 ∩ 퐵) with 퐴\퐵 = {푥, 푥 ∈ 퐴 푎푛푑 푥 ∉ 퐵} is the set difference of the sets A and B. (5)푐푎푟푑(퐴 × 퐵) = 푐푎푟푑(퐴)푐푎푟푑(퐵) with 퐴 × 퐵 = {(푥, 푦), 푥 ∈ 퐴 푎푛푑 푦 ∈ 퐵} is the Cartesian product of the sets A and B. 푛 푛 (6) 퐴푖 ∩ 퐴푗 = ∅, for : 푖 ≠ 푗 , 푖, 푗 = 1,2, … , 푛 ⇒ 푐푎푟푑(⋃푖=1 퐴푖) = ∑푖=1 푐푎푟푑(퐴푖)

Definition4: Let, for 푘 ∈ ℕ: ℙ푘 = {푝 ∈ ℙ, 푝 ≤ 푘} and 휋(푘) = 푐푎푟푑(ℙ푘) (which is the number of the elements in the finite set ℙ푘). The function 푘 → 휋(푘) is called the prime-counting function.

Definition5: We have:푓 = 푂(푔) in the neighborhood of +∞ ⇔ ∃퐴 > 0∃퐵 ∈ ℝ such that: 푡 ≥ 퐵 ⇒ |푓(푡)| ≤ 퐴|푔(푡)|

Proposition3: (The Schoenfeld inequality) (See [1], [5], [10], [12]) 푛 푑푡 √푛ln (푛) ∀푛 ≥ 2657 |휋(푛) − ∫ | ≤ 0 ln (푡) 8휋 Where : 푛 푑푡 푛 푑푡 ∗ ∫ = ∫ + 푅 0 ln (푡) 2 ln (푡) * 푅 = 1.045163780117492784844588889194131365226155781512 … (voir [14]) 푛 푑푡 I.e. with the notation of definition 5 : 휋(푛) − ∫ = 푂(√푛 ln (푛)) in the neighborhood of +∞. 0 ln (푡) Proposition4 : We have : (i) lim 푎푛 = 푠 ∈ ℝ ⇔ ∀휖 > 0∃푁 ∈ ℕ such that∀푛 ≥ 푁: |푎푛 − 푠| < 휖 푛→+∞ (ii) lim 푎푛 = +∞ ⇔ ∀휖 > 0∃푁 ∈ ℕ such that∀푛 ≥ 푁: 푎푛 > 휖 푛→+∞ (iii)lim 푓(푥) = 푓(푎) ⇔ ∀휖 > 0∃훿 > 0: |푥 − 푎| < 훿 ⇒ |푓(푥) − 푓(푎)| < 휖 푥→푎

Proposition5: (See [4], p 31 and [9], p 4) Recall that for any sequences (푥푛) and (푦푛) of ℝ, we have : (1)The number:limsup(푥푛) = 푖푛푓푛∈ℕ(sup({푥푘; 푘 ≥ 푛}) = lim ( sup { 푥푘, 푘 ≥ 푛}) exits always in [−∞, +∞] 푛→+∞ (2)The number:liminf (푥푛) = 푠푢푝푛∈ℕ(inf({푥푘, 푘 ≥ 푛}) = lim (inf {푥푘, 푘 ≥ 푛}) exists always in [−∞, +∞] 푛→+∞ (3)(i) limsup(푥푛) = −liminf (−푥푛) (ii)liminf푥푛 ≤ limsup푥푛 (4) lim 푥푛 = limsup(푥n) = liminf(푥푛) 푛→+∞ (5) If 푦푛 converges : (i) limsup(푥푛 + 푦푛) = limsup(푥푛) + lim 푦n 푛→+∞ (ii) liminf(푥푛 + 푦푛) = liminf(푥푛) + lim 푦n 푛→+∞ (6) If (푥푛) is bounded⊂ [푎, 푏] ⊂ ℝ, we have: 푎 ≤ liminf푥푛 ≤ limsup푥푛 ≤ 푏

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[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 ∀휖 > 0∃푛 ∈ ℕ∀푘 ≥ 푛 푥푘 < 푠 + 휖 (7)(푖)limsup푥푛 = 푠 ∈ ℝ ⇔ { ∀휖 > 0∀푛 ∈ ℕ∃푘 ≥ 푛 푥푘 > 푠 − 휖 ∀휖 > 0∃푛 ∈ ℕ∀푘 ≥ 푛 푥푘 > 푠 − 휖 (ii) liminf푥푛 = 푠 ∈ ℝ ⇔ { ∀휖 > 0∀푛 ∈ ℕ∃푘 ≥ 푛 푥푘 < 푠 + 휖

푎 ≤ 푏 Proposition6 : ∀푎, 푏, 푥, 푦 ∈ ℝ : { ⇒ 푎푥 ≤ 푏푦 0 ≤ 푥 ≤ 푦

Proposition7 if 푦푛 ≥ 0 ∀푛, we have : liminf (푥푛푦푛) ≥ liminf(푥푛) liminf(푦푛)

Proof : (of proposition7) By proposition 6 and the definition of « the inf.», we have successively : 0 ≤ 푖푛푓푛≥푝푦푛 ≤ 푦푛 ∀푛 ≥ 푝: { ⇒ ∀푛 ≥ 푝: 푖푛푓푛≥푝푥푛푖푛푓푛≥푝푦푛 ≤ 푥푛푦푛 ⇒ 푖푛푓푛≥푝푥푛푖푛푓푛≥푝푦푛 ≤ 푖푛푓푛≥푝(푥푛푦푛) ⇒ 푖푛푓푛≥푝푥푛 ≤ 푥푛 liminf푥푛liminf푦푛 = lim 푖푛푓푛≥푝푥푛 lim 푖푛푓푛≥푝푦푛 = lim 푖푛푓푛≥푝푥푛푖푛푓푛≥푝푦푛 ≤ lim 푖푛푓푛≥푝(푥푛푦푛) = 푝→+∞ 푝→+∞ 푝→+∞ 푝→+∞ liminf(푥푛푦푛)

Proposition8 : (i) Any non empty part 퐸of ℕ has a minimal element: min (퐸). min (퐸) is characterized by :min(퐸) ∈ 퐸, ∀푥 ∈ 퐸: 푥 ≥ min (퐴), min(퐴) = 0 or min(퐴) − 1 ∉ 퐸 (ii)Any non empty part 퐸 of ℕ, bounded above, has a maximal element :max(퐸). max (퐸) is characterized by : max(퐸) ∈ 퐸, ∀푥 ∈ 퐸: 푥 ≤ max (퐸) and max(퐸) + 1 ∉ 퐸.

Proposition9 : if 퐼 ⊂ ℝ is an interval, 푎 ∈ 퐼 and 푓: 퐼 → ℝ is a function, we have : (1)푓 continuous in 푎 ⇔ lim 푓(푥) = 푓(푎) 푥→푎

(2) 푓 continuous in 훼 ⇔ ((∀(훼푘) a sequence: lim αk = α) ⇒ ( lim 푓(훼푘) = 푓(훼))) k→+∞ 푘→+∞ (3)푓 continuous on 퐼 ⇔ 푓 continuous in any a ∈ 퐼

Proposition10 : (Intermediate value theorem) (See [15]) Let [푎, 푏](푎 < 푏) an interval of ℝ and 푓:[푎, 푏]→ ℝ a continuous function. If 푓(푎)푓(푏) < 0, then :∃푐 ∈]푎, 푏[ such that : 푓(푐) = 0.

Proposition11 : Any sequence (푡푘) ⊂ [푎, 푏] (an interval of ℝ), has a convergente subsequence, denoted also (푡푘), with : lim 푡푘 = 푡 ∈ [푎, 푏]. 푘→+∞

THE PROOF OF THE GOLDBACH BINARY CONJECTURE: Theorem: ∀푛 ≥ 2∃(푝, 푞) ∈ ℙ2푛 × ℙ2푛 such that 2푛 = 푝 + 푞

Proof: (of the Theorem) *Let: 퐴2푛 = {(푝, 푞) ∈ ℙ2푛 × ℙ2푛, 푝 + 푞 = 2푛} for an integer 푛 ≥ 2 퐵푘 = {(푝, 푞) ∈ ℙ푘 × ℙ푘, 푝 + 푞 ≤ 푘} for an integer k≥ 4. *Show that: 퐴2푛 is ≠ ∅. *It is evident that the theorem is proved if we show that: 푐푎푟푑(퐴2푛) > 0. *The proof of the theorem will be deduced from the lemmas below.

Lemma1 : We have :

(1) 푐푎푟푑(퐵2푛) = ∑푝∈ℙ2푛 휋(2푛 − 푝)

(2) 푐푎푟(퐴2푛) = ∑푝∈ℙ2푛−1 휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)

Proof: (of lemma1) { } { } (1)*We have:퐵2푛 = ⋃푝∈ℙ2푛({푝} × ℙ2푛−푝) with: ( 푝 × ℙ2푛−푝) ∩ ( 푞 × ℙ2푛−푞) = ∅ for: 푝 ≠ 푞. *So, by proposition2, we have:

푐푎푟푑(퐵2푛) = ∑ 푐푎푟푑({푝} × ℙ2푛−푝) = ∑ 푐푎푟푑(ℙ2푛−푝) = ∑ 휋(2푛 − 푝) 푝∈ℙ2푛 푝∈ℙ2푛 푝∈ℙ2푛

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[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 (2)* Noting that 2푛 being not prime, for 푛 ≥ 2, we have: ℙ2푛 = ℙ2푛−1 *But: 퐵2푛 = 퐴2푛 ∪ 퐵2푛−1 with: 퐴2푛 ∩ 퐵2푛−1=∅.

*So :푐푎푟푑(퐴2푛)=푐푎푟푑(퐵2푛) − 푐푎푟푑(퐵2푛−1) = ∑푝∈ℙ2푛 휋(2푛 − 푝) − ∑푝∈ℙ2푛−1 휋(2푛 − 1 − 푝)

=∑푝∈ℙ2푛−1 휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)

Lemma2: We have: ∀푛 ≥ 2658 1 1 − 휋(2푛 − 2658)√2푛 − 2 ln(2푛 − 2) ≤ 0 ≤ ∑ (휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) ≤ 휋(2푛 − 4휋 푝∈ℙ2푛−1\ℙ2푛−2658 4휋 2658)√2푛 − 2 ln(2푛 − 2))

Proof: (of lemma2) *We have: 2푛−1 2657 ∑푝∈ℙ2푛−1\ℙ2푛−2658(휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) ≤ ∑푘=2푛−2657(휋(2푛 − 푘) − 휋(2푛 − 푘 − 1)) = ∑푘=1 휋(푘) − 휋(푘 − 1) = 휋(2657) − 휋(0) =휋(2657) *We have:

푛 ≥ 2658 ⇒ 0 ≤ ∑푝∈ℙ2푛−1\ℙ2푛−2658(휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) ≤ 휋(2657) = 휋(2758) = 384 ≤ 1 1 휋(2658)√5314 ln(5314) = 19108.3526 < 휋(2푛 − 2658)√2푛 − 2ln (2푛 − 2) 4휋 4휋 *So: 1 1 − 휋(2푛 − 2658)√2푛 − 2 ln(2푛 − 2) ≤ ∑ (휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) ≤ 휋(2푛 − 4휋 푝∈ℙ2푛−1\ℙ2푛−2658 4휋 2658)√2푛 − 2 ln(2푛 − 2))

Lemma3: We have, ∀ 푛 ≥ 2658: 1 푑푡 푐푎푟푑(퐴 ) = ∑ ∫ + 푂(휋(2푛 − 2658)√2푛 − 2ln (2푛 − 2)) 2푛 푝∈ℙ2푛−2658 0 ln (푡+2푛−푝−1) 푂(휋(2푛−2658) 2푛−2 ln(2푛−2)) 1 With:∀푛 ≥ 2658 | √ | ≤ 휋(2푛−2658)√2푛−2ln (2푛−2) 2휋

Proof: (of lemma3) *2푛 − 2658 ≥ 2 ⇔ 2푛 ≥ 2660 ⇔ 푛 ≥ 1330 *By the assertion (2) of lemma1, we have:

푐푎푟푑(퐴2푛) = ∑푝∈ℙ2푛−1(휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) = ∑푝∈ℙ2푛−2658(휋(2푛 − 푝) − 휋(2푛 − 푝 −

1)) + ∑푝∈ℙ2푛−1\ℙ2푛−2658(휋(2푛 − 푝) − 휋(2푛 − 푝 − 1) ) *By the Schoenfeld inequality (See proposition 3), we have, for any prime integer 푝 such that 2푛 − 푝 − 1 ≥ 2657 ⇔ 푝 ≤ 2푛 − 2658 ⇔ 푝 ∈ ℙ2푛−2658: 2푛−푝 푑푡 √2푛−푝ln (2푛−푝) 2푛−푝 푑푡 √2푛−푝ln (2푛−푝) ∫ − ≤ 휋(2푛 − 푝) ≤ ∫ + 0 ln (푡) 8휋 0 ln (푡) 8휋 { 2푛−푝−1 푑푡 √2푛−푝−1ln (2푛−푝−1) 2푛−푝−1 푑푡 √2푛−푝−1ln (2푛−푝−1) − ∫ − ≤ −휋(2푛 − 푝 − 1) ≤ − ∫ + 0 ln (푡) 8휋 0 ln (푡) 8휋 2푛−푝 푑푡 √2푛−푝ln (2푛−푝) 2푛−푝 푑푡 √2푛−푝ln (2푛−푝) ⇒ ∫ − ≤ 휋(2푛 − 푝) − 휋(2푛 − 푝 − 1) ≤ ∫ + 2푛−푝−1 ln (푡) 4휋 2푛−푝−1 ln (푡) 4휋 2푛−푝 푑푡 1 ⇒ ∑ ∫ − ∑ √2푛 − 푝ln (2푛 − 푝) ≤ ∑ (휋(2푛 − 푝) − 휋(2푛 − 푝 − 푝∈ℙ2푛−2658 2푛−푝−1 ln (푡) 4휋 푝∈ℙ2푛−2658 푝∈ℙ2푛−2658 2푛−푝 푑푡 1 1)) ≤ ∑ ∫ + ∑ √2푛 − 푝ln (2푛 − 푝) 푝∈ℙ2푛−2658 2푛−푝−1 ln (푡) 4휋 푝∈ℙ2푛−2658 *Then, by the variable change 푡 = 푢 + 2푛 − 푝 − 1 in the involved integals and by noting that:∑푝∈ℙ2푛−2658 √2푛 − 푝 ln(2푛 − 푝) ≤ 푐푎푟푑(ℙ2푛−2658)√2푛 − 2 ln(2푛 − 2) = 휋(2푛 − 2658)√2푛 − 2ln (2푛 − 2), we have: 1 − 휋(2푛 − 2658)√2푛 − 2 ln(2푛 − 2) ≤ ∑ (휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) − 4휋 푝∈ℙ2푛−2658 1 푑푡 1 ∑ ∫ ≤ 휋(2푛 − 2658)√2푛 − 2ln (2푛 − 2)) 푝∈ℙ2푛−2658 0 ln (푡+2푛−푝−1) 4휋 *So, by the assertion (2) of lemma1 and by lemma2, we have successively, for 푛 ≥ 2658:

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[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 1 푑푡 1 (∑ ∫ − (휋(2푛 − 2658)√2푛 − 2 ln(2푛 − 2))) ≤ 푐푎푟푑(퐴 ) = ∑ (휋(2푛 − 푝∈ℙ2푛−2658 0 ln(푡+2푛−푝−1) 2휋 2푛 푝∈ℙ2푛−1

푝) − 휋(2푛 − 푝 − 1)) = ∑푝∈ℙ2푛−2658(휋(2푛 − 푝) − 휋(2푛 − 푝 − 1)) + ∑푝∈ℙ2푛−1\ℙ2푛−2658(휋(2푛 − 푝) − 휋(2푛 − 1 푑푡 1 푝 − 1) ≤ (∑ ∫ + (휋(2푛 − 2658)√2푛 − 2 ln(2푛 − 2))) 푝∈ℙ2푛−2658 0 ln(푡+2푛−푝−1) 2휋 *Finally, with the notation of definition3, we have: 1 푑푡 푐푎푟푑(퐴 ) = ∑ ∫ + 푂(휋(2푛 − 2658)√2푛 − 2 ln(2푛 − 2)) 2푛 푝∈ℙ2푛−2658 0 ln (푡+2푛−푝−1)

휋(2푛−2658) 1 푑푡 휋(2푛−2658) Lemma4: We have: (1)0 < ≤ ∑ ∫ ≤ ln (2푛−2) 푝∈ℙ2푛−2658 0 ln (푡+2푛−푝−1) ln (2657) 1 푑푡 1 푑푡 ∑푝∈ℙ ∫ ∑푝∈ℙ ∫ (2) lim 2푛−2658 0 ln (푡+2푛−푝−1) = 0(⇔ lim 2푛−2658 0 ln (푡+2푛−푝−1) = +∞) 푛→+∞ 휋(2푛−2658)√2푛−2ln (2푛−2) 푛→+∞ 휋(2푛−2658)√2푛−2ln (2푛−2)

Proof: (of lemma4)

(1)∀푛 ≥ 2658 ∀푡 ∈ [0,1]∀푝 ∈ ℙ2푛−2658, we have: ln(2657) ≤ ln (푡 + 2푛 − 푝 − 1) ≤ ln (2푛 − 2) 1 1 1 1 1 푑푡 1 ⇔ ≤ ≤ ⇒ 0 < ≤ ∫ ≤ ln (2푛−2) ln (푡+2푛−푝−1) ln (2657) ln (2푛−2) 0 ln (푡+2푛−푝−1) ln (2657) 휋(2푛−2658) 1 푑푡 휋(2푛−2658) ⇒ 0 < ≤ ∑ ∫ ≤ ln (2푛−2) 푝∈ℙ2푛−2658 0 ln (푡+2푛−푝−1) ln (2657) 1 푑푡 ∑푝∈ℙ ∫ 1 (2)*So:0 < 2푛−2658 0 ln (푡+2푛−푝−1) ≤ 휋(2푛−2658)√2푛−2ln (2푛−2) √2푛−2 ln(2푛−2)ln (2657) 1 푑푡 1 ∑푝∈ℙ ∫ And lim = 0 ⇒ lim 2푛−2658 0 ln (푡+2푛−푝−1) = 0 푛→+∞ √2푛−2 ln(2푛−2)ln (2657) 푛→+∞ 휋(2푛−2658)√2푛−2ln (2푛−2)

Lemma5: We have: 푐푎푟푑(𝐴 ) 푂(휋(2푛−2658) 2푛−2 ln(2푛−2)) 1 0 ≤ liminf 2푛 = liminf √ ≤ π(2n−2658)√2n−2ln (2n−2) π(2n−2658)√2n−2ln (2n−2) 2π

Proof: (of lemma5) 푐푎푟푑(𝐴 ) *The inequalities follow by use of the assertion (6) of proposition 5, because: 2푛 ≥ 0 and π(2n−2658)√2n−2ln (2n−2) 푂(휋(2푛−2658) 2푛−2 ln(2푛−2)) 1 √ ≤ ∀n ≥ 2658 π(2n−2658)√2n−2ln (2n−2) 2π *The equalities follow by use of lemma3, the relations (i), (ii) of the assertion (5) of proposition 5 and the relation (2) of lemma4.

Lemma6: ∃푁 ≥ 2658 such that:∀푛 ≥ 푁 푐푎푟푑(퐴2푛) > 0

Proof : (of lemma6)

card(A2n) First case : if liminf 1 dt = +∞ ∑ ∫ p∈ℙ2n−2658 0 ln (t+2n−p−1) card(A2n) *We have, by the assertion (4) of proposition5 : lim 1 dt = +∞ 푛→+∞ ∑ ∫ p∈ℙ2n−2658 0 ln (t+2n−p−1) *So, by the assertion (ii) of proposition 3 :∃푁 ≥ 2658 such that ∀푛 ≥ 푁 푐푎푟푑(퐴2푛) > 0

card(A2n) Second case : if liminf 1 dt = s < +∞ ∑ ∫ p∈ℙ2n−2658 0 ln (t+2n−p−1) *By the assertion 5(ii) of proposition 5, the assertion (8) of proposition5, lemma3 and lemma5, we have successively: card(A2n) card(A2n) s − 1 = liminf( 1 dt ) − 1 = liminf( 1 dt − 1) ∑ ∫ ∑ ∫ p∈ℙ2n−2658 0 ln(t+2n−p−1) p∈ℙ2n−2658 0 ln(t+2n−p−1) O(π(2n−2658)√2n−2 ln(2n−2)) O(π(2n−2658)√2n−2 ln(2n−2)) π(2n−2658)√2n−2 ln(2n−2) = liminf ( 1 dt ) = liminf( × 1 dt ) ∑ ∫ π(2n−2658)√2n−2 ln(2n−2) ∑ ∫ p∈ℙ2n−2658 0 ln(t+2n−p−1) p∈ℙ2n−2658 0 ln(t+2n−p−1)

http: // www.gjaets.com/ © Global Journal of Advance Engineering Technology and Sciences [5]

[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 O(π(2n − 2658)√2n − 2 ln(2n − 2)) π(2n − 2658)√2n − 2 ln(2n − 2) ≥ liminf ( ) liminf( ) 1 dt π(2n − 2658)√2n − 2 ln(2n − 2) ∑ ∫ p∈ℙ2n−2658 0 ln(t + 2n − p − 1) 푐푎푟푑(𝐴2푛) π(2n−2658)√2n−2 ln(2n−2) =liminf ( ) liminf( 1 dt ≥ 0 π(2n−2658)√2n−2 ln(2n−2) ∑ ∫ p∈ℙ2n−2658 0 ln(t+2n−p−1) 푐푎푟푑(𝐴 ) Remark: We have, in this case, necessarily: liminf 2푛 = 0, π(2n−2658)√2n−2 ln(2n−2) π(2n−2658)√2n−2 ln(2n−2) 푐푎푟푑(𝐴2푛) because: liminf( 1 dt = +∞ ⇒ (if: liminf ≠ 0), ∑ ∫ π(2n−2658)√2n−2 ln(2n−2) p∈ℙ2n−2658 0 ln(t+2n−p−1) card(A2n) that : liminf 1 dt = +∞. ∑ ∫ p∈ℙ2n−2658 0 ln (t+2n−p−1) *That is : 1 ≤ 푠 < +∞ 1 *So, by the assertion 7(ii) of proposition 5 (written for : 휖 = ), we have : 2 푐푎푟푑(퐴 ) 1 1 1 ∃푁 ≥ 2658 ∀푛 ≥ 푁: 2푛 ≥ s − ≥ 1 − = > 0 π(2n − 2658)√2n − 2 ln(2n − 2) 2 2 2 *That is : ∃푁 ≥ 2658 ∀푛 ≥ 푁 such that : 푐푎푟푑(퐴2푛) > 0.

Lemma7 : ∀푛 ∈ {2,3,4,5, … ,2657} 푐푎푟푑(퐴2푛) > 0 and 푐푎푟푑(퐴0) = 푐푎푟푑(퐴2) = 0

Proof : (of lemma7) See Wims [17] which is an online site devoted, by the Nice University (France), to giving, for very heigh values of even positive integers, all there Goldbach-decompositions in two prime positive integers.

Lemma8 : ∀푛 ≥ 2 푐푎푟푑(퐴2푛) > 0.

Proof : (of lemma8) Consider the subsets 퐴, 퐵 and 퐶 of ℕ defined by : 퐴 = {푁 ∈ ℕ, ∀푛 ≥ 푁 푐푎푟푑(퐴2푛) > 0} 퐵 = {푁 ∈ ℕ, ∃푛 ≥ 푁 푐푎푟푑(퐴2푛) = 0} 퐶 = {푁 ∈ ℕ, 푐푎푟푑(퐴2(푁)) = 0}

Claim1 : The set 퐴 has a minimal element : min(퐴) = 푚 ≥ 2

Proof : (of claim1) *By lemma6, the subset 퐴 = {푁 ∈ ℕ, ∀푛 ≥ 푁 푐푎푟푑(퐴2푛) > 0} of ℕ is ≠ ∅ (non empty). *So, by proposition 8(i), 퐴 has a minimal element 푚 = min (퐴). *By lemma7 : 푐푎푟푑(퐴0) = 푐푎푟푑(퐴2) = 0 ⇒ 푚 ≥ 2

Claim2 : 퐵 = ℕ\퐴

Proof : (of claim2) *We have : 푁 ∈ ℕ\퐴 ⇔ 푁 ∉ 퐴 ⇔ ∃푛 ≥ 푁 such that : 푐푎푟푑(퐴2푛) = 0 ⇔ 푁 ∈ 퐵 *So the result follows.

Claim3 : 퐵 is finite with max(퐵) = 푚 − 1

Proof : (of lemma3) *By claim1 :min(퐴) = 푚 ≥ 2.

*So, because : 푚 − 1 ∉ 퐴, we have : 푐푎푟푑(퐴2(푚−1)) = 0 and : 푚 − 1 ∈ 퐵. *Because : 푚 ∈ 퐴, we have : ∀푛 ≥ 푚 푐푎푟푑(퐴2푛) > 0. *Suppose that ∃푁 ∈ 퐵 such that : 푁 > 푚 − 1 i .e. 푁 ≥ 푚

*By definition of 푁 :∃푝 ≥ 푁 such that : 푐푎푟푑(퐴2푝) = 0 * But : 푚 ∈ 퐴 ⇒ ∀푛 ≥ 푚 푐푎푟푑(퐴2푛) > 0. *So : 푝 ≥ 푁 ≥ 푚 ⇒ 0 < 푐푎푟푑(퐴2푝) = 0

[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 *This being contradictory, we have : ∀푁 ∈ 퐵 푁 ≤ 푚 − 1 *The result follows

Claim4 : (i)∀푁 ∈ ℕ : 푁 ∈ 퐴 ⇒ 푁 + 1 ∈ 퐴 (ii)So :퐴 = {푚, 푚 + 1, 푚 + 2, … } = {푁 ∈ ℕ, 푁 ≥ 푚}

Proof : (of claim4) (i)Let 푁 ∈ 퐴, we have : ∀푛 ≥ 푁: 푐푎푟푑(퐴2푛) > 0 ⇒ ∀푛 ≥ 푁 + 1 ≥ 푁: 푐푎푟푑(퐴2푛) > 0 i.e. 푁 + 1 ∈ 퐴. (ii)So : 푚 = min(퐴) ⇒ 퐴 = {푚, 푚 + 1, 푚 + 2, … }

Claim5 : 퐵 = {0,1, … , 푚 − 1} = {푁 ∈ ℕ, 0 ≤ 푁 ≤ 푚 − 1}

Proof : (of claim5) The result is obtained by combination of claim2 and the assertion (ii) of claim4.

Claim6 : We have :

퐵 = {0,1, … , 푚 − 1} = {푁 ∈ ℕ, ∃푛 ≥ 푁 푐푎푟푑(퐴2푛) = 0} ⊃ 퐶 = {푁 ∈ ℕ, 푐푎푟푑(퐴2(푁)) = 0}

Proof : (of claim6) *If 푁 ∈ 퐶, we have :푐푎푟푑(퐴2푁) = 0. So :∃푛 = 푁 ≥ 푁 such that :푐푎푟푑(퐴2푛) = 0. *That is :퐶 ⊂ 퐵

Claim7 : We have max(퐶) = 푚 − 1

Proof : (of lemma7)

*We have :푐푎푟푑(퐴2(푚−1)) = 0 ⇒ 푚 − 1 ∈ 퐶. *We have : 푐푎푟푑(퐴2푚) > 0 ⇒ 푚 ∉ 퐶. *We have : 퐶 ⊂ 퐵 = {0,1, … , 푚 − 1} ⇒ ∀푁 ∈ 퐶: 푁 ≤ 푚 − 1. *This shows that : max(퐶) = 푚 − 1

Claim8 : we have 푚 = 2

Proof : (of lemma8) *We have :푚 ≥ 2. suppose contrarily that 푚 ≥ 3. *The claim 8 will be deduced from the under-claims below.

Under-claim1 : We have : 푚 − 1 ∈ 퐶 ⇔ ∀푝, 푞 ∈ ℙ2(푚−1)|2(푚 − 1) − 푝 − 푞| > 0

Proof : (of the under-claim1) The result follows by definition of 퐶 = {푁 ∈ ℕ, 푐푎푟푑(퐴2푁) = 0} and by definition of 퐴2푁 = {(푝, 푞) ∈ ℙ2푁 × ℙ2푁, 2푁 = 푝 + 푞}

Under-claim2 : We have ∗ (i)Let: 푛 is a positive integer ≤ 푚 − 1, 푎, 푏 ∈ ℙ2푛, 푝, 푞 ∈ ℙ2(푚−1) and 푘, 푟 ∈ ℕ . (1) if : |2푛 − 푎 − 푏| ≥ |2(푚 − 1) − 푝 − 푞|, we have : |2푛 − 푎 − 푏| > 0 (2) if : |2푛 − 푎 − 푏| < |2(푚 − 1) − 푝 − 푞|, then : 1 2 1 (a)∃휃(푛, 푘, 푟, 푎, 푏, 푝, 푞) = 훼 ∈] , − [ suchthat : 푘,푟 2 3 푟 2 2 1 −훼 1 −훼 훼 (|2푛 − 푎 − 푏| + )3 푘,푟 = (1 − 훼 )(|2(푚 − 1) − 푝 − 푞| + )3 푘,푟 푘,푟 푘 푘,푟 푘 1 2 (b)∃훼 = 훼 ∈] , [ such that : 푟 2 3 2 2 −훼 −훼 훼(|2푛 − 푎 − 푏|)3 = (1 − 훼)(|2(푚 − 1) − 푝 − 푞|)3 (ii)∀푎, 푏 ∈ ℙ2푛 |2푛 − 푎 − 푏| > 0 (iii)∀0 ≤ 푛 ≤ 푚 − 1 : 푐푎푟푑(퐴2푛) = 0

[Ghanim et al., 5(5): May, 2018] ISSN 2349-0292 Impact Factor 3.802 (iv)퐶 = 퐵

Proof : (of the under-claim2) (i) (1) the result follows by the under-claim1. 1 2 1 (2)(a) Consider, on [ , − ], the continuous function 푓 defined by : 2 3 푟 푘,푟 2 2 1 −푡 1 −푡 푓 (푡) = 푡(|2푛 − 푎 − 푏| + )3 − (1 − 푡)(|2(푚 − 1) − 푝 − 푞| + )3 푘,푟 푘 푘 *By the hypothesis «|2푛 − 푎 − 푏| < |2(푚 − 1) − 푝 − 푞| », we have : 1 1 1 1 1 1 ∗∗ 푓 ( ) = ((|2푛 − 푎 − 푏| + )6 − (|2(푚 − 1) − 푝 − 푞| + )6) < 0 푘,푟 2 2 푘 푘 2 1 2 1 1 1 1 1 1 1 **푓 ( − ) = ( − )(|2푛 − 푎 − 푏| + )푟 − ( + ) (|2(푚 − 1) − 푝 − 푞| + )푟 > 0, for a great positive 푘,푟 3 푟 3 푟 푘 3 푟 푘 1 integer 푟, because, by the assertion (i) of proposition 4 (written for : 휖 = ), we have : 6 2 1 2 1 1 2 1 1 1 1 lim 푓푘,푟( − ) = − = ⇒ ∃푁 ∈ ℕ ∀푟 ≥ 푁: 푓푘,푟 ( − ) ≥ − = > 0 푟→+∞ 3 푟 3 3 3 3 푟 3 6 6 *So, by the intermediate value theorem (See proposition 10), we have : 1 2 1 ∃훼 ∈] , − [ such that 푓 (훼 ) = 0 푘,푟 2 3 푟 푘,푟 푘,푟 *The result follows. 1 2 1 (b) and (ii) *By proposition 11, the bounded sequence(훼 ) ⊂] , − [ (for a fixed 푟 and a variable 푘), has a 푘,푟 푘 2 3 푟 1 2 1 convergente subsequence, denoted also(훼푘,푟)푘, such that : lim 훼푘,푟 = 훼푟 = 훼 ∈ [ , − ]. 푘→+∞ 2 3 푟 2 2 −훼 −훼 * By proposition 9 : lim 푓푘,푟(훼푘,푟) = 0 ⇔ 훼(|2푛 − 푎 − 푏|)3 = (1 − 훼)(|2(푚 − 1) − 푝 − 푞|)3 > 0 ⇔ 푘→+∞ 푐푎푟푑(퐴2푛) = 0 ⇔ 푛 ∈ 퐶

1 Remark : 훼 ≠ , because of our hypothesis «|2푛 − 푎 − 푏| < |2(푚 − 1) − 푝 − 푞| » 2 (iii) The result follows by combination of (i), (ii) of the under-claim2. (iv)The result follows by combination of claim6, claim7 and the assertion (iii) of the under-claim 2.

Under-claim3 : We have :푚 = 2

Proof : (of under-claim3) *By lemma 7, we have : 2 ∉ 퐶. *So, by the assertion (iv) of the under-claim 2 : 푚 − 1 = max(퐶) < 2 ⇔ 푚 < 3. *This contradicting our hypothesis « 푚 ≥ 3 », we have effectively showed that : 푚 = 2.

CONCLUSION I have showed the Goldbach binary conjecture via the Schoenfeld inequality, using elementary mathematical analysis tools, notably the intermediate value theorem.

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