The 2 Goldbach's Conjecture by Proof Mantzantzakouras Nikos

Abstract:The1741Goldbach[1]madehismostfamouscontributionto mathematicswith the conjecture that all even numbers can be expressed as thesumofthetwoprimes(currentlyConjecturereferredtoas"alleven numbersgreaterthan 2canbeexpressedasthesum-twoprimes).Yet,no proof of Goldbach's Conjecture has been found. This assumption seems to be rightforalargeamountofnumbersusingnumericalcalculations.Some examples are 10 = 3 + 7, 18 = 7 + 11, 100 + 97 = 3, and so on. But there are amultitudesumsprimesmeetinganevennumberwhichIwouldsayis irregular,butincreaseswiththesizeoftheevennumber.Tosamehappens and with an odd number as example 25, ie. 25 = 3 + 3 + 19 = 3 + 5 + 17 = = 3 + 11 + 11 = 5 + 7 + 13 = 7 + 7 + 11. We say only 5 cases and only those that meettheconstantsumof25.Firstturnsin Theorem3thattheremaybeat least a pair of primes, such that their sum is equal to every even number. But atthesametimerevealsthemethodoffindingallpairsthatsatisfythis condition.AccordingtoTheorem4,thesecondguessswitchestoformfirst guess and this is primarily elementary, to prove the truth of 1.The alternative and the side stream assistant, to prove Goldbach's Conjecture in this research wastheMathematicaprogramwhichisthe maintoolfordatacollection,but also for finding all the steps in order to demonstrate each part of the proof. We mustmention that Vinogradov proved to1937, that for every sufficiently large number can be expressed, that is the sum of the three primes. And finally the Chinese mathematician Chen Jing Run proved big first for a constant number thatisthesumofthefirstthreein1966[2].Finally,theinvestigationof Goldbach'sConjecturehasactedasacatalystforthecreationand development of many methods that are useful, with many theorems that help and other areas of mathematics. Definition. A natural number p > 1 is called prime number if and only if the only divisors of is (+/-) 1 and (+/) p. A natural number n > 1 which is not prime, will be called composite. Therefore a positive integer p > 1 is Prime if and only if for each 1 ) , ( = n p ,orp n p = ) , (i.e the related prime to the n or p divides n. The 2 is first but outside of 2 is odd numbers. The 25 first primes numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 67, 71, 73, 79, 83, 89, 97.We note that the number 1 is not considered neither prime nor composite. Lemma 1. Every prime number greater than 1, it is odd and is written in the form1 2 + kand therefore the sum of 2 primes is an even number. Proof .. Let p, prime in N. Supposekp , + kp by N e k, ,k kp p >+then we will have1 ) ( 2 1 2 + + + + = ++k k p pk k) 1 2 ( 2 + + = k to be even as easily seen. Lemma 2. The sum of 3 odd numbers it is an odd number. Proof..If p, prime in N ,natural numbers. Assume,kp , + + k kp p , , , , N e kk k kp p p > >+ + then we'll have= + + + + + + + = + ++ +1 2 2 1 2 2 1 2 k k k p p pk k k 1 ) 1 3 ( 2 + + + + = kthat's odd as easy it seems. Theorem.1 If we have a integerN h h n e = , 2even number, asking to find the upper and lower limit of the sum of two odd, which equal with n. Proof.. From Lemma 1 we have: i). N hnh n h n h h n e >> > = + = , 0 ,4 21 2 / 2 1 2 / 2 ) 1 2 ( 2 .Ii).The largest odd should be less than3 n . Suppose therefore that 2 43 1 2s s +nh n h. So the limits of the number of all cases, it will be2 44 2 s < nhn. TheExistence Theorems 2 or 3primes, constant sum. Theorem 2.ForN w h h n n e = > , , 2 , 2 and2 1p p n + =,which 2 1, p pare prime numbers, and there is a positive integerj withw j 2 =if2 / nis oddand 1 2 + =w j if2 / nis even, thenj n 2 /andj n + 2 /is odd and possibly also prime numbers.[8]. Proof . .. i) We assume that we have 2 oddand possibly primesc = 2 /1n pand + = 2 /2n p with,to be positive numbers. Because c c = + + = + = 2 / 2 /2 1n n p p n .ii) Previouslyby (i), we assume that thereis aninteger,0 > jso c = = j ,andthen implies that 2 1p p n + = withj n p = 2 /1 andj n p + = 2 /1.We can now transform the relationship 2 1p p n + =in accordance with previously to1 2 2) 2 / ( 2 / p p n n p n = = and also 2 2) 2 / ( 2 / p n p n = + . From the mathematical logic is clear that if w j 2 =with2 / nodd and1 2 + =w jif2 / n is even, the 2 1, p presults always odd numbers. Of course if may be primes, as you know it will always be odd numbers. Thereforethere isalwaysapositivejinteger such thatj n 2 /and j n + 2 /is odd numbers and possibly primes numbers and apply that) 2 / ( ) 2 / ( j n j n n + + = . We define as (x) is the function that determines the number of primes that are less than the equal by x. Specificallyapplicablefor the sieve of Eratosthenes that ] 2 [ x k t =integer part ofx 2 . We prove then that there may be at least a couple of primes such that there the sum is equal to n. TheFirst Conjecture of Goldbachs . Theorem 3.Given an even,N w h h n n e = > , , 2 , 2 ,2 ), ( > = k n k t and kpis prime correspondingto an index k.Then there are positive integersjso in order to apply in each case conditions: a. If2 / nis odd. thenw j 2 =and b. If2 / n is even, then1 2 + =w j so that there is at least a pair, and for which each of the pairj n 2 / j n + 2 /to be prime and not divisible by any of the primes2, 3, 5, 7,..,. kp . Proof.. We start with theN w h h h n n e >= = > , , 0 , 2 , 2from Theorem.1, and Theorem.2 occursthattherethenj n 2 / andj n + 2 / isoddandpossiblyalsoprime numbers withw j 2 = .In this case the maximum number of valuesN we will occurfrominequality 3 2 / 2 ) 3 2 / (213 2 2 / s . s s + n w n w n w n .Itis obviousastotalsolutionsthatrepresentallcasesforann/2=oddis } 3 2 / , , , 6 , 4 , 2 , 0 { = n S whilewhenn/2=evenis } 3 2 / , , , 7 , 5 , 3 , 1 { = n S .Theset thatwegetandtheinequalitymod2,i.e.Accordancewiththesieveof Eratosthenes is) 2 (mod 0 2 / = j n that so) ( ] 2 / 2 [ n n k t t = =. Must will first needfrom the setS , remove or reject all cases of individual systems modulo ... 1s) 2 (mod 0 2 /) (mod 0 2 /2= = j np j n 2s) 2 (mod 0 2 /) (mod 0 2 /3= = j np j n .. ..... ..... fs) 2 (mod 0 2 /) (mod 0 2 /= = j np j nk The comprehensive solution of all systems, isfrom the Union of the individual solutions of each system fs s s s , , , , ,3 2 1, 1 1 s s k f and is the unique solution of systems, with the respective setsfS , i.e ifiS R1 == . The clear solution such as we seek will be the difference of setsSand R , i.e R S = O . So the solutionO that we are interestedcoincides with the solution of the systemunequally mod i.e.

) (mod 0 2 /. ........... ..........) (mod 0 2 /. ........... ........... ..........) (mod 0 2 /) (mod 0 2 /21kvp j np j np j np j n= = = = for which each of the pair j n 2 / andj n + 2 / is primes and not divisible by any of the primes2,3,5,7, , vp ,...,kp ,with k v s s 1 . By this logic therefore determine that the possibility thatj n 2 / andj n + 2 /be prime numbers, it is now safe to say there may be at least a pair) , (2 1 p pof primes, such that their sum equalsn , because the setO is never empty set becauseSis never empty set andS R = withS R c .If} : { } ( O e = Oi ix x Nwith {i = 1,2,3, ..., } where the cardinality of the setO, then theN ()define the elements of the set O .The pairs of primes arising from the setO, given by the relations ... i ii ix n px n p+ = =2 / '2 /11n p pi i= + '1 1

Further to previous we symbolizedto)} ' , ),...( ' , ),( ' , {( ) (1 1 2 1 2 1 1 1 1 1 p p p p p p P = Othe setof pairs that correspond to the set O,which is derived from the system unequally mod (MJ) and constitutes the first part of the set of solutions of couples) ' , (1 1 i ip p with {i = 1, 2,, } so that the sum to equalsnand cardinality . Finally we define as T = {2,5,7, , kp } to set the of primes so that the couplesj n 2 /andj n + 2 / who are primes not divisible by any of the primes2,3,5, , kp , as definedmultitudek.More specifically define the setof primes v v vp p p T P2 2 2): ' , {( ) ( = T e and) /(2 2 v v vm p p u + , mZ e : . = e ) (mod 0 2 / {v vp j n u )} 2 (mod 0 2 / = j n } with2 , 1 > s < k k v,where vuthe residues arising from the system )} 2 (mod 0 2 / ) (mod 0 2 / { = . = j n p j nv with cardinalityo.Finally the complete set of pairs of the primes to sumnwill be the set) ( ) (2T P P LO =as defined above, with cardinality o +. TheSecondConjectureof Goldbachs Theorem 4. ForN h h n n e + = > , 1 2 , 2 and 3 2 1p p p n + + = ,with 3 2 1, , p p p prime numbers. Namely every odd number is obtained as the sum of 3 prime numbers. Proof..Accordingtotherelationship 3 2 1 3 2 1p n p p p p p n = + + + = .Ifcall i i ip p p n3 2 1+ + =and if we have many different valuesip3 and other primes, then wecancall i i i ip n p p S3 2 1 2 = + = thesumsof2primes iS2.Accordingto Theorem 3, we know the cardinality of cases areiS2,with i i o +, in each case ip n3 of different values ip3 according to each preset selection .The method of finding all the cases is precisely from selected values that are ip3 which is the firstoftheprimeslessofn,tothegreatestprimeofthetwoisclosertothe value3 / n .The fact leads the order of the Triad 3 2 1, , p p p in relationship to3 / n .i) In this case, if the value of the lower limit of the selected values ip3will be the value 3p of that is obviously the second greater price series during the first of prime numbers after i.e ) 3 / (n Round . In the mathematica language will define this number]]] 3 / [ [ Pr [ Pr n Round ime Next ime Next . ii) If3 2 1 3 2 1 3 2 13 / 3 / 3 / p n p p p n p p p n p p s < < v < < s v < < < . In this case the value of the lower limit of the selected values ip3will be the value 3p of which is obviously a next-largest number first value after) 3 / (n Round .In language of mathematicawill define this figure as]] 3 / [ [ Pr n Round ime Next .Iii) If. 3 /3 2 1p n p p = = = The value of the lower limit of the selected values ip3will be the value of 3p which is obviously the ). 3 / ( 3 / n Round n = And for the three cases in relation to each device, we symbolized as the prime minimum mino . Finally, the biggest ip3being the first of the first below, will be denoted by the language of mathematica as 1 , 6 ] , [ Pr > s n n ime Next

and will be the largest prime number beforen, less than or equal to n-6, which symbolizes maxo .If you now call the cardinality of the primes, o o o t = ]) , ([max min we will know how many sums of 2 will have, i.e {i = 1,2,3, ., }. Therefore the set of summation oftotal solutions of 2 primes will be } ,..., , {2 2 2 1 2 3 oS S S L = . If we now call i i i + = with {i = 1,2,3, ..., } by Theorem 3, the final number of the individual triadstie. the final number 2 1 t ... + + + = k k will be the total number of pairs of primes, for sums of2 ,i i i ip n p p S3 2 1 2 = + = , with cardinality and hence the total number of triads equalt ,using multitude sums with two primes, making -multiple use of the method Theorem 3. .The permissible cases o ,..., 2 , 1 ,2= i Si therefore the will is the total .. } ,..., , {2 2 2 1 2 3 oS S S L = . Namely every odd number is obtained as the sum of3 of primes numbers after respectively switched to multiple totals 2 primes, multitude,the existence of which fully ensures the Theorem 3.a) For example sum 2 primes, let's say when m = n / 2=159is odd and 83 . 17 = ntherefore i.e get to mod, the first 2,3,5,7,11,13,17 .The corresponding program with mathematica and according to the theory would be:# Program 1 #... n/2 := 159;m:=n/2; Cases[Reduce[ Mod[m + j, 17]0 && Mod[m - j, 17] 0 &&Mod[m + j, 13] 0 && Mod[m - j, 13] 0 && Mod[m + j, 11] 0 && Mod[m - j, 11] 0 && Mod[m + j, 7]0 && Mod[m - j, 7] 0 && Mod[m + j, 5] 0 && Mod[m - j, 5] 0 &&Mod[m + j, 3] 0 && Mod[m - j, 3] 0 &&Mod[m + j, 2] 0 && Mod[m - j, 2] 0 &&0 j (m - 3), {j}, Integers], Except[j]] Results: j=8,j=20,j=22,j=32,j=52,j=70,j=80,j=92,j=98,j=112,j=118,j=122} are the values for j. The} 122 , 118 , 112 , 98 , 92 , 80 , 70 , 52 , 32 , 22 , 20 , 8 { } : { } ( = O e = Oi ix x Nwith =12. Therefore the set of pairs will be = = O )} ' , ),...,( ' , ),( ' , {( ) (1 1 2 1 2 1 1 1 1 1 p p p p p p P{(37,281),(41,277),(47,271),(61,257),(67,251),(79,239),(89,229),(107,211),(127,191),(137,181),(139,179),(151,167)}. withThereafter the T = {2,5,7, ..., 17} with k = 7 and shuts off after looking at how many primes values we receive from all T. The corresponding program will be mathematica according to relevant theory # Program 2 #..n:= 159 2; m:=n;c:=IntegerPart[m];Cases[Table[Reduce[ Mod[m-p',k]==0 &&Mod[m+p',2] 0 && Mod[m-p',2] 0 && p=2 && p12 && p3>2 && p1