GOLDBACH CONJECTURE Goldbach conjecture: ”Each even number greater than six, could be written as sum of two primes” This conjecture is almost experimentally proven on computer , but we still miss theoretical proof. I’ll try to reach it by the method showed below. Let define set S: the S is set of primes a1,a2,a3,a4,a5…an ,such that a1<a2<a3<a4<a5<…<an. Now we need interval of numbers L (0,a1*a2*a3*a4*a5*…*an). We have to prove that in interval L exists one more prime number a_(n+1): If we take number m=a1*a2*a3*a4*a5…*an, and increase it for 1,we got the number n=m+1=a1*a2*a3*a4*a5…an+1.Number n is odd and it’s obvious that it cannot be combination of the elements of the set S. (a2)^p(a3)^q(a4)^r(a5)^s(a6)^t=a1a2a3a4a5a6a7+1 (a2)^p(a3)^q(a4)^r(a5)^s(a6)^t-a1a2a3a4a5a6a7=1 a2* ((a2)^(p-1)(a3)^q(a4)^r(a5)^s(a6)^t-a1a3a4a5a6a7)=1 but it’s obvious that difference between is >=1 so we have the proof that in interval L has to be one more prime at least to form the number n,so we can write that: a1*a2*a3*a4*a5*…*a_(n-1)>a_n ........................................(i) That means a1*a2>a3 a1*a2*a3>a4 so the inequality i could be wriiten as a1*a2*(a1*a2)*(a1*a2*a1*a2.....=(2*3)^(2^(n-2))>a_n or 6^(2^(n-2))>an if we approximate 6 with 2^3 then 2^(3*2^(n-2))>an if we approximate 3 with 2^2 we go that 2^(2^(n))>an.......(D) according to 2^k>k^2 for k>5 the largest value an can have to satisfy inequality (D) is an<k^2.....(U) because (D) is satisfied for all k so that means 2^(2n)>an and again repeat (U) (2n)^2>an 4n^2>an n^2>an/4 and it is easy to prove that a1*a2*a3*a4*a5*...a_(n-1)>6an: or a3*a4*a5*...*a_(n-1)>an...........(G): Proof is the same as to prove (A): t=a3*a4*a5*a6*a7+1 isn't divisible by the elements of the set B= (a3,a4,a5,a6,a7) because if it is divisible then t could be written as (a3)^q(a4)^r(a5)^s(a6)^t=a3a4a5a6a7+1 (a3)^q(a4)^r(a5)^s(a6)^t-a3a4a5a6a7=1 a3* ((a3)^(q-1)(a4)^r(a5)^s(a6)^t-a4a5a6a7)=1 it's obvious that expression in the brackets is higher or equal 1. end of proof. But that doesn't mean that t isn't divisible by 6. And it's obvious that the smallest number which is divisible by set B is a3*a4*a5*a6*a7+a3 because a3* ((a3)^(q-1)(a4)^r(a5)^s(a6)^t-a4a5a6a7)=a3 or which is possible ((a3)^(q-1)(a4)^r(a5)^s(a6)^t-a4a5a6a7)=1 a3=5 odd number isn't divisible by 2 so the number t+1(or +3) isn't divisible by 2 because number t-1 is odd. So if t+1 is divisible by 3 then it could be written as 3^k and t+3 is 3^k+2 and it cannot be divisible by 3 and as it is odd it cannot been divisible by 2 so a3*a4*a5*a6>a7 or for all n a3*a4*a5*...*a_(n-1)>an..............(G) End of proof. according to (G) 4n^2>6a_n or n^2>3*a_n/2 to simplify inequality we it is allowed to say: n^2>a_n....................(Z) now We have to determine number of numbers that can be sum of k pairs of different primes. For example 24=13+11=17+7=23+1.the number of that combinations is n*(n+1)/2 when we sub it from n^2 we get that of n primes could be at least formed n(n-1)/2 now let sub of an, n (because there are n prime numbers in interval a_n) and then divide it by 2 to eliminate odd numbers and using (Z) (n^2-n)/2>(a_n-n)/2 (n^2-n)/2=n(n-1)/2>(a_n-n)/2..............(R) where (a_n-n)/2 is actually number od even numbers in interval (0,an] so the GoldBach conjecture is proven.