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Radic

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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.

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.