Definitive General Proof of Goldbach's conjecture
DDefinitive General Proof of Goldbach’sConjecture
Kenneth A. Watanabe, PhDDecember 6, 2018
The Goldbach conjecture states that every even integer is the sum of twoprimes. This conjecture was proposed in 1742 and, despite being obviouslytrue, has remained unproven. To prove this conjecture, I have identified asubset of the even numbers that have relatively few prime pairs comparedto the other even numbers. This subset is the set of all n such that n = 2 p where p is prime. An equation was derived that determines the number ofprime pairs for n = 2 p . It is then proven that the equation never goes tozero for any n , and as n increases, the number of prime pairs also increases,thus validating Goldbach’s conjecture. On June 7th, 1742, the German mathematician Christian Goldbach [1] wrotea letter to Leonhard Euler in which he proposed that every even integergreater than 2 can be expressed as the sum of two primes. From the timeit was proposed until now, the conjecture was suspected to be true as Eulerreplied to Goldbach "That ... every even integer is the sum of two primes,I regard as a completely certain theorem, although I cannot prove it." [2].However, the conjecture has remained unproven for over 250 years. In March2000, Faber and Faber Co. offered a $1 million prize to anyone who can proveGoldbach’s conjecture. Unfortunately, the offer expired in 2002 and the prizewent unclaimed. 1 a r X i v : . [ m a t h . G M ] D ec graphic representation of the Goldbach partitions by Cunningham andRingland [3] (Figure 1) shows all the prime pairs for even n from n = 4 to n = 96 . It appears that in general, the number of prime pairs increaseswith increasing n, but there are many exceptions. Note that where n = 2 p where p is a prime number, the number of prime pairs are fewer than when n = 2 × × p or n = 2 × × × p . For example, 94 has 5 prime pairs, but96 has 7 prime pairs and 90 has 9 prime pairs. Same with 62 versus 66 or60. In fact, it turns out that the values where n has the fewest prime pairsoccur for n = 2 i × p where i is an integer from 1 to log ( p ) or n = 2 j where jis an integer > n =2 p , n = 2 × p (or n = 6 p ), n = 2 × p (or n = 10 p ) and n = 2 × × p (or n = 30 p ) where p is a prime number (Figure 2). Note that the numberof prime pairs is significantly fewer when n = 2 p (blue line) compared tothe other values of n . The more prime factors of n , the higher the numberof prime pairs as can be seen by the other curves n = 2 × p (orange line), n = 2 × p (gray line) and n = 2 × × p (green line). Note that all thecurves have a positive slope, i.e. the number of prime pairs increases withincreasing n , and the gap between the curves also increases with increasing n . Goldbach’s conjecture was shown to be valid for all values of n up to n = 4 × by Silva et al [4] and the curves in their research look similar tothose in Figure 2. This of course does not mean that there is no possibilitythat somewhere far out at n = 10 or n = 10 or even higher that thecurve will turn around and go down to zero or that maybe there is a singlevalue for n somewhere that cannot be expressed as the sum of two primes.The following proof confirms that Goldbach’s conjecture is true for all evenvalues for n . Before we get into the proof, let me define a couple of functions that arenecessary for the proof.Let the function l ( x ) represent the largest prime number less than x . Forexample, l (10) = 7 , l (20) = 19 and l (19) = 17 .2igure 1: Goldbach partitions of even numbers from 4 to 96. The open circleson the intersections of the purple and pink lines represent the prime pairsthat total the n value on the vertical column. [3]3igure 2: There are a fewer number of prime pairs when n = 2 p (blue line)versus n = 6 p (orange line), n = 10 p (gray line) and n = 30 p (green line).Number of prime pairs for (A) even integers n less than 5,000 and (B) evenintegers n less than 50,000.Let the function g ( x ) represent the next higher prime number greaterthan x . For example, g (10) = 11 , g (20) = 23 and g (23) = 29 .Let capital P represent all pairs ( x, y ) such that x + y = n and both x and y are odd integers greater than 1. The values of x or y need not be prime. To prove Goldbach’s conjecture, we have to show that for any given even in-teger n there is at least one pair ( p x , p y ) such that both p x , and p y are primeand p x + p y = n . Since all primes except 2 are odd, we will only consider oddintegers and we will exclude any pair containing 1 since 1 is not considereda prime number. The first step is to find the set of all pairs of odd integers ( x, y ) such that x + y = n and x and y are both greater than 2. Then we mustcalculate the number of pairs ( u, v ) such that either u or v is evenly divisibleby a number > 2 (i.e. not prime), and that u + v = n . If we can show that4he number of pairs ( x, y ) , is greater than the number of pairs ( u, v ) , thenwe know that there exists at least one pair ( x (cid:48) , y (cid:48) ) such that x (cid:48) + y (cid:48) = n and x (cid:48) and y (cid:48) are prime numbers.For every even integer n , there are n/ pairs of odd numbers ( x, y ) such that x + y = n . Excluding pairs (1 , n − and ( n − , , there are P = ( n/ − pairs of odd numbers ( x, y ) such that x + y = n and x and y are both greaterthan 1.The following table shows some examples of the values of n and the num-ber of pairs. n Table 1. Relation between n and number of pairsThe next step is to count all pairs ( u, v ) where u + v = n and either u isnot prime or v is not prime. To do this, we must find all pairs where the x or y coordinate is divisible by 3 but x (cid:54) = 3 and y (cid:54) = 3 . Then find all pairswhere x or y is divisible by 5, then 7, then 11 etc. until we reach a primenumber p , such that there are no pairs divisible by p . Then we sum up all thenon-prime pairs. If for all even integers n , the total of non-prime pairs is lessthan the total number of pairs, then we have proven Goldbach’s conjecture. There are two cases for calculating the number of pairs divisible by 3:5ase 1: n is an even integer and n is evenly divisible by 3.In this case, there are exactly ( P + 2) / pairs where x or y is divisible by 3.That is because for pair (3 , n − , n − is evenly divisible by 3. For pair (9 , n − , both the 9 and n − are divisible by 3. For any odd integer i ,the pair (3 i, n − i ) has both x and y coordinates divisible by 3. For pair ( n − , , n − is evenly divisible by 3.For example, let’s take n = 990 where P = (990 / − pairs.There are (493 − / pairs such that the x coordinate is divisible by3 and x (cid:54) = 3 . For the case where x = 3 , (3,987), the y coordinate 987 isdivisible by 3. So in total, there are (( P + 2) /
3) = 165 pairs where either thex or y coordinate is divisible by 3 for n = 990 .Case 2: n is an even integer and n is not evenly divisible by 3In this case, there are ×(cid:98) ( P − / (cid:99) pairs where x or y is divisible by 3. Forany n not divisible by 3, with P pairs of odd numbers, there are (cid:98) ( P − / (cid:99) pairs where the x coordinate is divisible by 3. We have to subtract 1 from P since we do not count the pair (3 , n − . Likewise, for the y coordinate,there are (cid:98) ( P − / (cid:99) pairs where the y coordinate is divisible by 3 excluding ( n − , . For even integer n not divisible by 3, the total number of pairs ( x, y ) where x + y = n and either the x or y coordinate is divisible by 3 is × (cid:98) ( P − / (cid:99) .For example, let’s take n = 994 where P = (994 / − pairs.There are (cid:98) (495 − / (cid:99) = 164 pairs such that the x coordinate is divisibleby 3. (9,985), (15,979),(21,973), (27,967), ... (x,n-x),(x+6,n-(x+6)),(x+12,n-(x+12)), ... (981,13),(987,7). There is (cid:98) (495 − / (cid:99) = 164 pairs such thatthe y coordinate is divisible by 3. (985,9), (979,15),(973,21), (967,27), ...(13,981),(7,987). For n = 994 and P = 498 , there are
164 + 164 = 328 pairssuch that the x coordinate or y coordinate is divisible by 3. That means thereare −
328 = 170 pairs where both x and y coordinates are not divisibleby 3. As can be seen, for a very large n not divisible by 3, the number ofpairs evenly divisible by 3 approaches the following equation: Pairs divisible by 3 = × (cid:98) ( P − / (cid:99) lim n →∞ = P × (2 / For low values of n, this approximation may overestimate the number ofpairs evenly divisible by 3. But this errs on the side of caution since we will6e subtracting this number from the total number of pairs to calculate thenumber of prime pairs.When n = 994 there are 328 pairs where x or y is divisible by 3 comparedto when n = 990 there are only 165 pairs where x or y is divisible by 3. Thereare twice as many pairs divisible by 3 when n is not divisible by 3. There are two cases for calculating the number of pairs divisible by 5, justas there were for pairs divisible by 3.Case 1: n is an even integer and n is evenly divisible by 5 and P = ( n/ − .In this case, there are exactly ( P + 2) / pairs where x or y is divisible by 5.The reasoning is very similar a that for the case where n is divisible by 3.The pair (5 , n − , n − is evenly divisible by 5. For (15 , n − , both the xand y are divisible by 5. For any odd i , the pair (5 i, n − i ) has both x andy divisible by 5. For the pair ( n − , , n − is evenly divisible by 5.For example, let’s take n =990 where P = (990 / − pairs. Thereis (cid:98) (493 − / (cid:99) = 98 pairs such that the x coordinate is divisible by 5. Thereis (cid:98) (493 − / (cid:99) = 98 pairs such that the y coordinate is divisible by 5. Butbecause all of the pairs where x and y are greater than 5, every pair wherex is divisible by 5, y is also divisible by 5. (5,985), (15,975), (25,965), ... Forthe pair (5,985), y is divisible by 5, and pair (985,5) x is divisible by 5. Soin total, there are 99 pairs where either x or y is divisible by 5 for n = 990.Of these 99 pairs, 33 pairs of these numbers are also divisible by 3.Case 2: n is even and is not evenly divisible by 5 and P = ( n/ − .In this case, there are × (cid:98) ( P − / (cid:99) pairs where x or y is divisible by 5. Forany n not divisible by 5, with P pairs of odd numbers, there are (cid:98) ( P − / (cid:99) pairs where the x coordinate is divisible by 5. We have to subtract 2 from Psince we do not count the first two pairs (3 , n − and (5 , n − . For any n not divisible by 5, with P pairs of odd numbers, there are (cid:98) ( P − / (cid:99) pairswhere the y coordinate is divisible by 5. The total number of pairs whereeither the x or the y coordinate is divisible by 5 is × (cid:98) ( P − / (cid:99) where n is not divisible by 5.For example, let’s choose an n that is not divisible by 5 to maximize the7umber of pairs that have an x or y that is divisible by 5. So we cannotuse n = 990 as in our previous example. Let’s choose n = 994 which has P = 495 pairs. The number 994 is not evenly divisible by 3 or 5. There are × (cid:98) (495 − / (cid:99) = 196 pairs such that the x or y coordinate is divisibleby 5. There are (cid:98) (495 − / (cid:99) = 98 pairs such that the x coordinate isdivisible by 5. (15,979), (25,969),(35,959), (45,949), ... (975,19),(985,9).Likewise for the y coordinate, there are (cid:98) (495 − / (cid:99) = 98 pairs divisibleby 5. (979,15), (969,25),(959,35), (949,45), ... (19,975),(9,985). So in total,there are × (cid:98) (495 − / (cid:99) or 196 pairs where either x or y is divisible by 5.When n = 994 there are 196 pairs where x or y is divisible by 5 comparedto when n = 990 there are only 99 pairs where x or y is divisible by 5. Thereare twice as many pairs divisible by 5 when n is not divisible by 5.Out of these 196 pairs, there are 132 pairs (two-thirds) that have an xor y coordinate that is divisible by 3. This is because for every x-coordinatedivisible by 5 starting with (15,979), every third pair where x is divisible by5 is also divisible by 3 (yellow numbers) and starting with (25,969), everythird pair the y coordinate is divisible by 3 (orange numbers).( 15 ,979),(25, 969 ),(35,959),( 45 ,949),(55, 939 ),(65,929),( 75 ,919),(85, 909 ),(95,899),( 105 ,889),(115, 879 )So only about one third (64 pairs) of the 196 pairs where x or y is divisibleby 5 are not divisible by 3. For very large n not divisible by 5, the numberof pairs divisible by 5 and not 3 approaches the following equation: Pairs divisible only by 5 lim n →∞ = P × (1 / × (2 / Let’s just look at the case where n is not divisible by 7 because if n wasdivisible by 7, the pairs where the x coordinate is divisible by 7 will overlapwith the cases where the y coordinate is divisible by 7. If n is divisible by 7,then only about / th of the pairs would be divisible by 7, whereas if n is notdivisible by 7, then / th of the pairs would be divisible by 7. In fact, any n = 2 qp where q and p are prime numbers and < q < p , the pairs wherethe x coordinate is divisible by q , the y coordinate will also be divisible by q . This overlap reduces the number of non-prime pairs thus increasing thenumber of prime pairs. The goal is to find the values of n with the fewestnumber of prime pairs to prove that there are no values of n with zero prime8airs. Thus, the values of n with the fewest prime pairs would be n = 2 j for any integer j greater than 2 or n = 2 i p where i is an integer from 1 to log ( p ) . These values have no prime number q which evenly divides both thex and y coordinate.For any even n not divisible by 7, with P pairs of odd numbers, thereare (cid:98) ( P − / (cid:99) pairs where the x coordinate is divisible by 7. We have tosubtract 3 from p since we do not count the first three pairs (3 , n − , (5 , n − and (7 , n − . Likewise for the y coordinate, there are (cid:98) ( P − / (cid:99) pairs divisible by 7. The total number of pairs where either the x or the ycoordinate is divisible by 7 for n not divisible by 7 is × (cid:98) ( P − / (cid:99) . Thegeneral formula for the number of pairs of odd numbers divisible by a primenumber p for n not divisible by p is: × (cid:98) P − (( p − / /p (cid:99) .For example, let’s choose an n = 998 which has 497 pairs. The number998 is evenly divisible only by prime numbers 499 and 2. Thus, 998 takesthe form n = 2 p . Values of n = 2 p have the maximum number of non-primepairs and thus represent the lower bound on the number of prime pairs.There are × (cid:98) (497 − / (cid:99) = 330 pairs such that the x or y coordinateis divisible by 3.There are × (cid:98) (497 − / (cid:99) = 198 pairs such that the x or y coordinateis divisible by 5.Of these 198 pairs, there are (198 / − pairs where the x coordinateis divisible by 3 and an equal number where the y coordinate is divisible by3 giving a total of 130 pairs divisible by both 3 and 5. Subtracting the130 pairs that are divisible by 3 gives 68 pairs divisible by 5 and not 3.This number closely resembles P × (1 / × (2 / which in this case equals × (1 / × (2 /
5) = 66 . .There are × (cid:98) (497 − / (cid:99) = 140 pairs such that the x or y coordinateis divisible by 7. Subtracting the pairs that are also divisible by 3 or 5 gives28 pairs divisible only by 7. For very large n , the number of pairs divisibleby 7 and not 5 or 3 approaches the following equation: Pairs divisible only by 7 lim n →∞ = P × (1 / × (3 / × (2 / Number of pairs divisible by a prime > 7
For any n not divisible by prime p , with P pairs of odd numbers, there are (cid:98) [ P − (( p − / /p (cid:99) pairs where the x coordinate is evenly divisible by p .Likewise for the y coordinate, there are (cid:98) [ P − (( p − / /p (cid:99) pairs that areevenly divisible by p . The total number of pairs where either the x or the y9oordinate is divisible by p is × (cid:98) [ P − (( p − / /p (cid:99) where n is not evenlydivisible by p.There are × (cid:98) (497 − / (cid:99) = 88 pairs such that the x or y coordinate isdivisible by 11. Subtracting the pairs that are also divisible by 3, 5 or 7 gives12 pairs divisible only by 11. For very large n , the number of pairs divisibleby 11 and not 7, 5 or 3 approaches Pairs divisible only by 11 lim n →∞ = P × (1 / × (3 / × (5 / × (2 / There are × (cid:98) (497 − / (cid:99) = 74 pairs such that the x or y coordinateis divisible by 13. Subtracting the pairs that are also divisible by 3, 5, 7 or11 gives 8 pairs divisible only by 13. For very large n , the number of pairsdivisible by 13 and not 11, 7, 5 or 3 approaches Pairs divisible only by 13 lim n →∞ = P × (1 / × (3 / × (5 / × (9 / × (2 / The general formula for number of pairs divisible by p and not any lowerprime approaches the following equation as n gets very large is as folows: Pairs divisible only by p lim n →∞ = P × (1 / × (3 / × (5 / × (9 / × × ( l ( p ) − /l ( p ) × (2 /p ) Continuing on, the number of pairs divisible by prime numbers for n =998 is summarized in the Table 2.Note that the last prime number that had at least 1 pair where x or ywas divisible by it was 31. Also notice that l ( √ also equals 31.When calculating the number of pairs divisible by a prime number p i ,we only need to check if the x or y coordinate is divisible by prime numbersless than or equal to p i = l ( √ n ) . For any prime p i < l ( √ n ) , excludingpair ( p i , n − p i ) , the next pair where x is a multiple of p i is (3 p i , n − p i ) ,but this pair is excluded because p i is divisible by 3. The next pair is (5 p i , n − p i ) , but this pair is also excluded because it is divisible by 5.Same with p i , p i , p i etc. So all pairs will be excluded until we reach ( p i × p i , n − p i × p i ) . If p i × p i is greater than n, then the y coordinate willbe less than 0. Thus, there are no pairs that are evenly divisible by p i larger10 p pairs divisible byprimes less than p p Table 2. Number of non-prime pairs ( x, y ) such that x + y = 998 .than l ( √ n ) . The pairs only need to be checked if they are divisible by primesless than or equal to l ( √ n ) .For example, n = 100 , √
100 = 10 and l (10) = 7 . Out of all the oddpairs that total 100, there are only seven pairs where x coordinate is evenlydivisible by 7, (7,93), (21,79), (35,79), (49,51), (63,37), (77,23) and (91,9).Excluing the pair where x = 7 and pairs divisible by 3 or 5, leaves only 1pair remaining (77,23).If p i = 11 , the next higher prime over 7, there are only 4 pairs where x isevenly divisible by 11, (11,89), (33,67), (55,45) and (77,23). Ignoring pairswhere x = 11 and pairs divisible by 3, 5 or 7 leaves no pairs where the xcoordinate is evenly divisible by 11 and not 3, 5 or 7. The total number of pairs that contain an x or y that is evenly divisible bya prime is as follows:
Non-prime pairs = (cid:80) ( p = l ( √ n ))( p =3) pairs evenly divisible by p and no lower prime For very large n = 2 p where p is prime, the total number of pairs that eitherx or y is divisible by a prime number (i.e. not prime) is as follows:11 × (2 /
3) + P × (2 / × (1 /
3) + P × (2 / × (1 / × (3 /
5) + P × (2 / × (1 / × (3 / × (5 /
7) + P × (2 / × (1 / × (3 / × (5 / × (9 /
11) + P × (2 /l ( √ n )) × (1 / × (3 / × (5 / × (9 / × × ( l ( l ( √ n )) − /l ( l ( √ n )) where P is the number of pairs such that x + y = n .This can be expressed as the following equation: Number of Non-prime Pairs = P × ( (cid:80) ( l ( √ n ))( p =3) (2 /p ) (cid:81) l ( p ) q =3 ( q − /q where the summation and products are over prime numbers.Let us define the function W(x) as the following: W ( x ) =(1 /
3) +(1 / × (1 /
3) +(1 / × (1 / × (3 /
5) +(1 / × (1 / × (3 / × (5 /
7) +(1 / × (1 / × (3 / × (5 / × (9 /
11) +(1 /x ) × (1 / × (3 / × (5 / × (9 / × × ( l ( x ) − /l ( x ) This can be expressed as the following equation: W ( x ) = (cid:80) xp =3 1 p (cid:16)(cid:81) l ( p ) q =3 ( q − q (cid:17) Using this expression for W(x), the expression for number of pairs that con-tain a non-prime number can be simplified to
Number of non-prime pairs = P × W ( l ( √ n )) Subtracting the number of non-prime pairs from the total number of pairs P gives us the number of prime pairs. Number of prime pairs = P − P × W ( l ( √ n )) = P [1 − W ( l ( √ n ))] Thus to prove Goldbach’s conjecture, we have to prove that the number ofprime pairs is greater than 0 as defined by Equation 1 below.12igure 3: The Equation P(1-2W) Accurately Predicts n gets large. The equation P(1-2W) conservativelyunderestimates the number of prime pairs (A.) for values of n less than 5000,but as n increases (B.) the number of prime pairs approaches P(1-2W). Equation 1: P (1 − W ( l ( √ n ))) > . or in terms of n ( n/ − × (1 − W ( l ( √ n ))) > .Equation 1 should equal the number of prime pairs for very large n = 2 p .To confirm that no error was made in the derivation of this equation and todetermine how close the equation approximates the number of prime pairs, Iplotted all values for P (1 − W ( l ( √ n ))) versus n for n = 10 to 5,000 and for n = 10 to 50,000 and compared them to the actual number of prime pairs.As can be seen in Figure 3A, the curve for P (1 − W ( l ( √ n ))) (red curve)conservatively underestimates the number of prime pairs compared with theactual number of prime pairs for n = 2 p (blue curve) for n less than 5,000.But as n gets large, (Figure 3B) the actual number of prime pairs does indeedapproach P (1 − W ( l ( √ n ))) .To utilize proof by induction, we must get (1-2W(p)) in terms of W(l(p)).To do this, we must look at the actual values of × W ( p ) . × W (3) = (2 / × W (5) = (2 /
3) + (2 / × (1 / × W (7) = (2 /
3) + (2 / × (1 /
3) + (2 / × (1 / × (3 / × W (11) = (2 /
3) + (2 / × (1 /
3) + (2 / × (1 / × (3 /
5) + (2 / × (1 / × / × (5 / Etc ...Therefore, the values of − × W ( p ) are as follows: − W (3) = 1 - (2/3) = 1/3 − W (5) = [1 - (2/3)] - (2/5)(1/3) = (1/3) (3/5) − W (7) = [1 - (2/3) - (2/5)(1/3)] - (2/7)(1/3)(3/5) = (1/3)(3/5)(5/7) − W (11) = [1 - (2/3) - (2/5)(1/3) - (2/7)(1/3)(3/5)] - (2/11)(1/3)(3/5)(5/7)= (1/3)(3/5)(5/7) 9/11)Notice the value of 1-2W(p) (yellow) can be substituted into the green partof 1-2W(g(p)). Therefore, these equations can be simplified to: Equation 2: [1 − × W ( p )] = [( p − /p ] × [1 − × W ( l ( p ))] Another way to think about how we get to equation 2 is by cutting awaypieces from a pie.The pie has a value of 1. We cut away / rds from the pie leaving 1/3 .Now from this piece , we cut / ths away leaving / ths of 1/3 .Now from this piece , we cut / ths away leaving / ths of the last piece .Now from this piece , we cut / ths away leaving / ths of the last piece .For each iteration, we cut away /p ths leaving ( p − /p of the previous piece,thus resulting in equation 2.To prove Equation 1, we will use proof by induction. Since we know P isgreater than 0, we must prove that 1- 2W(p) > 0.The base case p = 3 . − W ( p ) = 1 − W (3) = 1 − / / which is greater than 0.Now assuming that Equation 1 is true [1 - 2W(p) > 0], we must prove that1 - 2W(g(p)) > 0.Assumption: 1 - 2W(p) > 0Prove: 1 - 2W(g(p)) > 0Substituting g(p) into Equation 2 gives: − W ( g ( p )) = [( g ( p ) − /g ( p )] × [1 − × W ( p )] Since we assumed that [1 − × W ( p )] is greater than 0, and [(g(p)-2)/g(p)] isgreater than 0, then the product of [(g(p)-2)/g(p)] and [1 − × W ( p )] must14e greater than 0. Thus the equation holds true for g(p) and the GoldbachConjecture has been proven for n = 2 p . Since n = 2 p represents the lowerbound on the number of prime pairs for n , the Goldbach Conjecture has beenproven for all even integers n . n Increases
We can also prove that as n increases, the number of prime pairs also in-creases. Suppose we have P pairs for even number n and we have P [1 − W ( l ( √ n ))] prime pairs. If we approximate P = n/ for large n , we get thefollowing equation: P [1 − W ( l ( √ n ))] = ( n/ − W ( l ( √ n ))] For prime number p = l ( √ n )) , for large n we can approximate n ≈ p .Actually, n will be at least p + 1 , but approximating n ≈ p errs on the sideof caution. Substituting for n we get P [1 − W ( l ( √ n ))] = ( p / − W ( p )] The number of prime pairs for n = p is : ( p / − W ( p )] .The number of prime pairs for n = g ( p ) is : ( g ( p ) / − W ( g ( p ))] =( g ( p ) / g ( p ) − /g ( p )] × [1 − W ( p )] = Using equation 2 g ( p )[( g ( p ) − / × [1 − × W ( p )] Subtracting the number of prime pairs for n = p from n = g ( p ) givesus the following expression: g ( p )[( g ( p ) − / × [1 − W ( p )] − ( p / − W ( p )] =[(1 − W ( p )) / × [ g ( p )( g ( p ) − − p ] Since we have just proven that (1 - 2W(p)) > 0, and g(p) is greater thanp, the value [ g ( p )( g ( p ) − − p ] is always greater than 0. Thus the productis greater than 0. This proves that the number of prime pairs for n = g ( p ) is greater than the number of prime pairs for n = p . Thus, the number ofprime pairs increases with increasing n .15 I have shown that even integers n=2p, where p is prime, have relatively fewerprime pairs than even integers n (cid:54) = 2 p . I have shown that for even integersn=2p, as n gets large, the number of prime pairs approaches the equation: Number of prime pairs = (( n/ − − W ( l ( √ n ))) where l ( √ n ) is the largest prime number less than √ n and W(x) is definedas W ( x ) = (cid:80) xp =3 1 p (cid:16)(cid:81) l ( p ) q =3 ( q − q (cid:17) I have shown by proof by induction, that the above equation for number ofprime pairs does not go to zero and in fact increase indefinitely as n increases.It is also shown that even integers n = 2 p is the lower bound on the numberof prime pairs compared to even integers n (cid:54) = 2 p , thus validating Goldbach’sconjecture for all even integers n .
11 Future Directions
Future work will involve applying this technique of pairing numbers to provethe Twin Prime Conjecture and Polignac’s Conjecture [5]. Polignac’s Con-jecture states that there is an infinite number of prime pairs ( p , p ) suchthat | p − p | = 2 i where i is an integer greater than 0. The Twin PrimeConjecture is the case where i = 1 .To prove the Twin Prime conjecture, we need to find the number of twinprimes less than an integer n , ( π ( n ) ) . To do this, we first pair odd numbers ( x, y ) such that x +2 = y and y < = n . For example, (3,5),(5,7),(7,9),(9,11)...,(n-4,n-2),(n-2,n). Then by eliminating pairs that are divisible by 3, 5, 7, 11 etc,the remaining pairs are twin primes.The number of twin primes less than n will approach the following equationas n gets large: π ( n ) = P (1 − W ( l ( √ n ))) This equation is identical to Equation 1 of the Goldbach proof. What thismeans is, that for large values of n, the number of twin primes less than n16ill approach the number of prime pairs ( p , p ) such that p + p = n andthe proof of the Twin Prime Conjecture is reduced to the proof of Goldbach’sConjecture for n = 2 p .For other cases of Polignac’s Conjecture, for example primes separated by6, 10 or 30, are cases of the Goldbach Conjecture for n = 6 p, n = 10 p or n = 30 p . Thus, Polignac’s Conjecture can be proven.Applying this technique to other prime number conjectures will lead to fur-ther proofs.
12 ReferencesReferences [1] Goldbach, C.
Letter to Euler, Correspondance mathématique et physiquede quelques célèbres géomètres du XVIIIème siècle (Band 1), St.-Pétersbourg 1843, pp. 125-129[2] Ingham, AE. "Popular Lectures" (PDF). Archived from theoriginal (PDF) on 2003-06-16. Retrieved 2009-09-23. https://web.archive.org/web/20030616020619/http:/claymath.org/Popular_Lectures/U_Texas/Riemann_1.pdf [3] Adam Cunningham and John Ringland
Goldbach partitions of the evenintegers from 4 to 96 https://commons.wikimedia.org/wiki/File:Goldbach_partitions_of_the_even_integers_from_4_to_96.svg [4] Tomás Oliveira e Silva, Siegfried Herzog and Silvio Pard
Empirical ver-ification of the even Goldbach conjecture and computation of prime gapsup to × Mathematics of Computation, 83 (2014), 2033-2060[5] Alphonse de Polignac
Recherches nouvelles sur les nombres premiers
Comptes Rendus des Séances de l’Académie des Sciences (1849)
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