Corrected Proof of Goldbach's Conjecture

Discussion in 'Number Theory' started by ddinunno, Mar 29, 2023.

  1. ddinunno

    ddinunno

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    Goldbach's Conjecture: Every 2n (n > 1) is a sum of two primes.

    Proof:
    A prime number is a positive integer p greater than 1 that has only 1 and itself as product factors.

    Some prime pairs (p1, p2) are easy to identify, such that p1 + p2 = 2n for n > 1. For example, if n = 2 then 2n = 4; and if p1=p2=2, then 2 + 2 = 4 = 2n. Also, since 2n = n + n, then when n=p is prime, then 2n=2p and p1=p2=p; So, again p1 + p2 = 2n. Therefore, any prime p is its own solution to p1 + p2 = 2n.

    Also, any twin primes (primes separated by only one integer such as 3,5, 11,13, 17,19, or 41,43), also has an easy solution. Let n be the integer between the twin pair with p1 = (n + 1) and p2 = (n – 1). Then, p1 + p2 = (n +1) + (n-1) = 2n.

    In general, for any integer ‘x’ (0 <= x < n), when (n – x) and (n + x) are prime numbers, then the pair p1 = n + x and p2 = n – x is a solution to p1 + p2 = 2n.

    For all 2n < 2, the integer pairs of 2n can be viewed as a circle with odd integers paired together as shown in figure 1 (and more generally in figure 2). Integer pairs will always be “horizontal” in this view and equidistant to 2n (also to n.). Twelve, i.e., 2(6), is familiar circle (a clock with 0=12). The prime pair (7, 5) is the solution 7 + 5 = 12 = 2(6).

    upload_2023-3-28_20-8-47.png
    Figure 1: Prime-pair solutions to 2n.


    For every 2n, the sum of each integer pair = 2n. And, the midpoint (at the bottom) becomes a “pair” itself [e.g. (2,2), (3,3), or (5,5) in figure 1 ]. So, while the midpoint n = 2n/2 is a solution for n=2 and 3, Eight has a solution (5, 3) a twin prime; Ten has two solutions (5, 5) and (7, 3); and Twelve has a solution (7, 5) as another twin prime.

    For any 2n, these pairs are generalized as (2n-1 ,1), (2n-2 ,2), (2n-3, 3) … (n, n) as shown in figure 2.
    upload_2023-3-28_20-9-22.png
    Figure 2: Circle of Integer Pairs of 2n.

    From figure 2 we can see that the pairs (2n-x, x) for (0 <= x <= n) are the set of all possible pairs of integers that sum to 2n. And, p1 and p2 must be less than 2n. The mid-point of the circle is 2n/2, (or just n).

    Since (2n, 0) and (2n-2, 2) can never be prime pairs, because they are even numbers. (2n is even whether n is even or odd.) And, (2n-1, 1) is paired with 1, and 1 is not a prime. Therefore, as 2n’s increase, the first possible occurrence of a prime pair for a given 2n is (2n-3, 3), and the last occurrence is (n, n).

    Each side (i.e., right side: 0 to n and left side: n to 2n) must be a mix of composites and primes, because Bertrand's postulate, proven in 1852, states that, ”If n > 1, then there is always at least one prime p such that n < p < 2n.”

    Let (2n-k) be that prime p. Then for any 2n (n>1), (2n-k) is one (or more) of 2n-3, 2n-5, 2n-7, 2n-11, 2n-13, 2n-17, 2n-19, …2n-p (for p < n), and (2n-k) is prime. Then (2n-k)(k) and any of the products (2n-3)(3), (2n-5)(5), (2n-7)(7), … (n, n) is a product of primes and can be expressed as a difference of squares.
    (2n - k)(k) = (n+n - k)(n-n + k) = (n + (n-k))(n - (n-k)) = n^2 – (n-k)^2

    Let x = n - k, then from above, (2n - k)(k) = n^2 - x^2 which equals (n-x)(n+x). Because k and 2n-k are primes, then (n-x) and (n+x) are both prime numbers.
    Proof: Given k and (2n-k) are prime and by definition are equidistant from n (and from 2n), then on either right or left side n = (n-x) + x = k + x (See figure 3.) Since x = n - k then (n – x) = k which is prime, and since x = (2n-k) - n then n+x = n + 2n-k – n = (2n-k) which is prime. Therefore, n-x and n+x are both prime and (n-x) + (n+x) = 2n.

    upload_2023-3-28_20-9-46.png
    Figure 3: (n-x) and (n+x) are prime.

    Therefore, k (or k’s) and (2n-k) guarantee that an x exists as an offset from n, for p1 and p2 with n-x + n+x = p1 + p2 = 2n, for some x (1 < x < n). (Note: when x=1 a prime pair is the solution, and for (n, n) when n is a prime, it is its own solution.)

    Therefore, for any n > 1 there exist a 2n with a prime pair p1, p2 such that p1+p2 = 2n.
    QED
     

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    ddinunno, Mar 29, 2023
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  2. ddinunno

    ddinunno

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    Above figure 1, I meant to say, "For all 2n > 2," - not less than.
     
    ddinunno, Mar 29, 2023
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  3. ddinunno

    ddinunno

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    Just because (2n-k) is prime does not necessarily mean k is prime. I need to prove that at least one of (2n-3, 2n-5, 2n-7, etc. must be a prime pair. Then I can call it (2n-k) and the rest is good.
     
    ddinunno, Mar 29, 2023
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  4. ddinunno

    ddinunno

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    I have not (yet) proved that k is prime given the existence of prime (2n-k), but I do have an algorithm that checks for a 2n with no prime pairs as n increases which will return a FALSE response if no prime pairs exist.

    I "just" need to determine how to translate this algorithm into a mathematical statement. I don't anticipate this will be completed easily or soon.
     
    ddinunno, Mar 31, 2023
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