Fermat Factoring and the Goldbach Conjecture


Goldbach's Conjecture: Every even number greater than 6, can be expressed as the sum of two primes.

There is a lot of evidence that this is a true statement. So far, no counter example has been found. No proof is available either. In my attempts to factor large composite numbers, I found the following facts with respect to the Fermat factoring method.
Assume that we bend the Integer Number Line back onto itself at an arbitrary number. The numbers then make an additive pair that result in the same number. If we bend at N, then the additive constant is 2N. It is also proven that there lies at least one prime between 1 and N and N and 2N.

1 9
2 8
3 7
4 6
5 5
The numbers all add to 10 and the two primes 3 and 7 satisfy the Goldbach Conjecture.
PQPQN2-A2
1 9 = 9 = 52-42
2 8 = 16 = 52-32
3 7 = 21 = 52-22
4 6 = 24 = 52-12
5 5 = 25 = 52-02
The numbers when multiplied together, form the basis for the Fermat Factoring method.
where PQ = N2-A2, N is the number where the integer line is folded and A is the relative distance from N.


The following examples reveal more information about the number alignment and the predictability of solving the factorization. Since most of the numbers that interest us are odd, except for the place the line folds, I'll show only the odd number pairs.

Pmod 4Qmod 4PQmod 4mod 9A2mod 9
1 1 27 3 27 3 0 121 4
3 3 25 1 75 3 3 100 1
5 1 23 3 115 3 7 81 0
7 3 21 1 147 3 3 49 4
9 1 19 3 171 3 0 25 7
11 3 17 1 187 3 7 9 0
13 1 15 3 195 3 6 1 1
14 2 14 2 196 0 7 0 0
Pmod 4Qmod 4PQmod 4mod 9A2mod 9
1 1 29 1 29 1 2 196 7
3 3 27 3 81 1 0 144 0
5 1 25 1 125 1 8 100 1
7 3 23 3 161 1 8 64 1
9 1 21 1 189 1 0 36 0
11 3 19 3 209 1 2 16 7
13 1 17 1 221 1 5 4 4
15 3 15 3 225 1 0 0 0
Pmod 4Qmod 4PQmod 4mod 9A2mod 9
1 1 31 3 31 3 4 225 0
3 3 29 1 87 3 6 169 7
5 1 27 3 135 3 0 121 4
7 3 25 1 175 3 4 81 0
9 1 23 3 207 3 0 49 4
11 3 21 1 231 3 6 25 7
13 1 19 3 247 3 4 9 0
15 3 17 1 255 3 3 1 1
16 0 16 0 256 0 4 0 0
Pmod 4Qmod 4PQmod 4mod 9A2mod 9
1 1 33 1 33 1 6 256 4
3 3 31 3 93 1 3 196 7
5 1 29 1 145 1 1 144 0
7 3 27 3 189 1 0 100 1
9 1 25 1 225 1 0 64 1
11 3 23 3 253 1 1 36 0
13 1 21 1 273 1 3 16 7
15 3 19 3 285 1 6 4 4
17 1 17 1 289 1 1 0 0
Pmod 4Qmod 4PQmod 4mod 9A2mod 9
1 1 35 3 35 3 8 289 1
3 3 33 1 99 3 0 225 0
5 1 31 3 155 3 2 169 7
7 3 29 1 203 3 5 121 4
9 1 27 3 243 3 0 81 0
11 3 25 1 275 3 5 49 4
13 1 23 3 299 3 2 25 7
15 3 21 1 315 3 0 9 0
17 1 19 3 323 3 8 1 1
18 2 18 2 324 0 0 0 0

Here is all I have found out so far. I hope this takes some of the mystery out of it. In all of this is the answer to both the factoring of numbers and the Golbach Conjecture. The conjecture can be stated in a few ways:
For every Integer N there are two primes that satisfy the equation
N2 - A2 = PQ
For every Integer N there lie at least two primes P, Q equaidistant from N
As yet, nothing has jumped out and hit me.

Here are the observations from the tables, assuming we are trying to factor PQ, and PQ does not have a factor of 3:
When PQ = 1 mod 4, N is odd and A is even.
When PQ = 3 mod 4, N is even and A is odd.

When PQ = 1 mod 9, N2 = 1 mod 9, A2 = 0 mod 9
When PQ = 4 mod 9, N2 = 4 mod 9, A2 = 0 mod 9
When PQ = 7 mod 9, N2 = 7 mod 9, A2 = 0 mod 9

When PQ = 2 mod 9, N2 = 0 mod 9, A2 = 7 mod 9
When PQ = 5 mod 9, N2 = 0 mod 9, A2 = 4 mod 9
When PQ = 8 mod 9, N2 = 0 mod 9, A2 = 1 mod 9

Following are formulae for generating perfect squares with specific mod 9 properties. Where k equals zero to n of the Integers.

9(9k2+2k)+1=1 mod 9 and,
9(9k2+16k)+64=1 mod 9
9(9k2+4k)+1=4 mod 9 and,
9(9k2+14k)+49=4 mod 9
9(9k2+8k)+16=7 mod 9 and,
9(9k2+10k)+25=7 mod 9
9k2=0 mod 9


Last updated: 1/29/2001
Webmaster: Ulrich (Ulie) Sondermann usondermann@earthlink.net
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