This is a follow up to a previous blog entry.
As previously conjectured:
Where p is prime, 1 < a1 < a2 +...+ < am < p, if there exists
{n, a1, a2, ..., am, p, x} that can satisfy (a1n + a2n +...+ amn) / pn-1 = x, then there exists, with the same number of terms (a) and exponent (n), {n, a1, a2, ... , am, b} that can satisfy a1n + a2n +...+ amn = bn.
An example of a {n, a1, a2, a3, a4, p, x} that satisfies
(a15 + a25 + a35 + a45) / p4 = x, is {5, 31, 40, 43, 51, 61, 45}.
(315 + 405 + 435 + 515) / 614 = 45
Therefore, there exists {n, a1, a2, a3, a4, b} that can satisfy
a15 + a25 + a35 + a45 = b5.
Suppose that if there are solutions to either equation, then there are solutions to both. And that all solutions to either type correlate to solutions of the other.
The example in the conjecture is a 5th power. Therefore, I'll be covering 5th powers first, then 4th powers, and 6th powers.
5th Powers
The lowest 5th power solution having the sum of 4 terms that equal one term is
275 + 845 + 1105 + 1335 = 1445 (Lander and Parkin, 1967)
and has a modular correlation with
(315 + 405 + 435 + 515) / 614 = 45
144 - 40 - 43 = 0 mod 61
133 - 31 = 0 mod 51
110 + 61 -51 = 0 mod 40
84 + 45 = 0 mod 43
27 + 61 - 43 = 0 mod 45
The only other solution currently known is
555 + 31835 + 289695 + 852825 = 853595 (R. Frye 2004)
and correlates with
(345 + 1795 + 1855 + 2225) / 2234 = 380
85359 - 223 = 0 mod 34
85282 + 222 - 34 = 0 mod 185
28969 + 222 + 179 - 380 = 0 mod 223
3183 + 34 + 185 + 222 - 223 = 0 mod 179
55 - 34 - 179 - 222 = 0 mod 380
There are a total of 9 equations where p <= 1249.
(315 + 405 + 435 + 515) / 614 = 45
(345 + 1795 + 1855 + 2225) / 2234 = 380
(675 + 1295 + 2405 + 2925) / 3134 = 308
(205 + 235 + 3125 + 3435) / 3734 = 398
(1105 + 2345 + 5295 + 7095) / 7394 = 742
(365 + 3755 + 3935 + 7195) / 7614 = 623
(75 + 3045 + 4345 + 4615) / 10194 = 36
(145 + 6085 + 8685 + 9225) / 10194 = 1152
(1165 + 6395 + 7065 + 9805) / 11534 = 671
Each of the 9 equations can correlate with multiple 5th power solutions. Additional 5th power solutions exist that have yet to be discovered.
4th Powers
The first 4th power solution discovered having the sum of 3 terms that equal one term is
26824404 + 153656394 + 187967604 = 206156734 (N. Elkies 1986)
and correlates with
(244 + 874 + 1094) / 197 3 = 26
20615673 + 24 + 97 = 0 mod 26
18796760 + 24 + 26 - 87 = 0 mod 109
15365639 + 24 + 197 = 0 mod 87
2682440 + 109 = 0 mod 197
The lowest 4th power solution is
958004 + 2175194 + 4145604 = 4224814 (R. Frye 1988)
also correlates with
(244 + 874 + 1094) / 197 3 = 26
422481 + 197 + 26 - 87 - 24 = 0 mod 109
414560 - 87 = 0 mod 197
217519 - 197 - 26 = 0 mod 24
95800 + 197 - 109 = 0 mod 26
There are a total of 25 equations where p <= 1249.
(244 + 874 + 1094) / 1973 = 26
(514 + 1754 + 1974) / 2233 = 221
(24 + 1074 + 3304) / 4193 = 163
(644 + 3194 + 3514) / 4393 = 302
(94 + 404 + 3184) / 4493 = 113
(704 + 2554 + 3484) / 4493 = 209
(104 + 3014 + 3984) / 4793 = 303
(804 + 2234 + 3414) / 5033 = 126
(714 + 3224 + 4154) / 5573 = 234
(434 + 3874 + 5754) / 5993 = 613
(1004 + 1554 + 5364) / 6733 = 273
(244 + 4794 + 5874) / 7273 = 446
(2094 + 2114 + 5194) / 7973 = 151
(6384 + 7014 + 7224) / 7973 = 1341
(3744 + 4694 + 5444) / 8233 = 279
(1284 + 1504 + 6274) / 8393 = 263
(1054 + 2364 + 5884) / 8633 = 191
(84 + 1904 + 8814) / 9293 = 753
(1204 + 6534 + 8494) / 9413 = 842
(914 + 8224 + 9064) / 9473 = 1331
(3874 + 7604 + 7974) / 10493 = 658
(2454 + 5914 + 10454) / 10693 = 1079
(1894 + 6354 + 7384) / 11513 = 302
(1954 + 6614 + 9174) / 12173 = 499
(1674 + 4884 + 9234) / 12493 = 402
Many 4th power solutions have been discovered. Certainly there are many more.
6th Powers
Sum of six terms of the sixth power:
(96 + 156 + 196 + 206 + 286 + 346) / 375 = 31 (John Y, Project Primality, 2009)
(386 + 496 + 946 + 1306 + 1376 + 1476) / 1795 = 121 (John Y, Project Primality, 2011)
Indicates that solutions exist for a16 + a26 + a36 + a46+ a56 + a66 = b6. As of this writing none have been discovered.
Sum of five terms of the sixth power:
(2786 + 3736 + 7986 + 9546 + 10046) / 13035 = 543 (John Y, Project Primality, 2011)
(586 + 2126 + 5776 + 8936 + 9786) / 14395 = 230 (John Y, Project Primality, 2011)
These are the only two examples where p<=1439.
The integers may be large enough to assist in finding a solution for
a16 + a26 + a36 + a46+ a56 = b6.
Contact Info: projectprimality@yahoo.comLast update: May 12, 2012
Wednesday, August 17, 2011
Wednesday, September 15, 2010
Collatz conjecture
The Collatz conjecture is also known as the 3n + 1 conjecture. Take any natural number n. If n is even, divide it by 2 to get n / 2, if n is odd multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1.
For instance, starting with n = 19, generates the sequence 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
Multiplying any odd integer by 3, then adding 1, always results in an even integer. The sequence can be shortened using (3n + 1) / 2 when n is odd. Which produces the sequence 19, 29, 44, 22, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1.
Let the repeating process be (3n + 1) / 2 when n is odd, and maintain n / 2 when n is even.
Dividing random even integers by 2, results in a 50/50 chance of being even or odd.
Starting any sequence with a random integer, at any given point within the sequence, if there are as many even integers as odd, the value for the next integer will be lower than the starting value of (n). The longer the sequence, the greater allowable disparity between the number of even and odd integers for (n) to be lower than its original value. This demonstrates a natural decay. Since each integer has a 50/50 chance of being even or odd, coupled with the natural decay, all values for (n) will eventually reach 1.
Scratching my head, wondering why it's called a conjecture...
© Copyright 2010 ProjectPrimality.blogspot.com
Last update: September 16, 2010
For instance, starting with n = 19, generates the sequence 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
Multiplying any odd integer by 3, then adding 1, always results in an even integer. The sequence can be shortened using (3n + 1) / 2 when n is odd. Which produces the sequence 19, 29, 44, 22, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1.
Let the repeating process be (3n + 1) / 2 when n is odd, and maintain n / 2 when n is even.
Dividing random even integers by 2, results in a 50/50 chance of being even or odd.
Starting any sequence with a random integer, at any given point within the sequence, if there are as many even integers as odd, the value for the next integer will be lower than the starting value of (n). The longer the sequence, the greater allowable disparity between the number of even and odd integers for (n) to be lower than its original value. This demonstrates a natural decay. Since each integer has a 50/50 chance of being even or odd, coupled with the natural decay, all values for (n) will eventually reach 1.
Scratching my head, wondering why it's called a conjecture...
© Copyright 2010 ProjectPrimality.blogspot.com
Last update: September 16, 2010
Monday, March 8, 2010
Simple Method for Fermat's Last Theorem and Beyond
Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the Diophantine equation an + bn = cn for any integer value of n greater than two.
All variables denote positive integers.
Where p is a prime number, n = 2, 1 < a < b < p,
(an + bn) can be evenly divided by pn-1.
Examples:
(22 + 42) / 51 = 4
(32 + 42) / 51 = 5
(22 + 102) / 131 = 8
Therefore, since (a2 + b2) can be evenly divided by p1, a2 + b2 = c2 has solutions.
Where p is a prime number, n > 2, 1 < a < b < p,
(an + bn) is not evenly divisible by pn-1. Therefore, it is reasonable to assume an + bn = cn cannot be satisfied when n > 2.
Adding terms to the left of the equation:
Let the variables in equation an + bn = cn be renamed to a1n + a2n = bn.
Adding two additional terms to the left of the equation would be:
a2n + a2n + a3n + a4n = bn.
Conjecture:
Where p is prime, 1 < a1 < a2 +...+ < am < p, if there exists
{n, a1, a2, ..., am, p, x} that can satisfy (a1n + a2n +...+ amn) / pn-1 = x, then there exists, with the same number of terms (a) and exponent (n), {n, a1, a2, ... , am, b} that can satisfy a1n + a2n +...+ amn = bn.
An example of a {n, a1, a2, a3, a4, p, x} that satisfies
(a15 + a25 + a35 + a45) / p4 = x, is {5, 31, 40, 43, 51, 61, 45}.
(315 + 405 + 435 + 515) / 614 = 45
Therefore, there exists {n, a1, a2, a3, a4, b} that can satisfy
a15 + a25 + a35 + a45 = b5.
© Copyright 2010 ProjectPrimality.blogspot.com
Last update: August 9, 2010
All variables denote positive integers.
Where p is a prime number, n = 2, 1 < a < b < p,
(an + bn) can be evenly divided by pn-1.
Examples:
(22 + 42) / 51 = 4
(32 + 42) / 51 = 5
(22 + 102) / 131 = 8
Therefore, since (a2 + b2) can be evenly divided by p1, a2 + b2 = c2 has solutions.
Where p is a prime number, n > 2, 1 < a < b < p,
(an + bn) is not evenly divisible by pn-1. Therefore, it is reasonable to assume an + bn = cn cannot be satisfied when n > 2.
Adding terms to the left of the equation:
Let the variables in equation an + bn = cn be renamed to a1n + a2n = bn.
Adding two additional terms to the left of the equation would be:
a2n + a2n + a3n + a4n = bn.
Conjecture:
Where p is prime, 1 < a1 < a2 +...+ < am < p, if there exists
{n, a1, a2, ..., am, p, x} that can satisfy (a1n + a2n +...+ amn) / pn-1 = x, then there exists, with the same number of terms (a) and exponent (n), {n, a1, a2, ... , am, b} that can satisfy a1n + a2n +...+ amn = bn.
An example of a {n, a1, a2, a3, a4, p, x} that satisfies
(a15 + a25 + a35 + a45) / p4 = x, is {5, 31, 40, 43, 51, 61, 45}.
(315 + 405 + 435 + 515) / 614 = 45
Therefore, there exists {n, a1, a2, a3, a4, b} that can satisfy
a15 + a25 + a35 + a45 = b5.
© Copyright 2010 ProjectPrimality.blogspot.com
Last update: August 9, 2010
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