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 2015-07-11, 14:54 #1 irina   Jul 2015 22 Posts Algebraithm for calculating primes For prime number A, there is only one value B, such that what А + В2 = С2 В = (А-1)/2 С = (А+1)/2 А = С2 – В2 = (С-В)*(С+В) С – В = 1 If the number of semiprime A = k1 * k2, then there are at least two values, such that А + В2 = С2 1. 1) А = С2 – В2 = (С-В)*(С+В); k1 = C-B; k2= C + B B2 = (n + trunc (sqrt (A))2 – A; n – natural number [1; +∞); C = n + trunc (sqrt (A)) 2. 2) А = С2 – В2 = (С-В)*(С+В); С – В = 1 В = (А-1)/2 (B- maximum) С = (А+1)/2 Example, А = 21 B1 = (А-1)/2 = (21-1)/2 = 10; С = (А+1)/2 = (21+1)/2 = 11 21 + 102 = 112 If semiprime A, then there is at least one value В2< B1: Sqrt (21) = 4,58257.. Trunc (4,58257) = 4 B2 = (n + 4)2 – 21 for n =1 B2 = (1+4)2 – 21 = 4; B = 2; C = n+ 4 = 1 + 4 =5 A = С2 – В2 = (С-В)*(С+В) = (5 – 2)*(5+2) = 3*7 А=k1 * k2= 3*7 Last fiddled with by Batalov on 2015-07-11 at 16:02 Reason: fixed formatting for squares (only that)
 2015-07-11, 17:57 #2 xilman Bamboozled!     "𒉺𒌌𒇷𒆷𒀭" May 2003 Down not across 242338 Posts
 2015-07-11, 21:16 #3 danaj   "Dana Jacobsen" Feb 2011 Bangkok, TH 38A16 Posts For even more awesomeness you could use Deterministic M-R with ceil(n/4) as the limit. (I really wish the previously referenced task could be renamed, as far too many people think what is described is actually AKS) Tempted to post the primality regex... Last fiddled with by danaj on 2015-07-11 at 21:17
2015-07-13, 12:24   #4
R.D. Silverman

Nov 2003

746010 Posts

Quote:
 Originally Posted by irina For prime number A, there is only one value B, such that what А + В2 = С2 В = (А-1)/2 С = (А+1)/2 А = С2 – В2 = (С-В)*(С+В) С – В = 1 If the number of semiprime A = k1 * k2, then there are at least two values, such that А + В2 = С2 1. 1) А = С2 – В2 = (С-В)*(С+В); k1 = C-B; k2= C + B B2 = (n + trunc (sqrt (A))2 – A; n – natural number [1; +∞); C = n + trunc (sqrt (A)) 2. 2) А = С2 – В2 = (С-В)*(С+В); С – В = 1 В = (А-1)/2 (B- maximum) С = (А+1)/2 Example, А = 21 B1 = (А-1)/2 = (21-1)/2 = 10; С = (А+1)/2 = (21+1)/2 = 11 21 + 102 = 112 If semiprime A, then there is at least one value В2< B1: Sqrt (21) = 4,58257.. Trunc (4,58257) = 4 B2 = (n + 4)2 – 21 for n =1 B2 = (1+4)2 – 21 = 4; B = 2; C = n+ 4 = 1 + 4 =5 A = С2 – В2 = (С-В)*(С+В) = (5 – 2)*(5+2) = 3*7 А=k1 * k2= 3*7
This is nothing but trivial algebra. Where is the ALGORITHM?
You have failed to specify any kind of procedure. All you have done is
assert the existence of some values satisfying some relations.

2015-07-13, 12:27   #5
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

7·292 Posts

Quote:
 Originally Posted by R.D. Silverman This is nothing but trivial algebra. Where is the ALGORITHM?
Algebraithm for calculating primes?

2015-08-16, 01:55   #6
alpertron

Aug 2002
Buenos Aires, Argentina

2·3·223 Posts

Quote:
 Originally Posted by irina For prime number A, there is only one value B, such that what А + В2 = С2 В = (А-1)/2 С = (А+1)/2 А = С2 – В2 = (С-В)*(С+В) С – В = 1 If the number of semiprime A = k1 * k2, then there are at least two values, such that А + В2 = С2 1. 1) А = С2 – В2 = (С-В)*(С+В); k1 = C-B; k2= C + B B2 = (n + trunc (sqrt (A))2 – A; n – natural number [1; +∞); C = n + trunc (sqrt (A)) 2. 2) А = С2 – В2 = (С-В)*(С+В); С – В = 1 В = (А-1)/2 (B- maximum) С = (А+1)/2 Example, А = 21 B1 = (А-1)/2 = (21-1)/2 = 10; С = (А+1)/2 = (21+1)/2 = 11 21 + 102 = 112 If semiprime A, then there is at least one value В2< B1: Sqrt (21) = 4,58257.. Trunc (4,58257) = 4 B2 = (n + 4)2 – 21 for n =1 B2 = (1+4)2 – 21 = 4; B = 2; C = n+ 4 = 1 + 4 =5 A = С2 – В2 = (С-В)*(С+В) = (5 – 2)*(5+2) = 3*7 А=k1 * k2= 3*7
It appears that you suggest to select n=1, 2, 3, ... until you get the factorization. This is just Fermat's method.

 2018-05-28, 13:50 #7 irina   Jul 2015 22 Posts Mersenne prime number search algorithm ? Last fiddled with by irina on 2018-05-28 at 14:00

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