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Old 2011-01-05, 19:40   #12
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Quote:
Originally Posted by kurtulmehtap View Post
Not Really, I am still not sure if a composite Mersenne number can have more than 1 pair for x^2 + 27y^2.
well if I did the math right it should come out to 4*x^2 +27y^2 but I guess you simplied 4*x^2->x^2 ?
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Old 2011-01-05, 19:45   #13
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well if I did the math right it should come out to 4*x^2 +27y^2 but I guess you simplied 4*x^2->x^2 ?
That's right
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Old 2011-01-05, 23:08   #14
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Theorem checks out up to p=73, though thats not very far. Mersenne primes have 1 solution and composite numbers have 0 or 2:


Code:
p	(x,y) so 2p-1=4*x2+27*y2

5	1,1
7	4,1
11	no solution
13	23,15
17	181,1
19	149,127
23	no solution
29	no solution
31	23081,783
37	142357,45695 and 185341,1119
41:	no solution
43:	no solution
47:	no solution
53:	no solution
59:	no solution
61:	752652049,38443119
67:	4922679991,1369547633 and 5053371809,1297114833
71:	no solution
73:	no solution

Last fiddled with by ATH on 2011-01-05 at 23:11
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Old 2011-01-06, 15:55   #15
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I wish finding primes in lucas sequences were easy lol, if so we could rely on the fact that mersenne numbers are U(3,2) if I remember correctly.

Last fiddled with by science_man_88 on 2011-01-06 at 16:06
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Old 2011-01-07, 13:35   #16
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(2x)^2+ 3(3y)^2 could be transformed to:

(Qx)^2 + P(Py)^2

which can technically at least in this case be transformed to:

(Qx)^Q + P(Py)^Q

which may be transformed further I believe but I can't remember enough math right now to do that anyone else up to looking at this ?
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Old 2011-01-07, 14:53   #17
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Quote:
Originally Posted by science_man_88 View Post
(2x)^2+ 3(3y)^2 could be transformed to:

(Qx)^2 + P(Py)^2

which can technically at least in this case be transformed to:

(Qx)^Q + P(Py)^Q

which may be transformed further I believe but I can't remember enough math right now to do that anyone else up to looking at this ?
Q^Q*x^Q + P^(Q+1)*P*(y^Q) = Q^Q*x^Q + P^(Q+2)*(y^Q) = 4*x^2 +81*y^2 which can only equal the other one if y=0, I think I went to far.

Last fiddled with by science_man_88 on 2011-01-07 at 15:01
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Old 2011-01-07, 15:19   #18
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The conjecture is: Mq is a prime if and only if there exists only one pair (x, y) such that: Mq = (2x)^2+ 3(3y)^2.

So you can not make the 2 and 3 into new variables P and Q. Then its not this conjecture anymore, and if P and Q can be anything, then the conjecture is most likely not true anymore, since there is probably more solutions for different P and Q.

Last fiddled with by ATH on 2011-01-07 at 15:20
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Old 2011-01-07, 15:24   #19
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Quote:
Originally Posted by ATH View Post
The conjecture is: Mq is a prime if and only if there exists only one pair (x, y) such that: Mq = (2x)^2+ 3(3y)^2.

So you can not make the 2 and 3 into new variables P and Q. Then its not this conjecture anymore, and if P and Q can be anything, then the conjecture is most likely not true anymore, since there is probably more solutions for different P and Q.
P and Q are variables for lucas sequences! and they match up for mersenne numbers for where I placed them. maybe if we got this to work for mersenne primes we could use the same basic formula for other lucas sequences ?

Last fiddled with by science_man_88 on 2011-01-07 at 15:50
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Old 2011-01-07, 15:52   #20
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I see where I went wrong above in my expansion I multiplied by p in 2 places not one doh.
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Old 2011-01-07, 20:24   #21
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Quote:
Originally Posted by http://en.wikipedia.org/wiki/Lucas_sequence
Among the consequences is that Ukm is a multiple of Um, implying that Un can be prime only when n is prime. Another consequence is an analog of exponentiation by squaring that allows fast computation of Un for large values of n. These facts are used in the Lucas–Lehmer primality test.
wrong then why in the Fibonacci numbers do 2 3 and 5 go back to back to back ? 3 primes in a row where ?
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Old 2011-01-08, 03:12   #22
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Quote:
Originally Posted by science_man_88 View Post
wrong then why in the Fibonacci numbers do 2 3 and 5 go back to back to back ?
What's the contradiction? 1 | 2, 2 | 2; 1 | 3, 1 | 3, 3 | 3; 1 | 5, 5 | 5.
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