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2011-01-05, 19:40   #12
science_man_88

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Jul 2009
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203008 Posts

Quote:
 Originally Posted by kurtulmehtap 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 ?

2011-01-05, 19:45   #13
kurtulmehtap

Sep 2009

1001002 Posts

Quote:
 Originally Posted by science_man_88 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

 2011-01-05, 23:08 #14 ATH Einyen     Dec 2003 Denmark 3,313 Posts 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
 2011-01-06, 15:55 #15 science_man_88     "Forget I exist" Jul 2009 Dumbassville 203008 Posts 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
 2011-01-07, 13:35 #16 science_man_88     "Forget I exist" Jul 2009 Dumbassville 20C016 Posts (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 ?
2011-01-07, 14:53   #17
science_man_88

"Forget I exist"
Jul 2009
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Quote:
 Originally Posted by science_man_88 (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

 2011-01-07, 15:19 #18 ATH Einyen     Dec 2003 Denmark CF116 Posts 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
2011-01-07, 15:24   #19
science_man_88

"Forget I exist"
Jul 2009
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26×131 Posts

Quote:
 Originally Posted by ATH 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

 2011-01-07, 15:52 #20 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts I see where I went wrong above in my expansion I multiplied by p in 2 places not one doh.
2011-01-07, 20:24   #21
science_man_88

"Forget I exist"
Jul 2009
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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 ?

2011-01-08, 03:12   #22
CRGreathouse

Aug 2006

3×1,993 Posts

Quote:
 Originally Posted by science_man_88 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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