[QUOTE=kurtulmehtap;244684]Not Really, I am still not sure if a composite Mersenne number can have more than 1 pair for x^2 + 27y^2.[/QUOTE]

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 ?

[QUOTE=science_man_88;244719]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 ?[/QUOTE]

That's right

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 2[sup]p[/sup]-1=4*x[sup]2[/sup]+27*y[sup]2[/sup]

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
[/CODE]

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.

(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 ?

[QUOTE=science_man_88;244931](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 ?[/QUOTE]

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.

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.

[QUOTE=ATH;244946]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.[/QUOTE]

P and Q are variables for [URL="http://en.wikipedia.org/wiki/Lucas_sequence"]lucas sequences[/URL]! 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 ?

I see where I went wrong above in my expansion I multiplied by p in 2 places not one doh.

[QUOTE="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.[/QUOTE]
wrong then why in the Fibonacci numbers do 2 3 and 5 go back to back to back ? 3 primes in a row where ?

[QUOTE=science_man_88;244984]wrong then why in the Fibonacci numbers do 2 3 and 5 go back to back to back ?[/QUOTE]

What's the contradiction? 1 | 2, 2 | 2; 1 | 3, 1 | 3, 3 | 3; 1 | 5, 5 | 5.