Properties of Mersenne numbers

[science_man_88]

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 ?

[kurtulmehtap]

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

[ATH]

Theorem checks out up to p=73, though thats not very far. Mersenne primes have 1 solution and composite numbers have 0 or 2:

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

[science_man_88]

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.

[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 ?

[science_man_88]

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.

[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.

[science_man_88]

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 ?

[science_man_88]

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

[science_man_88]

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 ?

[CRGreathouse]

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.