Properties of Mersenne numbers
 

In Tony Reix's Properties of Mersenne and Fermat numbers online paper
you see:
Mq is a prime if and only if there exists only one pair (x, y) such that:
Mq = (2x)^2+ 3(3y)^2.
The proof is missing. Can anybody provide a proof?
By numerical testing different Mq values I have found that if Mq is composite there is no pair (x,y) that satisfies the condition.
Is it possible that if Mq is composite there can be 2 or more pairs?
Thanx in advance...

[QUOTE=kurtulmehtap;241966]In Tony Reix's Properties of Mersenne and Fermat numbers online paper
you see:
Mq is a prime if and only if there exists only one pair (x, y) such that:
Mq = (2x)^2+ 3(3y)^2.
The proof is missing. Can anybody provide a proof?
[/QUOTE]

I have not verified that the result is true. I will assume that it is.

I will sketch a proof. This result has very little to do with Mersenne
primes.

Let Q = (2x)^2 + 3(3y)^2. Q is prime iff this representation is unique.


Now, follow the (standard!) proof that an integer that is 1 mod 4 is prime
iff it is the sum of two squares in a unique way. i.e. --Factor Q over
Q(sqrt(-3)) and observe that you are doing so in a UFD.


[QUOTE]

[QUOTE=R.D. Silverman;241969]I have not verified that the result is true. I will assume that it is.

I will sketch a proof. This result has very little to do with Mersenne
primes.

Let Q = (2x)^2 + 3(3y)^2. Q is prime iff this representation is unique.


Now, follow the (standard!) proof that an integer that is 1 mod 4 is prime
iff it is the sum of two squares in a unique way. i.e. --Factor Q over
Q(sqrt(-3)) and observe that you are doing so in a UFD.

[/QUOTE]

Pretty easy, huh? Good job.

[QUOTE=R.D. Silverman;241969]I have not verified that the result is true. I will assume that it is.

I will sketch a proof. This result has very little to do with Mersenne
primes.

Let Q = (2x)^2 + 3(3y)^2. Q is prime iff this representation is unique.


Now, follow the (standard!) proof that an integer that is 1 mod 4 is prime
iff it is the sum of two squares in a unique way. i.e. --Factor Q over
Q(sqrt(-3)) and observe that you are doing so in a UFD.
[/QUOTE]

Thanks a lot.
I'm stuck. If there is a unique pair (x,y) then Q is prime,however , if Q is composite, then can we assume that there are no (x,y) pairs or should we consider there are 2,3 or more pairs?
Thanks

[QUOTE=kurtulmehtap;242188][QUOTE=R.D. Silverman;241969]I have not verified that the result is true. I will assume that it is.

I will sketch a proof. This result has very little to do with Mersenne
primes.

Let Q = (2x)^2 + 3(3y)^2. Q is prime iff this representation is unique.


Now, follow the (standard!) proof that an integer that is 1 mod 4 is prime
iff it is the sum of two squares in a unique way. i.e. --Factor Q over
Q(sqrt(-3)) and observe that you are doing so in a UFD.

Thanks a lot.
I'm stuck. If there is a unique pair (x,y) then Q is prime,however , if Q is composite, then can we assume that there are no (x,y) pairs or should we consider there are 2,3 or more pairs?
Thanks[/QUOTE]

Hint: Composition of quadratic forms...

[QUOTE=R.D. Silverman;242204]Hint: Composition of quadratic forms...[/QUOTE]

Another hint: think third degree polynomial equations over Z.

So is the OPer satisfied?

[QUOTE=davar55;243127]So is the OPer satisfied?[/QUOTE]

Not Really, I am still not sure if a composite Mersenne number can have more than 1 pair for x^2 + 27y^2.

There is a thesis on this subject:
Mersenne primes of the form x^2+dy^2 by Bas Jansen at
[url]www.math.leidenuniv.nl/en/theses/31/[/url]

It has an entire section for the needed case d=27, but I still can't find the answer..

Please help.
I know that I am embarassing myself but I need the answer.

[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.

There is a thesis on this subject:
Mersenne primes of the form x^2+dy^2 by Bas Jansen at
[url]www.math.leidenuniv.nl/en/theses/31/[/url]

It has an entire section for the needed case d=27, but I still can't find the answer..

Please help.
I know that I am embarassing myself but I need the answer.[/QUOTE]

If the number is composite, there will be more than 1.

[QUOTE=R.D. Silverman;244687]If the number is composite, there will be more than 1.[/QUOTE]

Look up "idoneal".

[QUOTE=R.D. Silverman;244688]Look up "idoneal".[/QUOTE]

[url]http://www.google.ca/search?sourceid=chrome&ie=UTF-8&q=define:idoneal[/url]

[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.

[QUOTE=CRGreathouse;245063]What's the contradiction? 1 | 2, 2 | 2; 1 | 3, 1 | 3, 3 | 3; 1 | 5, 5 | 5.[/QUOTE]

[QUOTE]0, 1, 1, 2, 3, 5,[/QUOTE]

[QUOTE]Un can be prime only when n is prime.[/QUOTE]3 primes in a row but not 3 prime indexes in a row so it doesn't fit

but this is a thread on mersenne I guess I have no point then.

[QUOTE=science_man_88;245116]3 primes in a row but not 3 prime indexes in a row so it doesn't fit

but this is a thread on mersenne I guess I have no point then.[/QUOTE]

If you look at the actual divisibility property you'll see that it doesn't lead to a contradiction.

[QUOTE=CRGreathouse;245179]If you look at the actual divisibility property you'll see that it doesn't lead to a contradiction.[/QUOTE]

but have 3 primes in row does! as it says Un can be prime only if n is prime basically which if you notice means for 3 primes to be in a row we would need 3 primes with difference 1 to exist can you name those three ?

[QUOTE=science_man_88;244815]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.[/QUOTE]

The connection between Fib, Lucas, and the primes was alluded to in
a thread in this forum (over two years old) in which a nice conjecture
was first presented by me. I'll look for it later if you can't find it and
are interested.

[QUOTE=davar55;245292]The connection between Fib, Lucas, and the primes was alluded to in
a thread in this forum (over two years old) in which a nice conjecture
was first presented by me. I'll look for it later if you can't find it and
are interested.[/QUOTE]

[QUOTE=davar55;245288]There's a thread (over two years old) in Math or Puzzles here in which I
presented a conjecture of mine relating Fib and Lucas sequences.

I'll look for it later if you can't find it.

It needed a big numeric test that no one pursued here.
You might be interested.[/QUOTE]

I'm told not to post double and someone else does ?

[QUOTE=ATH;244741]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][/QUOTE]

The solution for 7 is (x,y) = (5,1) not (4,1).

[QUOTE=ATH;244741]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,[COLOR="Red"]45695[/COLOR] and 185341,[COLOR="Red"]1119[/COLOR]
41: no solution
43: no solution
47: no solution
53: no solution
59: no solution
61: 752652049,38443119
67: 4922679991,[COLOR="red"]1369547633[/COLOR] and 5053371809,[COLOR="red"]1297114833[/COLOR]
71: no solution
73: no solution
[/CODE][/QUOTE]

I see a link between the 2 that have 2 sets is this a bad thing ? look at sumdigits for the two sets of numbers highlighted

Looks like chance to me. Any reason to think otherwise?

[QUOTE=CRGreathouse;245362]Looks like chance to me. Any reason to think otherwise?[/QUOTE]

not with the data provided.I've done a little more research into the numbers highlighted by calculator and I found that the multiplication of the digits of each one then taking sumdigits gives a multiple of 9 which as you know should come back to 9 if repeated.

[CODE](20:17)>for(y=1,9,forstep(x=y,1000,9,print1(multiplydigits(x)%9","));print())
1,0,0,7,3,6,7,6,3,7,0,0,0,8,5,0,2,2,0,5,8,0,0,0,5,6,3,5,3,6,5,0,0,0,0,0,3,0,0,3,0,0,0,3,0,0,2,5,0,5,2,0,0,8,8,0,0,2,3,3,2,0,0,6,2,6,0,0,0,6,0,0,0,6,0,0,6,0,0,5,5,0,0,8,2,0,2,8,0,0,8,0,0,3,8,6,6,8,3,0,0,0,0,0,0,0,0,0,0,0,0,0,
2,1,0,0,6,1,3,3,1,6,0,0,0,0,7,3,6,7,6,3,7,0,0,0,7,1,0,4,4,0,1,7,0,0,0,3,0,0,3,0,0,3,0,0,0,0,6,4,3,3,4,6,0,0,4,0,0,7,4,0,4,7,0,0,1,1,0,0,6,0,0,6,0,0,0,6,0,0,0,3,1,3,0,0,1,6,6,1,0,0,7,7,0,0,4,1,0,1,4,0,0,0,0,0,0,0,0,0,0,0,0,
3,2,2,0,0,5,8,0,8,5,0,0,1,0,0,6,1,3,3,1,6,0,0,0,0,5,6,3,5,3,6,5,0,0,0,6,6,0,6,6,0,6,6,0,0,0,1,3,6,1,6,3,1,0,0,0,0,3,5,6,6,5,3,0,0,5,0,0,3,3,0,3,3,0,0,3,3,0,0,1,6,6,1,0,0,3,1,3,0,0,6,5,6,0,0,5,3,3,5,0,0,0,0,0,0,0,0,0,0,0,0,
4,3,4,3,0,0,4,6,6,4,0,0,2,2,0,0,5,8,0,8,5,0,0,2,0,0,3,2,6,6,2,3,0,0,0,0,3,0,0,3,0,0,3,0,0,0,5,2,0,8,8,0,2,5,0,0,0,8,6,3,8,3,6,8,0,0,0,0,0,6,0,0,6,0,0,0,6,0,0,8,2,0,2,8,0,0,5,5,0,0,5,3,3,5,0,0,6,5,6,0,0,0,0,0,0,0,0,0,0,0,0,
5,4,6,6,4,0,0,3,4,3,0,0,3,4,3,0,0,4,6,6,4,0,0,4,4,0,0,1,7,0,7,1,0,0,3,0,0,0,3,0,0,3,0,0,0,0,0,1,3,6,1,6,3,1,0,0,0,4,7,0,1,1,0,7,4,0,0,0,6,0,0,6,0,0,6,0,0,0,0,6,7,3,3,7,6,0,0,7,0,0,4,1,0,1,4,0,0,7,7,0,0,0,0,0,0,0,0,0,0,0,0,
6,5,8,0,8,5,0,0,2,2,0,0,4,6,6,4,0,0,3,4,3,0,0,6,8,6,0,0,8,3,3,8,0,0,6,6,0,0,6,6,0,6,6,0,0,4,0,0,6,4,3,3,4,6,0,0,0,0,8,6,3,8,3,6,8,0,0,0,3,3,0,3,3,0,3,3,0,0,0,4,3,6,4,6,3,4,0,0,0,0,3,8,6,6,8,3,0,0,8,0,0,0,0,0,0,0,0,0,0,0,0,
7,6,1,3,3,1,6,0,0,1,0,0,5,8,0,8,5,0,0,2,2,0,0,8,3,3,8,0,0,6,8,6,0,0,0,3,0,0,0,3,0,0,3,0,0,8,8,0,0,2,5,0,5,2,0,0,5,0,0,3,5,6,6,5,3,0,0,0,0,6,0,0,6,0,0,6,0,0,0,2,8,0,5,5,0,8,2,0,0,0,2,6,3,2,3,6,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
8,7,3,6,7,6,3,7,0,0,0,0,6,1,3,3,1,6,0,0,1,0,0,1,7,0,7,1,0,0,4,4,0,0,3,0,0,3,0,0,0,3,0,0,0,3,7,3,0,0,7,6,6,7,0,0,1,1,0,0,7,4,0,4,7,0,0,6,0,0,0,6,0,0,6,0,0,0,0,0,4,3,6,4,6,3,4,0,0,0,1,4,0,7,7,0,4,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,8,5,0,2,2,0,5,8,0,0,0,7,3,6,7,6,3,7,0,0,0,0,3,2,6,6,2,3,0,0,2,0,0,6,6,0,6,6,0,0,6,6,0,0,7,6,6,7,0,0,3,7,3,0,0,6,2,6,0,0,2,3,3,2,0,0,3,3,0,0,3,3,0,3,3,0,0,7,0,0,6,7,3,3,7,6,0,0,0,0,2,6,3,2,3,6,2,0,0,0,0,0,0,0,0,0,0,0,0,0,[/CODE]

okay that doesn't narrow things down.