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Old 2010-12-15, 15:16   #1
kurtulmehtap
 
Sep 2009

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Default 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...
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Old 2010-12-15, 15:35   #2
R.D. Silverman
 
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Quote:
Originally Posted by kurtulmehtap View Post
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?
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]
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Old 2010-12-15, 21:20   #3
davar55
 
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Quote:
Originally Posted by R.D. Silverman View Post
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.
Pretty easy, huh? Good job.
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Old 2010-12-16, 16:06   #4
kurtulmehtap
 
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Quote:
Originally Posted by R.D. Silverman View Post
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

Last fiddled with by wblipp on 2010-12-16 at 19:08 Reason: fix quotes
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Old 2010-12-16, 18:43   #5
R.D. Silverman
 
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[QUOTE=kurtulmehtap;242188]
Quote:
Originally Posted by R.D. Silverman View Post
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
Hint: Composition of quadratic forms...
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Old 2010-12-17, 19:41   #6
davar55
 
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Quote:
Originally Posted by R.D. Silverman View Post
Hint: Composition of quadratic forms...
Another hint: think third degree polynomial equations over Z.
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Old 2010-12-23, 21:48   #7
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So is the OPer satisfied?
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Old 2011-01-05, 14:15   #8
kurtulmehtap
 
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Quote:
Originally Posted by davar55 View Post
So is the OPer satisfied?
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
www.math.leidenuniv.nl/en/theses/31/

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.
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Old 2011-01-05, 15:20   #9
R.D. Silverman
 
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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.

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

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.
If the number is composite, there will be more than 1.
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Old 2011-01-05, 15:22   #10
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Quote:
Originally Posted by R.D. Silverman View Post
If the number is composite, there will be more than 1.
Look up "idoneal".
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Old 2011-01-05, 15:23   #11
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Quote:
Originally Posted by R.D. Silverman View Post
Look up "idoneal".
http://www.google.ca/search?sourceid...define:idoneal
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