Properties of Mersenne numbers

kurtulmehtap · 2010-12-15, 15:16

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...

R.D. Silverman · 2010-12-15, 15:35

Quote:

Originally Posted by kurtulmehtap

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]

davar55 · 2010-12-15, 21:20

Quote:

Originally Posted by R.D. Silverman

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.

kurtulmehtap · 2010-12-16, 16:06

Quote:

Originally Posted by R.D. Silverman

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

R.D. Silverman · 2010-12-16, 18:43

[QUOTE=kurtulmehtap;242188]

Quote:

Originally Posted by R.D. Silverman

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...

davar55 · 2010-12-17, 19:41

Quote:

Originally Posted by R.D. Silverman

Hint: Composition of quadratic forms...

Another hint: think third degree polynomial equations over Z.

davar55 · 2010-12-23, 21:48

So is the OPer satisfied?

kurtulmehtap · 2011-01-05, 14:15

Quote:

Originally Posted by davar55

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.

R.D. Silverman · 2011-01-05, 15:20

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.

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.

R.D. Silverman · 2011-01-05, 15:22

Quote:

Originally Posted by R.D. Silverman

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

Look up "idoneal".

science_man_88 · 2011-01-05, 15:23

Quote:

Originally Posted by R.D. Silverman

Look up "idoneal".

http://www.google.ca/search?sourceid...define:idoneal

2010-12-15, 15:16	#1
kurtulmehtap Sep 2009 2²×3² Posts	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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