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]