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Re: [PrimeNumbers] Pythagorean Triplets
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In a message dated 03/02/04 03:59:03 GMT Standard Time, harsh@... writes:
> How do I solve this equation. Find a and b for a given prime p.
This problem is not very well posed.
> What properties must p have for a solution to this equation to exist.
>
> a^2 + b^2 = 0 (mod p)
>
Solutions to that equation exist for /any/ prime p, as all it says is that p
divides the lhs!
I think you want the restrictions that a, b are in [0,p1], no?
In that case, either
(a) p = 2, or
(b) p = 1 mod 4.
Case (a) gives as the unique solution a=b=1.
Case (b) has many solutions.
Choose a = any integer in [1,p1];
then b = a*e mod p,
where e is one of the 2 solutions to the congruence
e^2 = 1 mod p.
I think there were recent posts to this group about how to take square roots
mod p.
Mike Oakes
[Nontext portions of this message have been removed] 0 Attachment
I would like to call primes of the form a^2+b^2=prime, pythagorean
primes.
Can anyone proove that there are infinite pythagorean primes?
How about the distribution of a prime p where a^2+b^2=p^2 ?
What if b=a+1, then do infinite pytahgorean primes exist?
I think something similar to how Euler prooved there are infinite
primes may work here but I am not sure.
Let me know!
Thanks,
Harsh Aggarwal
 In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
> How do I solve this equation. Find a and b for a given prime p.
> What properties must p have for a solution to this equation to
exist.
>
> a^2 + b^2 = 0 (mod p)
>
> and what if b= a+1
>
> Let me know!
>
> Thanks,
> Harsh Aggarwal 0 Attachment
 In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:> I would like to call primes of the form a^2+b^2=prime, pythagorean
Or you could just call them Gaussian primes, which is more or less
> primes.
>
> Can anyone proove that there are infinite pythagorean primes?
> How about the distribution of a prime p where a^2+b^2=p^2 ?
> What if b=a+1, then do infinite pytahgorean primes exist?
>
> I think something similar to how Euler prooved there are infinite
> primes may work here but I am not sure.
what they are. You should be able to find necessary information in any
introductory number theory book. Or on whatever websites there are,
probably.
As for b=a+1, well that changes it to just a single dimensional
problem, and is probably more difficult to answer. 0 Attachment
If both a and b are nonzero then z=a+bi, is a Gaussian prime iff
a^2+b^2 is an ordinary prime.
So there is actually no name for primes of the form a^2+b^2
Harsh Aggarwal
 In primenumbers@yahoogroups.com, "Andrew Swallow"
<umistphd2003@y...> wrote:>  In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
pythagorean
> > I would like to call primes of the form a^2+b^2=prime,
> > primes.
any
> >
> > Can anyone proove that there are infinite pythagorean primes?
> > How about the distribution of a prime p where a^2+b^2=p^2 ?
> > What if b=a+1, then do infinite pytahgorean primes exist?
> >
> > I think something similar to how Euler prooved there are infinite
> > primes may work here but I am not sure.
>
> Or you could just call them Gaussian primes, which is more or less
> what they are. You should be able to find necessary information in
> introductory number theory book. Or on whatever websites there are,
> probably.
>
> As for b=a+1, well that changes it to just a single dimensional
> problem, and is probably more difficult to answer. 0 Attachment
 In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:> I would like to call primes of the form a^2+b^2=prime, pythagorean
"Every prime of the form 4n+1 is the sum of two squares. Euler first
> primes.
>
> Can anyone proove that there are infinite pythagorean primes?
communicated the following elegant proof of this fact to Goldbach in
1749, two years after his original proof which was rathar vague on
this point ..."
Fermat's Last Theorem by Edwards, Springer Verlag, 1977.

ElevenSmooth: Distributed Factoring of 2^33264001
http://ElevenSmooth.com 0 Attachment
Elevensmooth,
http://www.math.uchicago.edu/~kobotis/media/163/kim.pdf
prooves there are infinite primes of the from a^2+b^2
So what about when b=a+1
Any thoughts on this
Harsh Aggarwal
 In primenumbers@yahoogroups.com, "elevensmooth"
<elevensmooth@y...> wrote:>  In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
pythagorean
> > I would like to call primes of the form a^2+b^2=prime,
> > primes.
first
> >
> > Can anyone proove that there are infinite pythagorean primes?
>
> "Every prime of the form 4n+1 is the sum of two squares. Euler
> communicated the following elegant proof of this fact to Goldbach in
> 1749, two years after his original proof which was rathar vague on
> this point ..."
>
> Fermat's Last Theorem by Edwards, Springer Verlag, 1977.
> 
> ElevenSmooth: Distributed Factoring of 2^33264001
> http://ElevenSmooth.com 0 Attachment
On Wed, 4 Feb 2004, eharsh82 wrote:> So what about when b=a+1
Then it becomes a univariate quadratic polynomial with interger
coefficients. No univariate polynomial of degree greater than has ever
been proved to produce an infinite number of primes when integers are
substituted for the variable. 0 Attachment
On Tue, 3 Feb 2004, Carl Devore wrote:> Then it becomes a univariate quadratic polynomial with interger
Should say "...degree greater than one has ever..."
> coefficients. No univariate polynomial of degree greater than has ever
> been proved to produce an infinite number of primes when integers are
> substituted for the variable.
 0 Attachment
I am not sure if this series has finite number of primes or not.
I think it has infinite primes.
I have found several primes in the 10000 digit category.
I used PFGW on (10^10000+$a)^2+ (10^10000+$a+1)^2
I'm also working on the series:
(2^$a)^2+(2^$a+1)^2
and
(2^$a)^2+(2^$a1)^2
It din't seem to have a problem with finding primes.
What do you think? let me know!
Harsh Aggarwal
 In primenumbers@yahoogroups.com, Carl Devore <devore@m...> wrote:> On Tue, 3 Feb 2004, Carl Devore wrote:
has ever
> > Then it becomes a univariate quadratic polynomial with interger
> > coefficients. No univariate polynomial of degree greater than
> > been proved to produce an infinite number of primes when integers
are
> > substituted for the variable.
>
> Should say "...degree greater than one has ever..." 0 Attachment
On Wed, Feb 04, 2004 at 05:31:24AM 0000, eharsh82 wrote:> I am not sure if this series has finite number of primes or not.
But that is a question you will never be able to answer, one way or
> I think it has infinite primes.
another, if all you're doing is search for primes of this type using
computer methods. So wouldn't it be more interesting to study the
abstract theory? Your original question was about Gaussian primes, or
primes congruent to 1 mod 4. That's all interesting and fairly basic
stuff. I would have thought that more informative answers would be found
in there.
Apologies if I'm talking rubbish. It just seems strange that on the one
hand you're interested in whether certain sets contain infinitely many
primes, yet on the other hand you're studying the sets using methods
guaranteed to not be able to answer the question, :)
Anyway, that's my morning rant out of the way...
Andy 0 Attachment
Here is my proof for infiniteness of these primes
if b=a+1
then we get 2*a^2+2*a+1 =p
solving this
a is an integer if there is a prime p such that 2*p1=m^2
or m^2+1/2 is a prime
The distribution of such primes would follow the distribution of
primes with the formula n^2+1
it is conjectured that such primes are infinite.

Taken from primepages.com
Are there infinitely many primes of the form n2+1?
There are infinitely many of the forms n2+m2 and n2+m2+1. A more
general form of this conjecture is if a, b, c are relatively prime, a
is positive, a+b and c are not both even,and b24ac is not a perfect
square, then there are infinitely many primes an2+bn+c [HW79, p19].

What do you all think?
Also the series I talked about, what do you think about's it
distribution.
a=2^n or 2^n1
b=a+1 = 2^n+1 or 2^n
the series reduce to ((2^n1)^2+1)/2 and ((2^n+1)^2+1)/2
What about the distribution of primes in such a series?
let me know!
Harsh Aggarwal
 In primenumbers@yahoogroups.com, Andy Swallow <umistphd2003@y...>
wrote:> On Wed, Feb 04, 2004 at 05:31:24AM 0000, eharsh82 wrote:
or
> > I am not sure if this series has finite number of primes or not.
> > I think it has infinite primes.
>
> But that is a question you will never be able to answer, one way or
> another, if all you're doing is search for primes of this type using
> computer methods. So wouldn't it be more interesting to study the
> abstract theory? Your original question was about Gaussian primes,
> primes congruent to 1 mod 4. That's all interesting and fairly basic
found
> stuff. I would have thought that more informative answers would be
> in there.
one
>
> Apologies if I'm talking rubbish. It just seems strange that on the
> hand you're interested in whether certain sets contain infinitely
many
> primes, yet on the other hand you're studying the sets using methods
> guaranteed to not be able to answer the question, :)
>
> Anyway, that's my morning rant out of the way...
>
> Andy 0 Attachment
Hi, I'm a newby so please be gentle.
OK
There is no integer solution for
X^2 + Y^2 = Z^2 when X and Y are prime.
(In other words, no two primes, squared and sumed can equal a perfect
square).
This came out of my Pythagorean triplets program
that seems to show that either X AND/OR Z are always prime
and Y is never prime.
I am not sophisticated enough to know whether the above is trivial.
But, I'm excited to find a group for prime numbers.
BTW, I have Visual Basic and or Excel demonstrations of
the statements above. No proofs, of course.
Thanks for any input.
pop_stack
leppart@... 0 Attachment
At 11:30 PM 2/9/2004, pop_stack wrote:>Hi, I'm a newby so please be gentle.
This holds if x and y are both odd (not just prime) because x^2+y^2 must be
>OK
>
>There is no integer solution for
>X^2 + Y^2 = Z^2 when X and Y are prime.
congruent to 2 mod 4 and z^2 must be congruent to 0 mod 4. It holds when
x=2 (more generally when x == 2 mod 4) and y is odd for similar reasons. 0 Attachment
I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n1)^2 + (2^n)
^2 are nothing but Aurifeuillian Factors.
They have some special properties that I have discovered.
1) Basically these numbers are 2^p+2^((p+1)/2)+1 and 2^p2^((p+1)/2)+1
2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p2^((p+1)/2)+1)
2^(2p)+1= M2p * L2p (Notation used in literature)
2) p must be prime so that either L or M can be a base 2PRP.
3) Either L or M is divisible by 5. When p%8=1 or 5 M is divisible by
5 and when p%8=3 or 7 L is divisible by 5. So only one candidate
remains for each prime.
4) So the factors of these numbers are of the form 4*p*k+1.
5) I think their distribution is really similar to mersenne primes,
as most of their properties. I have searched these numbers up to
p=35000 and am continuing to search higher. I have found them to
produce an equal number of primes as mersenne numbers. I think these
primes are the Gaussian equivalents of mersenne primes.
6) I think a top 20 list of these numbers can be started on the
primepages.org web page since these numbers are well known and have
been discussed in a lot of papers.
7) I am not sure if DWT can be used productively, with this series.
But if anyone knows how it can be used productively, please let me
know.
In order to speed up the search to higher n's, I am looking for a
sieve/ Trial factorer. Could someone with the required skill please
write me a program to sieve? I did try myself to write one but it is
not very fast. I am currently using that and sieving all numbers up
to 25G before moving to PRPing. (takes about 30 sec to take a
candidate to 25 G)
Let me know, if any one can help.
 Harsh Aggarwal
Here are the primes I have found so far.
2^1+2^((1+1)/2)+1
2^12^((1+1)/2)+1
 Complete Set 
2^3+2^((3+1)/2)+1
2^32^((3+1)/2)+1
 Complete Set 
2^5+2^((5+1)/2)+1
2^72^((7+1)/2)+1
2^11+2^((11+1)/2)+1
2^19+2^((19+1)/2)+1
2^29+2^((29+1)/2)+1
2^472^((47+1)/2)+1
2^732^((73+1)/2)+1
2^792^((79+1)/2)+1
2^1132^((113+1)/2)+1
2^1512^((151+1)/2)+1
2^157+2^((157+1)/2)+1
2^163+2^((163+1)/2)+1
2^1672^((167+1)/2)+1
2^2392^((239+1)/2)+1
2^2412^((241+1)/2)+1
2^283+2^((283+1)/2)+1
2^3532^((353+1)/2)+1
2^3672^((367+1)/2)+1
2^379+2^((379+1)/2)+1
2^4572^((457+1)/2)+1
2^997+2^((997+1)/2)+1
2^13672^((1367+1)/2)+1
2^30412^((3041+1)/2)+1
2^10141+2^((10141+1)/2)+1
2^14699+2^((14699+1)/2)+1
2^275292^((27529+1)/2)+1

 In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
> Here is my proof for infiniteness of these primes
>
> if b=a+1
> then we get 2*a^2+2*a+1 =p
> solving this
> a is an integer if there is a prime p such that 2*p1=m^2
> or m^2+1/2 is a prime
>
> The distribution of such primes would follow the distribution of
> primes with the formula n^2+1
>
> it is conjectured that such primes are infinite.
>
> 
> Taken from primepages.com
>
> Are there infinitely many primes of the form n2+1?
> There are infinitely many of the forms n2+m2 and n2+m2+1. A more
> general form of this conjecture is if a, b, c are relatively prime,
a
> is positive, a+b and c are not both even,and b24ac is not a
perfect
> square, then there are infinitely many primes an2+bn+c [HW79, p19].
>
> 
>
>
> What do you all think?
>
> Also the series I talked about, what do you think about's it
> distribution.
>
> a=2^n or 2^n1
> b=a+1 = 2^n+1 or 2^n
>
> the series reduce to ((2^n1)^2+1)/2 and ((2^n+1)^2+1)/2
>
> What about the distribution of primes in such a series?
>
> let me know!
>
>
> Harsh Aggarwal
>
>
>
>  In primenumbers@yahoogroups.com, Andy Swallow
<umistphd2003@y...>
> wrote:
> > On Wed, Feb 04, 2004 at 05:31:24AM 0000, eharsh82 wrote:
> > > I am not sure if this series has finite number of primes or not.
> > > I think it has infinite primes.
> >
> > But that is a question you will never be able to answer, one way
or
> > another, if all you're doing is search for primes of this type
using
> > computer methods. So wouldn't it be more interesting to study the
> > abstract theory? Your original question was about Gaussian
primes,
> or
> > primes congruent to 1 mod 4. That's all interesting and fairly
basic
> > stuff. I would have thought that more informative answers would
be
> found
> > in there.
> >
> > Apologies if I'm talking rubbish. It just seems strange that on
the
> one
> > hand you're interested in whether certain sets contain infinitely
> many
> > primes, yet on the other hand you're studying the sets using
methods
> > guaranteed to not be able to answer the question, :)
> >
> > Anyway, that's my morning rant out of the way...
> >
> > Andy 0 Attachment
In a message dated 26/02/04 01:58:11 GMT Standard Time, harsh@... writes:
> I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n1)^2 + (2^n)
You are confirming known results.
> ^2 are nothing but Aurifeuillian Factors.
>
> They have some special properties that I have discovered.
>
> 1) Basically these numbers are 2^p+2^((p+1)/2)+1 and 2^p2^((p+1)/2)+1
>
> 2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p2^((p+1)/2)+1)
> 2^(2p)+1= M2p * L2p (Notation used in literature)
>
> 2) p must be prime so that either L or M can be a base 2PRP.
>
> 3) Either L or M is divisible by 5. When p%8=1 or 5 M is divisible by
> 5 and when p%8=3 or 7 L is divisible by 5. So only one candidate
> remains for each prime.
>
> 4) So the factors of these numbers are of the form 4*p*k+1.
>
> 5) I think their distribution is really similar to mersenne primes,
> as most of their properties. I have searched these numbers up to
> p=35000 and am continuing to search higher. I have found them to
> produce an equal number of primes as mersenne numbers. I think these
> primes are the Gaussian equivalents of mersenne primes.
>
> 6) I think a top 20 list of these numbers can be started on the
> primepages.org web page since these numbers are well known and have
> been discussed in a lot of papers.
>
> 7) I am not sure if DWT can be used productively, with this series.
> But if anyone knows how it can be used productively, please let me
> know.
>
> In order to speed up the search to higher n's, I am looking for a
> sieve/ Trial factorer. Could someone with the required skill please
> write me a program to sieve? I did try myself to write one but it is
> not very fast. I am currently using that and sieving all numbers up
> to 25G before moving to PRPing. (takes about 30 sec to take a
> candidate to 25 G)
>
> Let me know, if any one can help.
>
I introduced these "Gaussian Mersennes" in a post to the Mersenne mailing
list in 2000:
http://www.mailarchive.com/mersenne@.../msg05162.html
Chris Caldwell already has a "Top20" page about these numbers:
http://primes.utm.edu/top20/page.php?id=41
You might also like to look at my post "Gaussian analogues of the Cullen and
Woodall primes" of Dec 2000:
http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&
O=T&T=1
Mike Oakes
[Nontext portions of this message have been removed] 0 Attachment
>
http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&O=T&T=1
I looked at your link and it is quite interesting. I have a couple of
comments though. You mention that G0(n) = n*(1+i)^n + 1 and is
related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
go on to show primes of that form. Am I missing something? n*(1+i)^n
+ 1 =/= n*2^(n/2) + 1. The same could be said of G2(n) and Woodalls.
You also have Ne(n) as n^2*2^n + 1, which predates the HyperCullen
search of Steven Harvey. He has noted your finds as he searches up to
200000.
Mark 0 Attachment
In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@...
writes:
> You mention that G0(n) = n*(1+i)^n + 1 and is
It is if n = 0 mod 8, which was (one of) the values I was talking about.
> related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
> go on to show primes of that form. Am I missing something? n*(1+i)^n
> + 1 =/= n*2^(n/2) + 1.
>
Remember: (1+i)^2 = 2*i.
Mike Oakes
[Nontext portions of this message have been removed]
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