## Re: Pythagorean Triplets

Expand Messages
• 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*p-1=m^2 or
Message 1 of 19 , Feb 4, 2004
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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*p-1=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 b2-4ac 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^n-1
b=a+1 = 2^n+1 or 2^n

the series reduce to ((2^n-1)^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
or
> primes congruent to 1 mod 4. That's all interesting and fairly basic
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
• 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
Message 2 of 19 , Feb 9, 2004
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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@...
• ... This holds if x and y are both odd (not just prime) because x^2+y^2 must be congruent to 2 mod 4 and z^2 must be congruent to 0 mod 4. It holds when x=2
Message 3 of 19 , Feb 9, 2004
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At 11:30 PM 2/9/2004, pop_stack wrote:
>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.

This holds if x and y are both odd (not just prime) because x^2+y^2 must be
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.
• I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n) ^2 are nothing but Aurifeuillian Factors. They have some special properties that I have
Message 4 of 19 , Feb 25, 2004
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I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^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^p-2^((p+1)/2)+1

2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p-2^((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 2-PRP.

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
2^1-2^((1+1)/2)+1
- Complete Set -
2^3+2^((3+1)/2)+1
2^3-2^((3+1)/2)+1
- Complete Set -
2^5+2^((5+1)/2)+1
2^7-2^((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
2^47-2^((47+1)/2)+1
2^73-2^((73+1)/2)+1
2^79-2^((79+1)/2)+1
2^113-2^((113+1)/2)+1
2^151-2^((151+1)/2)+1
2^157+2^((157+1)/2)+1
2^163+2^((163+1)/2)+1
2^167-2^((167+1)/2)+1
2^239-2^((239+1)/2)+1
2^241-2^((241+1)/2)+1
2^283+2^((283+1)/2)+1
2^353-2^((353+1)/2)+1
2^367-2^((367+1)/2)+1
2^379+2^((379+1)/2)+1
2^457-2^((457+1)/2)+1
2^997+2^((997+1)/2)+1
2^1367-2^((1367+1)/2)+1
2^3041-2^((3041+1)/2)+1
2^10141+2^((10141+1)/2)+1
2^14699+2^((14699+1)/2)+1
2^27529-2^((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*p-1=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 b2-4ac 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^n-1
> b=a+1 = 2^n+1 or 2^n
>
> the series reduce to ((2^n-1)^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
primes,
> or
> > primes congruent to 1 mod 4. That's all interesting and fairly
basic
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
• ... You are confirming known results. I introduced these Gaussian Mersennes in a post to the Mersenne mailing list in 2000:
Message 5 of 19 , Feb 26, 2004
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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^n-1)^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^p-2^((p+1)/2)+1
>
> 2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p-2^((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 2-PRP.
>
> 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.
>

You are confirming known results.

I introduced these "Gaussian Mersennes" in a post to the Mersenne mailing
list in 2000:
http://www.mail-archive.com/mersenne@.../msg05162.html

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

[Non-text portions of this message have been removed]
• ... 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
Message 6 of 19 , Feb 26, 2004
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>
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 pre-dates the Hyper-Cullen
search of Steven Harvey. He has noted your finds as he searches up to
200000.

--Mark
• In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@wi.rr.com ... It is if n = 0 mod 8, which was (one of) the values I was talking about.
Message 7 of 19 , Feb 26, 2004
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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
> 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.
>
It is if n = 0 mod 8, which was (one of) the values I was talking about.
Remember: (1+i)^2 = 2*i.

-Mike Oakes

[Non-text portions of this message have been removed]
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