Hello,

>     Did anybody look for primes of the form p#^2^n+1
> (or n!^2^n+1)? E.g., 1801#^16+1 is prime.

No, But!,
I did look at 2^n*n#-1 and have assembled my
fragments.
They have a good, neat looking form, but the form is
also flawed. For non-prime n, "proper" notation
would be 2^n*P(pi(n))#-1, ruining the repeating digits
effect in the presentation. When n is prime, the
notational collapse doesn't exist, so whatever these
primes should be called, n has to be prime for it to
be a perfect "whatever...".

There can be circumstances where taking the primorial
of n=composite can create several instances of the
same
number in an iterated run.  With this form, the
multipler 2^n ensures each is distinct.
Thus, is it flawful, lawful, awful to say
      2^16*16#-1 is prime?

Here's what I have found on these,
any one wants to take it farther,
be my guest.

-Dick

After n=16, they are all PRP's.
I never got around to testing them.

2^2*2#-1 is prime!
2^3*3#-1 is prime!
2^7*7#-1 is prime!
2^8*8#-1 is prime!
2^14*14#-1 is prime!
2^16*16#-1 is prime!
2^18*18#-1
2^20*20#-1
2^40*40#-1
2^42*42#-1
2^44*44#-1
2^53*53#-1
2^134*134#-1
2^154*154#-1
2^185*185#-1
2^187*187#-1
2^191*191#-1
2^197*197#-1
2^200*200#-1
2^201*201#-1
2^230*230#-1
2^235*235#-1
2^239*239#-1
2^244*244#-1
2^256*256#-1
2^282*282#-1
2^303*303#-1
2^358*358#-1
2^489*489#-1
2^536*536#-1
2^665*665#-1
2^684*684#-1
2^719*719#-1
2^1098*1098#-1
2^1204*1204#-1
2^1400*1400#-1
2^1516*1516#-1
2^1629*1629#-1
2^1903*1903#-1
2^1997*1997#-1
2^1999*1999#-1
2^2104*2104#-1
2^2477*2477#-1
2^3075*3075#-1
2^3676*3676#-1
2^3785*3785#-1
2^4115*4115#-1
2^5429*5429#-1
2^5808*5808#-1
2^6069*6069#-1
2^6276*6276#-1
2^9095*9095#-1
2^10423*10423#-1
2^10839*10839#-1
2^16181*16181#-1
2^17521*17521#-1
2^17734*17734#-1
2^20451*20451#-1
2^22560*22560#-1
2^29545*29545#-1
2^30069*30069#-1
2^33389*33389#-1
  ** Run Stopped at 34952

Here's the largest cofactor PRP's in the run
(2^12804*12804#-1)/15959
(2^15233*15233#-1)/263089
(2^18932*18932#-1)/116981
(2^27766*27766#-1)/1993067

And here are the caps of a couple Sophie Germains,
(no more of these at least to n=918).
2^(53-1)*53#-1 (digits:36 checksum:_8F526EDE_)
2^(201-1)*201#-1 (digits:143 checksum:_E1E2C9AF_)



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--- In primenumbers@y..., "Andrey Kulsha" <Andrey_601@t...> wrote:
>     Hello!
>
>     Did anybody look for primes of the form p#^2^n+1 (or n!^2^n+1)?
E.g., 1801#^16+1 is prime.
>
>     Best wishes,
>
>     Andrey

Check out

http://home.btclick.com/rw.smith/pp/page1.htm

and

http://ourworld.compuserve.com/homepages/hlifchitz/

--Mark

http://www.primepuzzles.net/conjectures/conj_008.htm

Proof of D. Andrica conjecture.

Two squares one with area P(n) the other P(n+1) where P(n+1) < 2*P(n) by proof
of "Bertrand's postulate" by Tschebycheff.

Subtract the area of the square P(n)/P(n) = 1: So the area
P(n+1)- P(n) = area D, the differance between the primes, is
(P(n+1)-P(n))/P(n) < 2-1 < 1.

So the sides have lengths of the squares are
sqrt(P(n+1)) < sqrt(2*p(n))and sqrt(P(n)) < sqrt(P(n+1). Or after dividing by
sqrt(P(n))
1 < sqrt(P(n+1))/sqrt(P(n)) < sqrt(2)
0 < sqrt(P(n+1))/sqrt(P(n))- 1 < sqrt(2)-1 < 1.

The reason we can divide by sqrt(P(n)) is P(n+1) - P(n) = D is the differance
between to squares.
D = (sqrt(P(n+1))- sqrt(P(n))) (sqrt(P(n+1))+sqrt(P(n)))
with
2*sqrt(P(n)) < sqrt(P(n+1))+sqrt(P(n))
making
D /sqrt(P(n))~(sqrt(P(n+1))-sqrt(P(n)))*(2 *sqrt(P(n)))/sqrt(P(n))
So QED

Do y'all  See any problems?



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

Let p and q be successive primes.
Andrica conjectures that
q - p < sqrt(q) + sqrt(p)
whereas Tschebysheff proved merely that
q - p < p
which is far weaker.

>>>Let p and q be successive primes.
Andrica conjectures that
q - p < sqrt(q) + sqrt(p)
whereas Tschebysheff proved merely that
q - p < p
which is far weaker.<<<

Andrica's conjecture doesn't seem very tight compared to the two:

x^2 < p < (x+1)^2

x^2 < p < x^2 + x

Andrica's conjecture looks comparable to the first one.

Can it be tightened?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

p, p(p+1)+1, p(p+1)(p+2)+1, p(p+1)(p+2)(p+3)+1

are all primes.

Alternatively,  p(p+1)-1 or p(p-1)+1 or p(p-1)-1

e.g.

2, 2.3+1=7

3, 3.4+1=13, 3.4.5+1=61


Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

At 03:57 PM 7/31/2002 +0000, Jack Brennen wrote:
>Phil's posting of an IMO problem from this year prompted me
>to go back and look up the USAMO (USA Math Olympiad) problems
>from the days when I competed on that test.
>
>I found the following gem from 1982.  I competed on the USAMO
>in 1982, and I was amazed to see this question, because I
>honestly don't remember it from 20 years ago, despite my
>rather intimate knowledge of the question nowadays:
>
>(4)  Prove that there exists a positive integer k such that
>      k*2^n+1 is composite for every positive integer n.
>
>I wish I could go back and see how I answered this one
>as a 16-year old kid.  :-)

*nod*

I took the first and second tests - the ones that aren't the olympiad, I
forget what they're called - and became the first kid in my district to be
allowed to take the second.

I don't think there was anything about primes, though, sadly - I might have
gotten the other ten points and been allowed to take the USAMO if there had
been.

This is drifting OT, but are kids supposed to be able to take the Olympiad
tests throughout high school?  I've heard things that seem to imply that,
here and elsewhere, but my school only allowed it for seniors who were in
accelerated math (perhaps 10-20 people a year).  Perhaps that's why I was
the first able to make it to the invitational (level 2) test?

Nathan

--- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>
>
> Do y'all  See any problems?

Maybe,

Bertrand/Tschebycheff: pn<(pn+1)<2*pn
Take: sqrt(pn) < sqrt(pn+1) < sqrt(2*pn)
divide by constant c: sqrt(pn)/c < sqrt(pn+1)/c < sqrt(2*pn)/c

A.) subtract lhs: 0 < (sqrt(pn+1)-sqrt(pn))/c < (sqrt(2*pn)-sqrt
(pn))/c

Regardless of the value of c, A.) must hold.
We need to get to Andrica's conjecture: sqrt(pn+1) - sqrt(pn)<1.
So, multiply A.) by c to get:
B.) 0 < sqrt(pn+1)-sqrt(pn) < sqrt(2*pn)-sqrt(pn)

We can use B.) to prove Andrica's Conjecture only
if it is always true that rhs <1 - i.e.:
  sqrt(2*pn)-sqrt(pn) < 1
divide by c=sqrt(pn)to get:
  sqrt(2)-1 < 1/sqrt(pn)
which is only true for n=1

QESh..

-Dick Boland

Nathan Russell wrote:
> I took the first and second tests - the ones that aren't the olympiad, I
> forget what they're called - and became the first kid in my district to be
> allowed to take the second.

The AHSME (Annual High School Math Exam) and the AIME (Annual Invitational
Math Exam).  I think that approximately 1% of the AHSME contestants get
invited to take the AIME.  That top 1% is also not evenly distributed --
some magnet schools and specialized math/science schools routinely get
30 to 50 students into the AIME every year.  Outside of these top-rung
high schools, probably 1 in 250 students advances to the AIME.  The USAMO
contestants are chosen based on the AHSME-AIME combined score, but it's not
a simple "make the cut" threshold -- non-seniors have it easier, and I
believe that every state of the US must be represented.

> This is drifting OT, but are kids supposed to be able to take the Olympiad
> tests throughout high school?  I've heard things that seem to imply that,
> here and elsewhere, but my school only allowed it for seniors who were in
> accelerated math (perhaps 10-20 people a year).  Perhaps that's why I was
> the first able to make it to the invitational (level 2) test?

The USA Olympiad is certainly open to students as young as 8th grade, perhaps
even younger.  I know that in many US high schools, only the "top-level" math
teacher knows anything about the AHSME-AIME-USAMO trilogy of tests.  Unless
that teacher seeks out precocious students in younger grades, they may never
be aware that they are eligible for the AHSME.  I took the AHSME for the
first time in 7th grade (12 years old), and I had to take the exam at a
different school, since my school knew nothing about the test.  I only knew
about it because I had been "discovered" by the county math team coach, who
insisted that I find a way to take the exam.  In 9th grade, I took the USAMO
for the first time -- then again in 10th and 12th grades, including a top-12
finish my senior year.

Jack:

> I had been "discovered" by the county math team coach,
> who insisted that I find a way to take the exam.
> In 9th grade, I took the USAMO for the first time
> -- then again in 10th and 12th grades, including
> a top-12 finish my senior year.

I have good friends in US, UK and Russia who have been
through such competitive mills and have clearly not been
harmed by them.

Even so, it seems to me that these contests
give the wrong idea about what good math
(and much else) truly requires.

Arete (the pursuit of excellence) is not the same
as javelin throwing...

Best

David

djbroadhurst wrote:
> I have good friends in US, UK and Russia who have been
> through such competitive mills and have clearly not been
> harmed by them.

I didn't find it to be harmful at all.  It did likely steer
me away from a career in mathematics, mainly because I learned
for the first time in my life that I wasn't the best at math.
I studied with and competed against people my age who could
"run circles around me" math-wise.  This could have been very
rough on me if my self-image was "great math whiz" -- but
that's not me.  Instead, it was liberating in a way, because
I no longer felt "obligated" to a career in mathematics.
And it definitely pushed me to improve my reasoning and
thinking skills, but that was certainly a good thing.

> Even so, it seems to me that these contests
> give the wrong idea about what good math
> (and much else) truly requires.
>
> Arete (the pursuit of excellence) is not the same
> as javelin throwing...

Well, the skills I learned and developed in training for
those Olympiads were quite relevant to good math.  In
order to score well on an Olympiad, one needs to have
aptitude for mathematics, a broad and comprehensive
understanding across a wide range of the field, and the
ability to express one's ideas and methods succinctly
and accurately.

On the other hand, the Olympiad is a closed-book exam;
if real mathematicians didn't have reference libraries
and the desire and ability to use them well, the progress
of mathematics would be severely impaired.  Similarly,
the Olympiad is an individual examination; real world
mathematicians (well, *respected* real world mathematicians)
have to collaborate with their peers on a regular basis --
besides joint papers, they have formal and informal peer review.
Lastly, an Olympiad competitor begins an Olympiad problem
with the knowledge that a solution exists which should take
him no more than about two hours to discover, formulate,
perfect, and "publish" -- real world mathematicians rarely
have that luxury.  :-)

I am wondering if this is positive or saying that it is not strong enough. I
also think I may be unclear so I restated it again below.

----- Original Message -----
From: "djbroadhurst" <d.broadhurst@...>
To: <primenumbers@yahoogroups.com>
Sent: Saturday, August 03, 2002 4:05 AM
Subject: [PrimeNumbers] Re: Andrica


> Let p and q be successive primes.
> Andrica conjectures that
> q - p < sqrt(q) + sqrt(p)
> whereas Tschebysheff proved merely that
> q - p < p
> which is far weaker.

rewriten as q < 2p
include p itself  p < q < 2p

take sqrt, sqrt (p) < sqrt (q) < sqrt(2)*sqrt(p) --- length of side with
square q
take out the factor sqrt (p), 1< sqrt (q) / sqrt(p) < sqrt(2)

subtract the area p from q by
take the sqrt, sqrt (p)
take out the factor sqrt (p), 1 --- length of side with square p
so
1-1 = 0 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1
1 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1 < 1

using the difference of two squares
q - p = (sqrt(q) + sqrt(p)) (sqrt(q) - sqrt(p))
(q - p) / (sqrt(q) + sqrt(p)) = (sqrt(q) - sqrt(p))
and
(sqrt (q) - sqrt(p)) < (sqrt(2) sqrt(p)) - (sqrt(p))
(sqrt (q) - sqrt(p))/sqrt(p) < (sqrt(2) - 1)(sqrt(p))/sqrt(p)
S = sqrt (q) / sqrt(p) -1 < sqrt(2) - 1 < 1
so the first number with p = 2 is < 1 and the limit S as p -> oo   = 0.
QED

I'm just a college student in physics so it would help if I could get some
feedback.

Thanks David.

John

I know that I have been using Tschebysheff proof. But is there somewhere on the
net a proof of this? I just would like to see how he did it.


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

What David is saying is that you have merely re-worked Tschebysheff's proof.

Without having put in anything extra into the system, all you have done is
find new expressions for existing formula.

As a Mathematician, your proof is lacking in several areas:

#rewriten as q < 2p
#include p itself  p < q < 2p

This could be written as: there exists a prime q, p<q<2p

#take sqrt, sqrt (p) < sqrt (q) < sqrt(2)*sqrt(p) --- length of side with
#square q
#take out the factor sqrt (p), 1< sqrt (q) / sqrt(p) < sqrt(2)

OK.

#subtract the area p from q by
#take the sqrt, sqrt (p)
#take out the factor sqrt (p), 1 --- length of side with square p
#so
#1-1 = 0 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1
#1 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1 < 1

Line 2 makes no sense, nor does line 3
Line 5 is OK, but Line6 contains obvious errors.

At this point what we have is sqrt(q)/sqrt(p) - 1 < 1, which you managed to
get to in your first proof, but seem to have failed to in this one.

From here, we can get sqrt(q)-sqrt(p)<sqrt(p), but this is far from
Andrica's conjecture.

#using the difference of two squares
#q - p = (sqrt(q) + sqrt(p)) (sqrt(q) - sqrt(p))
#(q - p) / (sqrt(q) + sqrt(p)) = (sqrt(q) - sqrt(p))
#and
#(sqrt (q) - sqrt(p)) < (sqrt(2) sqrt(p)) - (sqrt(p))
#(sqrt (q) - sqrt(p))/sqrt(p) < (sqrt(2) - 1)(sqrt(p))/sqrt(p)
#S = sqrt (q) / sqrt(p) -1 < sqrt(2) - 1 < 1
#so the first number with p = 2 is < 1 and the limit S as p -> oo   = 0.
#QED

With such a dodgy start, most Mathematicians would be rolling about on the
floor by now in tears, but persistence is to be encouraged.

Line 1 is good, but you can begin to see why this conjecture is considered
hard.
Line 2 is good.

After here, you need to explain where your subtitutions are coming from.
Also needed is a statement declaring how the algebra will lead to a proof of
the conjecture. This assists any reader in following your work, and takes
the pressure off the reader in trying to follow the work.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

#I know that I have been using Tschebysheff proof. But is there somewhere on
the net a proof of #this? I just would like to see how he did it.

I haven't seen a proof for this on the net, but I imagine it uses the PNT.

See:

http://www.users.globalnet.co.uk/~perry/maths/primedensity/primedensity.htm

Erdos proved that there are 2 primes between p and 2p:

http://www.maa.org/mathland/mathland_10_7.html

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com


-----Original Message-----
From: John W. Nicholson [mailto:johnw.nicholson@...]
Sent: 04 August 2002 09:26
To: primenumbers@yahoogroups.com
Subject: [PrimeNumbers] Tschebysheff


I know that I have been using Tschebysheff proof. But is there somewhere on
the net a proof of this? I just would like to see how he did it.


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


Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
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Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

> >     Did anybody look for primes of the form p#^2^n+1 (or n!^2^n+1)?
> E.g., 1801#^16+1 is prime.
[snip]
> Check out
>
> http://home.btclick.com/rw.smith/pp/page1.htm

     p#^2^n+1 isn't primoproth, it's primo-generalized-fermat.

     So, 1801#^16+1 appears to be the largest known such number (p#^2^n+1)?

     Best wishes,

     Andrey

--- Jon Perry <perry@...> wrote:
> With such a dodgy start, most Mathematicians would be rolling about
> on the
> floor by now in tears, but persistence is to be encouraged.

Jon,
I don't know if you're trying to be "Me Too" to David's "Big Dog",
   http://www.winternet.com/~mikelr/flame3.html ?
However, you're not really in a position to make snide comments about
proofs, particularly in such a patronising way, and especially
considering some analyses of your own proof style (
http://groups.yahoo.com/group/primenumbers/message/7546 )

Phil
(No bonus points for spotting the irony herein!)


=====
--
The good Christian should beware of mathematicians, and all those who make
empty prophecies. The danger already exists that the mathematicians have
made a covenant with the devil to darken the spirit and to confine man in
the bonds of Hell. -- Common mistranslation of St. Augustine (354-430)

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>     So, 1801#^16+1 appears to be the largest known such number (p#^2^n+1)?

     Such PRIME number, of course.

     Andrey

Please forgive me my spaming...

>     So, 1801#^16+1 appears to be the largest known such number (p#^2^n+1)?

789  7457#^16+1                   50805 p16  2000 Generalized Fermat

     I should be more attentive.

     Best wishes,

     Andrey

Into various permutations of the original:

http://www.users.globalnet.co.uk/~perry/maths/extendingtensquares/extendingt
ensquares.htm

including quadruples such that every combo of 3 is a square, e.g.:

1,5,7,24
1,8,45,91

and the new conjecture that:

For n>2, the simultaneous equations:

ab+1=x^n
ac+1=y^n
bc+1=z^n

have no solutions.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> For n>2, the simultaneous equations:
>
> ab+1=x^n
> ac+1=y^n
> bc+1=z^n
>
> have no solutions.

Oh, really.  I found a solution.  :-)

(a,b,c,x,y,z,n) == (2,171,25326,7,37,163,3)


To effectively search this, don't vary (a,b,c).  Instead choose n,
then vary (y,x,a).  y goes from 3 to LIMIT; x goes from 2
to y-1; a iterates over the divisors of gcd(x^n-1,y^n-1).

Pari/GP:

   n=3;for(y=3,10000,yt=y^n-1;for(x=2,y-1,xt=x^n-1;g=gcd(xt,yt);
   if(g>1,p=xt*yt;fordiv(g,a,if(a>1,w=p/(a^2)+1;
   z=round(w^(1/n));if(w==z^n,b=(x^n-1)/a;c=(y^n-1)/a;
   print("(",a,",",b,",",c,",",x,",",y,",",z,",",n,")")))))))

The solution above takes less than one second to find.

----- Original Message -----
From: "richard042" <richard042@...>
To: <primenumbers@yahoogroups.com>
Sent: Saturday, August 03, 2002 3:31 PM
Subject: [PrimeNumbers] Re: Proof of D. Andrica conjecture.


> --- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>
> >
> > Do y'all  See any problems?
>
> Maybe,
>
> Bertrand/Tschebycheff: pn<(pn+1)<2*pn
> Take: sqrt(pn) < sqrt(pn+1) < sqrt(2*pn)
> divide by constant c: sqrt(pn)/c < sqrt(pn+1)/c < sqrt(2*pn)/c
>
> A.) subtract lhs: 0 < (sqrt(pn+1)-sqrt(pn))/c < (sqrt(2*pn)-sqrt
> (pn))/c
>
> Regardless of the value of c, A.) must hold.
> We need to get to Andrica's conjecture: sqrt(pn+1) - sqrt(pn)<1.
> So, multiply A.) by c to get:
> B.) 0 < sqrt(pn+1)-sqrt(pn) < sqrt(2*pn)-sqrt(pn)
>
> We can use B.) to prove Andrica's Conjecture only
> if it is always true that rhs <1 - i.e.:
>  sqrt(2*pn)-sqrt(pn) < 1
> divide by c=sqrt(pn)to get:
>  sqrt(2)-1 < 1/sqrt(pn)
> which is only true for n=1
>
> QESh..
>
> -Dick Boland

Thanks to Dick and David Broadhurst,

You have me doing a change to even a better way of showing this:

Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2 where a, b,
and c are a real numbers.
Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we assume is
< c + a
Take the square root to find c
c = a + b
s^2 = 2ab + b^2 < a + b + a = 2a + b
Or 2ab + b^2 < 2a + b
So 0 < b < 1
Replace b with y = 1/b so 1 < y however
2a + y must be < a^2 by Bertrand/Tschebycheff:
So y < a^2 - 2a
By adding 1 on the right side we complete the square without going >= a^2
So y <= (a - 1)^2
This is the largest y value; the smallest is y => (a - (a - 1))^2
Let us replace (a-1) with a integer k that make up the complete residue set
for a that is not 0 == mod a.

So there is an integer 1 <= k <= a such that
b = 1/(a - k)^2, c = a + b, and s^2 = 2ab + b^2 <  2a + b

QED

Do I have it this time?

(a,b,c,x,y,z,n) == (2,171,25326,7,37,163,3)

Nice work. It's still unsolved for n>3 though.

See:

http://www.users.globalnet.co.uk/~perry/maths/extendingtensquares/extendingt
ensquares.htm

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

at:

http://www.users.globalnet.co.uk/~perry/maths/nproducingprimes/nproducingpri
mes.htm

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

--- jbrennen <jack@...> wrote:
> --- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> > For n>2, the simultaneous equations:
> >
> > ab+1=x^n
> > ac+1=y^n
> > bc+1=z^n
> >
> > have no solutions.
>
> Oh, really.  I found a solution.  :-)
>
> (a,b,c,x,y,z,n) == (2,171,25326,7,37,163,3)
>
>
> To effectively search this, don't vary (a,b,c).  Instead choose n,
> then vary (y,x,a).  y goes from 3 to LIMIT; x goes from 2
> to y-1; a iterates over the divisors of gcd(x^n-1,y^n-1).
>
> Pari/GP:
>
>   n=3;for(y=3,10000,yt=y^n-1;for(x=2,y-1,xt=x^n-1;g=gcd(xt,yt);
>   if(g>1,p=xt*yt;fordiv(g,a,if(a>1,w=p/(a^2)+1;
>   z=round(w^(1/n));if(w==z^n,b=(x^n-1)/a;c=(y^n-1)/a;
>   print("(",a,",",b,",",c,",",x,",",y,",",z,",",n,")")))))))
>
> The solution above takes less than one second to find.

Smart method.
n=4 (1352,9539880,9768370,337,339,3107,4)

Phil


=====
--
The good Christian should beware of mathematicians, and all those who make
empty prophecies. The danger already exists that the mathematicians have
made a covenant with the devil to darken the spirit and to confine man in
the bonds of Hell. -- Common mistranslation of St. Augustine (354-430)

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>Smart method.
>n=4 (1352,9539880,9768370,337,339,3107,4)

I agree, see same page for quote.

P.S. I hope you are all checking the case for 4 variables (nay 5...)....

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

One small but big change  See it near bottom.

>
> Thanks to Dick and David Broadhurst,
>
> You have me doing a change to even a better way of showing this:
>
> Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2 where a, b,
> and c are a real numbers.
> Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we assume
is
> < c + a
> Take the square root to find c
> c = a + b
> s^2 = 2ab + b^2 < a + b + a = 2a + b
> Or 2ab + b^2 < 2a + b
> So 0 < b < 1
> Replace b with y = 1/b so 1 < y however
> 2a + y must be < a^2 by Bertrand/Tschebycheff:
> So y < a^2 - 2a
> By adding 1 on the right side we complete the square without going >= a^2
> So y <= (a - 1)^2
> This is the largest y value; the smallest is y => (a - (a - 1))^2

Change the following
Let us replace (a-1) with a real number k.
So there is an real 0 < k < a such that
b = 1/(a - k)^2, c = a + b, and s^2 = 2ab + b^2 <  2a + b

  QED

  Do I have it this time?

Dear friends,

What is the biggest prime known so far?

What is the biggest known twin prime?

Where can I get more details about the ways through which(algorithms) prime
numbers are computed.


Regards,
S.R.Sudarshan Iyengar


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

--- "S.R.Sudarshan Iyengar" <sudarshansr@...> wrote:
> Dear friends,
>
> What is the biggest prime known so far?

   http://primepages.org/largest.html

2^13466917-1 4053946 Cameron, Woltman, Kurowski, GIMPS    2001
(notes)

> What is the biggest known twin prime?

   http://primepages.org/largest.html

318032361.2^107001±1  32220 Underbakke, Carmody, PrimeForm 2001

> Where can I get more details about the ways through
> which(algorithms) prime numbers are computed.

   http://primepages.org/

In particlular "Finding primes & proving primality" ->
   http://primepages.org/prove/index.html

Phil


=====
--
The good Christian should beware of mathematicians, and all those who make
empty prophecies. The danger already exists that the mathematicians have
made a covenant with the devil to darken the spirit and to confine man in
the bonds of Hell. -- Common mistranslation of St. Augustine (354-430)

__________________________________________________
Do You Yahoo!?
Yahoo! Health - Feel better, live better
http://health.yahoo.com

----- Original Message -----
From: "John W. Nicholson" <johnw.nicholson@...>
To: <primenumbers@yahoogroups.com>
Sent: Sunday, August 04, 2002 3:25 AM
Subject: [PrimeNumbers] Tschebysheff


> I know that I have been using Tschebysheff proof. But is there somewhere
on the net a proof of this? I just would like to see how he did it.
>
>
Is there any pages other than the ones that Jon sent? And is there anyone
who knows what this proof is good for? I mean What are all of the things
that have been found by using the fact that a prime is between p and 2*p?

I just want to understand the method that he used to find this proof.