I found this moving conversation going on over in sci.math.research and
knew you boys would want to have a go at it.

Go easy on them. Most of them probably can't tie a bowline or sew that
rip in the genoa.

I bet none of them can get a Yanmar 3GM running on a cold day when the
battery's dead.....though it might be interesting to watch them try....

One must admit, however, it's amazing to watch the genius mind running
at full throttle, though some of the results can result in armagheddon
for all of us.




Subject: my conjectu n! + prime(n) != m^k
From:
Newsgroups: sci.math.research

Hello,

Perhaps you may be interested to check out the conjecture, which I
have made lately ?


NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1
i.e.
n! + prime(n) != m^k


Kurt Foster responded to my post with the partial proof for k=2
see NMBRTHRY Archives - September 2008
http://listserv.nodak.edu/cgi-bin/wa...mbrthry&D=0&H=...

Prior to Kurt Foster's solution, David Harden together with Daniel
Berend responded to my post (on the SeqFan mail list) with the
similar
proof for the case of squares - see at the very bottom of the
attached
below email chain.:

Also see info on possible directions in proving my conjecture in
general - see attached below email chain.

Thanks,
Best Regards,
Alexander R. Povolotsky
---------- Forwarded message ----------
From: Alexander Povolotsky
Date: Aug 28, 2008 8:14 PM
Subject: remark from Noam Elkies
To: Florian Luca

Dear Florian,

Regarding C and c, which both were cited by you, and towards
refining
ABC ability to quantify upper n limit, Noam Elkies made the following
remark:

" If one could show that c0 is such, that the radical is always at
least c*C^(4/5), then I can give you an upper bound on n, such
that n! + prime(n) is a perfect power, and then what remains - it's
just a finite computation to show/verify that none of those remaining
candidates {n*} satisfy p_n* +n*! = m^k "

So (if I understood correctly what Noam Elkies meant in his remark as
"radical" ) then my question to you would be - is proving that

Rad(n!*p_n*m) = c*C^(4/5)

possible ?

Thanks,
Regards,
Alexander R. Povolotsky
=================================================
On Thu, Aug 28, 2008 at 9:45 AM, Florian Luca
wrote:
Dear Alex

Thanks for the conjecture. Igor mentioned linear forms in logarithms.
Here is
how they work in your case:
Write
n!=m^k-p_n
m and p_n are odd. The exponent of 2 in n! is roughly about n (it is
\ge
n-(log(n+1))/log 2).
Linear forms in logarithms tell you
that the exponent of 2 on the right is at most
C*log m *log p_n * log k.
Here C is a constant. Since log p_n is roughly log n
and log k cannot exceed log n either (because mn),
you get that the above bound is log m (log n)^2.
So,
log m n/(log n)^2, which in turn puts k(log n)^2.
So, you know now that k is not too large. You have an argument for 2
which
is very nice. Would it work for 3:
p_n is a cubic residue modulo all primes q=1 mod 3 n.
Is this enough to get a contradiction? Probably.
Then you can do it for 5, 7, up to all primes up to O((log n)^2).
Heuristically it should work at least under GRH and so on. Namely,
given
a prime p, the probability that a number which is not a pth power
looks like a pth power modulo q, a prime which is 1 mod p, is 1/p.
Assuming some independence over q, you multiply these probabilities
up
to x, to get (1/p)^{pi(x,p,1)}.
If p=O((log x)^2)), then the above amount is the reciprocal of
p^{pi(x,p,1)} which is about exp(x log p/p(log x)). Since p is a power
of
logarithm of x, this number is huge (it is at least exp(c x/(log x)^3)
with
some constant c.
So, you expect that there should be no p_n such that
p_n = m^p mod q for all primes q=1 mod p not exceeding n,
and all primes p=O((log n)^2) for large n.
The preprint that Igor sent could help to prove this under GRH.
Conditionally, it also follows immediately from ABC that there are no
solutions for large n, and the argument that I made above can be
immediately modified to prove that the set of n such that n!+p_n
is a perfect power is of asymptotic density zero. I am not sure if the
full
thing can be finished off unconditionally by these arguments only. But
maybe there is something that escapes me.

Cheers,
Florian
---------- Original Message -----------
From: "Alexander Povolotsky"
To:
Sent: Thu, 28 Aug 2008 05:00:47 -0400
Subject: conjectu n! + prime(n) != m^k

Dear Florian - Igor Shparlinski kindly recommended me to ask your
advise/opinion re proving of my conjecture.

Thanks,
Regards,
Alexander R. Povolotsky
---------- Forwarded message ----------
From: Igor Shparlinski
Date: Wed, Aug 27, 2008 at 10:10 PM
Subject: n! + prime(n) != m^k
To: Alexander Povolotsky

Hi,

I don't see why the case k = 2 is any special?
In fact the proof uses an variant of argument, which
has been used in studying so-called x-pseudosquares:
the claim is the p(n)-is an n-pseudosquare which is
impossible, by say Polya-Vinogradov. I'm attaching a
paper where you can find further references
Why can't you extend the proof to any k
(which is not too large).
On the other hand, when k is large
then form in logarithms may already give you something.
Actually they may already give everything you
need starting with k = 2.
I would suggest that you forward this question to
Florian Luca who would be the best person to ask.
Florian's email address is .

Best wishes,
Igor
At 8:53 PM -0400 27/8/08, Alexander Povolotsky wrote:

Hello Dear Igor,

Perhaps you may be interested to check out the conjecture,
which I have made lately ?

NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1 ?
The formal proof (from David Harden with further proof improvement
from Daniel Berend) so far only provided (in reply to my posting)
for k=2 (see below).

Regards,
Alexander R. Povolotsky
================================================== =======
David Harden wrote:

It is trivial to check this for n=3.
So we may assume that n = 4, which means n! is a multiple of 8 and
that p_n is odd.
Then n! + p_n = x^2
means p_n == x^2 (mod 8).
Since p_n is odd, x^2 is odd and, therefore, x^2 == 1 (mod 8) so p_n
== 1 (mod 8).
Let q be an odd prime with q = n.
(Note that q p_n.)
Then p_n == x^2 (mod q) so (using Legendre symbol notation)(p_n/q) =
1. Since p_n == 1 (mod 4), quadratic reciprocity tells us that
(q/p_n) = 1. Also, (2/p_n) = 1 because p_n == 1 (mod 8). This means
that the smallest prime quadratic nonresidue (equivalently, the
smallest positive quadratic nonresidue) modulo p_n is n ~
p_n/(log(p_n) - 1). This is very large; known effective bounds on
the smallest quadratic nonresidue modulo a prime fall well under
this. You have probably searched up to n large enough for these
bounds to apply and conclude the proof.


---- David

************************************************** **********************
***ð*****
From : berend daniel
To : David Harden , ,
Alexander Povolotsky
Subject :
Does any "n" exist, such that ( n! + prime(n) ) yields integral
square ? Date : Mon, Aug 11, 2008 03:21 AM

You don't need the bounds on quadratic non-residues.
Once you you know that all primes up to n are quadratic residues,
so are all numbers up to p_n, all of whose prime divisors do not
exceed n. This mean that most integers up to p_n are quadratic
residues.
But only half of them are. ... Contradiction.

Best,
Dani
---------- Forwarded message ----------
From: berend daniel
Date: Aug 12, 2008 5:28 AM
Subject: Does then NO "n" exist, such that ( n! + prime(n) )
yields integral m^k where k1 ?

--------- Forwarded message ----------
From: berend daniel
Date: Aug 12, 2008 5:28 AM
Subject: Does then NO "n" exist, such that ( n! + prime(n) )
yields integral m^k where k1 ?
To: Alexander Povolotsky
Cc: David Harden

Alexander Povolotsky wrote:
Does then NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1 ?
Thanks,
Regards,
Alexander R. Povolotsky
I certainly believe that's the case, but this doesn't seem to follow
from the same considerations.

Best,
Dani

"Larry" wrote in message
...
I found this moving conversation going on over in sci.math.research and
knew you boys would want to have a go at it.

Go easy on them. Most of them probably can't tie a bowline or sew that
rip in the genoa.

I bet none of them can get a Yanmar 3GM running on a cold day when the
battery's dead.....though it might be interesting to watch them try....

One must admit, however, it's amazing to watch the genius mind running
at full throttle, though some of the results can result in armagheddon
for all of us.



I hope to GOD they stay away from the Large Hadron Collider! LOL

--
"j" ganz @@
www.sailnow.com

"Capt. JG" wrote in
easolutions:

I hope to GOD they stay away from the Large Hadron Collider! LOL



The mathematicians created the atomic bombs. Without them it would have
been impossible. At the collider, they're working on black holes,
now....simply suck the opposing army to the New World Order into the black
hole, killing the overpopulation, prisoner problem and opposition to the
New World Order into oblivion at the same time.

"Larry" wrote in message
...
"Capt. JG" wrote in
easolutions:

I hope to GOD they stay away from the Large Hadron Collider! LOL



The mathematicians created the atomic bombs. Without them it would have
been impossible. At the collider, they're working on black holes,
now....simply suck the opposing army to the New World Order into the black
hole, killing the overpopulation, prisoner problem and opposition to the
New World Order into oblivion at the same time.



Ok. You convinced me. I'm going to vote for Huckabee!

--
"j" ganz @@
www.sailnow.com

27.

Easy.

--
Roger Long

On 2008-09-27 14:07:20 -0400, "Roger Long" said:

27.

Easy.

No, it's 42.

--
Jere Lull
Xan-à-Deux -- Tanzer 28 #4 out of Tolchester, MD
Xan's pages: http://web.mac.com/jerelull/iWeb/Xan/
Our BVI trips & tips: http://homepage.mac.com/jerelull/BVI/

"Roger Long" wrote in news:gblsoh$e4l$1
@registered.motzarella.org:

27.

Easy.

--
Roger Long





Damned engineers. Always have all the answers....

NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1
i.e.
n! + prime(n) != m^k


I assume "integral" really meant integer since no differential is given.

n=3

then

prime(3) = 2

substituting:

n! + prime(n)! = 3! + 2! = 8

2^3 = 8

Therefo

n! +prime(n)! = m^k

where n = 3, m =2 and k = 3

All integers and k1.

If I read you correctly, you just proved his conjecture is false. So all
those pointy headed mathematicians are running about trying to prove it to
be true. Should we tell them or let them run about? BTW, a simple, yet
elegant method on your part.

Mike

"Charles Momsen" wrote in message
...

NO "n" exist, such that
( n! + prime(n) )
yields integral m^k
where k1
i.e.
n! + prime(n) != m^k


I assume "integral" really meant integer since no differential is given.

n=3

then

prime(3) = 2

substituting:

n! + prime(n)! = 3! + 2! = 8

2^3 = 8

Therefo

n! +prime(n)! = m^k

where n = 3, m =2 and k = 3

All integers and k1.