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 k>1
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.exe?A1=ind0809&L=nmbrthry&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 <apovo...@...>
Date: Aug 28, 2008 8:14 PM
Subject: remark from Noam Elkies
To: Florian Luca <fl...@...>

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 c>0 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 <fl...@...>
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 m>n),
> 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" <apovo...@...>
> To: fl...@...
> Sent: Thu, 28 Aug 2008 05:00:47 -0400
> Subject: conjecture: 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 <i...@...>
>> Date: Wed, Aug 27, 2008 at 10:10 PM
>> Subject: Re: n! + prime(n) != m^k
>> To: Alexander Povolotsky <apovo...@...>

>> 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 fl...@....

>> 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 k>1 ?
>> > 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
>>
***************************************************************************­****\
*
>> >From : berend daniel <ber...@...>
>> To : David Harden <oddle...@...>, pev...@...,
>> Alexander Povolotsky <apovo...@...>
>> Subject : Re:
>> 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 <ber...@...>
>> Date: Aug 12, 2008 5:28 AM
>> Subject: Re: Does then NO "n" exist, such that ( n! + prime(n) )
>> yields integral m^k where k>1 ?

>> --------- Forwarded message ----------
>> From: berend daniel <ber...@...>
>> Date: Aug 12, 2008 5:28 AM
>> Subject: Re: Does then NO "n" exist, such that ( n! + prime(n) )
>> yields integral m^k where k>1 ?
>> To: Alexander Povolotsky <apovo...@...>
>> Cc: David Harden <oddle...@...>

>> Alexander Povolotsky wrote:
>> > Does then NO "n" exist, such that
>> > ( n! + prime(n) )
>> > yields integral m^k
>> > where k>1 ?
>> > 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