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

[Alexander Povolotsky]

It also appears that all integers i,
 which are generated as

i = n! + prime(n)

are all co-primes against each other ?

Thanks,
Regards,
Alexander R. Povolotsky
==========================================
On Tue, Aug 12, 2008 at 5:28 AM, berend daniel <berend at cs.bgu.ac.il> wrote:
> I certainly believe that's the case, but this doesn't seem to follow from
> the same considerations, as the function x\to x^k is sometimes onto.
>
> Best,
>
> Dani
>
> Alexander R. Povolotsky wrote:
>>
>> Does then NO "n" exist, such that
>>  ( n! + prime(n) )
>> yields integral m^k
>>  where k>1 ?
=================================================================
>> >From : berend daniel <berend at cs.bgu.ac.il>
>> To : David Harden <oddleehr at alum.mit.edu>, pevnev at juno.com
>> 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
>>
>> David Harden wrote:
************************************************************************************************
>> On Sun, Aug 10, 2008 at 10:01 AM, Alexander R. Povolotsky<apovolot at gmail.com
>> wrote:
>>
>> Hi,
>>
>> Does any "n" exist, such that ( n! + prime(n) ) yields integral square ?
>> I tried up to n=3000 to no avail (but may be my PARI program is faulty ;-)
>> ) .
>> I extended calculations to n<= 30,000 and still no result.
***********************************************************************************
>> 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