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

[Alexander Povolotsky]

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 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


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