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

Alexander Povolotsky apovolot at gmail.com
Wed Aug 13 16:35:10 CEST 2008

The answer to my question is NO, they are not coprimes
- see PARI program below

gp > j=[]; for(n=1,70, j=concat(j,(prime(n)+n!)); for(k=1, n-1,
if(gcd(j[n],j[k]) >1,print1
("n="n" k="k" j[n]="j[n]" j[k]="j[k]" gcd(j[n],j[k])="gcd(j[n],j[k]) " \n"))))

n=8 k=7 j[n]=40339 j[k]=5057 gcd(j[n],j[k])=13

n=10 k=4 j[n]=3628829 j[k]=31 gcd(j[n],j[k])=31

n=13 k=10 j[n]=6227020841 j[k]=3628829 gcd(j[n],j[k])=19

n=33 k=16 j[n]=8683317618811886495518194401280000137
j[k]=20922789888053 gcd(j[n],j[k])=47

n=37 k=33 j[n]=13763753091226345046315979581580902400000157
j[k]=8683317618811886495518194401280000137 gcd(j[n],j[k])=41

n=43 k=10 j[n]=60415263063373835637355132068513997507264512000000191
j[k]=3628829 gcd(j[n],j[k])=101

n=47 k=44
j[n]=258623241511168180642964355153611979969197632389120000000211
j[k]=2658271574788448768043625811014615890319638528000000193
gcd(j[n],j[k])=181

n=53 k=5
j[n]=4274883284060025564298013753389399649690343788366813724672000000000241
j[k]=131
gcd(j[n],j[k])=131

n=59 k=58
j[n]=138683118545689835737939019720389406345902876772687432540821294940160000000000277
j[k]=2350561331282878571829474910515074683828862318181142924420699914240000000000271
gcd(j[n],j[k])=491

n=63 k=5
j[n]=1982608315404440064116146708361898137544773690227268628106279599612729753600000000000307
j[k]=131
gcd(j[n],j[k])=131

n=63 k=53
j[n]=1982608315404440064116146708361898137544773690227268628106279599612729753600000000000307
j[k]=4274883284060025564298013753389399649690343788366813724672000000000241
gcd(j[n],j[k])=131

n=67 k=51
j[n]=36471110918188685288249859096605464427167635314049524593701628500267962436943872000000000000331
j[k]=1551118753287382280224243016469303211063259720016986112000000000233
gcd(j[n],j[k])=1229

n=70 k=7
j[n]=11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000349
j[k]=5057
gcd(j[n],j[k])=389

Thanks,
Regards,
Alexander R. Povolotsky
==========================================
On 8/13/08, Alexander Povolotsky <apovolot at gmail.com> wrote:
> 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.
> >>
> >> 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
>