Character
sumsand products of factorials
modulo p
par MOUBARIZ Z. GARAEV et FLORIAN LUCA
RÉSUMÉ. Dans cet article, on utilise des estimations de sommes
de charactères pour étudier les
produits
de factorielles modulo p.ABSTRACT. In this paper, we
apply
character sum estimates tostudy products
of factorials modulo p.1. Introduction In
[8],
A.Sarkozy proved
thefollowing
Theorem:Theorem 1.1. Let p be a
prime number,
u, v,S,
T beintegers
with1 u, v C p - l, 1 T p. Furthermore,
letCl, C2,
... ,Cu
andDi, D2,
... ,Dv
beintegers
withFor any
integers
n, letf(n)
denote the numbersof
solutionsof
Our first result in this paper is a modification of the above Theorem 1.1. In what
follows,
we use r > 1 for apositive integer.
We also use theVinogradov symbols
« and » and the Landausymbols
0 and o with theirusual
meanings.
Theorem 1.2. Let p be a
prime nurrtber, u,
v,S,
T beintegers
with1 u, v p - l, 1 T p. Furthermore, let Cl, C2,
... ,Cu
andDl, D2,
... ,Dv
beintegers satisfying (1.1)
and(1.2)
above.If f (n)
isthe
function defined
at(1.3) above,
we then haveFor
example,
when p islarge
and T is close topl/4,
thenletting
firstE > 0 be any fixed small
positive
realnumber,
and thenchoosing
r suchthat
p1/4r2 logp ~~,
we see that theright
hand side of(1.5)
is of order ofmagnitude
which is better than theright
hand sideof
expression (1.4).
We also
provide
oneapplication
to our Theorem 1.2. In[7],
Luca andStanica put, for every
non-negative integers s > t
andprime
number p,and asked for
optimal
choices for theparameters t
and s versus p, such thatPs,t (p)
covers all the non-zero residue classes modulo p. Theorem 1 in[7]
asserts the
following:
Theorem 1.3. Let E > 0 be
arbitrar y.
There exists acomputable positive
constant such that whenever p >
po(e),
thenPs,t (p) 2 71p for
all tand s such that t >
p~
and s - t >We use our Theorem 1.2 to
improve
the above Theorem 1.3 as follows:Theorem 1.4. Let E > 0 be
arbitrary.
There exists acomputable positive
constant such that whenever p >
po(e),
thenPs,t (p) 2 ZP for
all tand s such that t >
p~
and s - t >More information on
products
of factorials modulo p can be found in the recent papers[1, 3, 4, 5].
Acknowledgements
We thank the anonymous refereee for useful com-ments. Work
by
both authors waspartially supported by
Grant SEP-CONACyT
37259-E.2. The Proofs
The
Proof of
Theorem 1.2. Theproof
is based on Karatsuba’s method from[6].
In whatfollows, x
stands for anarbitrary
character modulo p, and xostands for the
principal
character.Clearly,
where
and
Clearly,
while
By
theCauchy-Schwartz inequality,
we haveNote that
where in the above estimate
(2.7)
we usedE’
for the sumonly
over thosesummation terms which are not zero, as well as the fact that
and is zero
otherwise, together
with the fact thatCz fi Cx, (mod p)
ifx’ are in
{1, ... , u}. Inserting (2.7)
and itsanalogue
for the set ofnumbers
1 ~ u} replaced by
the set of numbers1
yv}
into
(2.6),
weget
Thus,
from(2.5)
and(2.8),
we haveand the fact that
,
is
just Burgess’s
character sum estimate from[2]. Inequality (1.5)
nowfollows from
(2.1), (2.4), (2.9)
and(2.10).
0The
Proof of
Theoreml.l~.
We follow the method ofproof
of Theorem1 in
[7].
The idea there was to find a suitable list ofpositive integers
xl, x2, ... , xt
consisting
of many smallnumbers,
and each one of themrepeated
a suitable number oftimes,
such that we canmodify
the elementin
enough
ways so that to ensure that we can obtain all the congruence classes in7l.;.
The basicoperation by
which we canmodify
the element F shown at(2.11)
isgiven by
(M)
Assume thatil i2
...ij
and11 12
~ ~ ~lj
are twodisjoints
subsets
of
indices in{I, 2,
...,t}. Then,
Using (2.12)
with xll = ... =xh
=1, eliminating
the initial numberF,
and
taking
inverses in(2.12) above,
it sufhces to prove thefollowing
claimClaim 2.1. For any
non-negative integers
s > tsatisfying
thehypothesis of
Theorem1.4,
there existpositive integers
Xl, X2, .. - , xtsumming
up to s, such that every nonzero residue class modulo p can berepresented by
a number
of
theform
where the subset
of
indicesf il, i2, ... , of f 1, 2,
... ,t~
in(2.13)
can beany subset such that there exists another subset
of j
indicesf 11, l2,
... ,lj )
disjoint from f il, i2,
... ,ij I for
which Xlr = 1for
all r =1, 2, ... , j .
As in the
proof
of Theorem 1 in7,
we fix E >0,
and apositive integer
k
2 .
From now on, allpositive
constants c c ..., whichk k P m 2
will appear will be
computable
and willdepend only
on 1. Since p is assumed to belarge,
we suppose that p > 13. We show that if p islarge enough
withrespect to k,
we can then construct agood
sublist of numbers xl, x2, ... , xt in thefollowing
manner:(1)
We first take andrepeat exactly
two times each of theprime
numbers(2)
We thenadjoin
some even numbers zj , each one of them smaller thanpl/4+l/k
but such that thetotality
of those(counted
withmultiplici-
ties)
does not exceedcl log log
p.(3)
The numbers of the form(2.13),
where thezj’s
are from the lists 1and 2 and the maximum
length j
of aproduct
in(2.13)
is not morethan 21~ +
2ci log log p
cover the entireIt is clear that if we can prove the existence of a list
satisfying
1-3above,
then we are done.
Indeed,
we may firstadjoin
at the sublistconsisting
ofthe numbers
appearing
at 1 and 2 above a number of about2~+2ci log log
p valuesof x2
all of themequal
to 1. Thetotality
of all these numbers(the
ones from
1,
2 and these new valuesof x2
allequal
to1)
counted with theirmultiplicities,
sofar,
iscertainly
not more thanwhile their sum is at most
for
large
p. At this step, we mayfinally complete
the above list with several other values ofthe xi equal
to 1 until weget
a list withprecisely
t -1
numbers,
which ispossible by inequality (2.14) above,
and set the lastnumber of the list to be
equal
towhich is still
positive by inequality (2.15)
above.To prove the existence of a sublist with
properties
1-3above,
weproceed
as in the
proof
of Theorem 1 in[7].
We start with the set(2.16)
A:=n
and n isprime}.
The numbers from A will form the sublist mentioned at 1 above
but,
sofar,
we take each one of themexactly
once. Let,a., ...’ .,
We first notice that each value of n E A appears at most k times in an
arbitrary product
inBl.
We now show thatbl
:_#Bl
islarge. Indeed,
the set
Bl
willcertainly
contain all the numbers of the formwhere pi is an
arbitrary prime subject
to the conditionI I
Notice that the residue classes modulo p of the elements of the form
(2.18),
where the
primes
Pisatisfy
conditions(2.19),
are all distinct.Indeed,
thepoint
is that if two of the numbers of the form(2.18)
coincide modulop,
then,
aftercancelling
the factor of2-k,
weget
two residue classes ofintegers
which coincide modulo p. Now each one of these twointegers
issmaller than p, therefore if
they
coincide modulo p, thenthey
mustbe,
infact, equal.
Now the fact thatthey
are all distinct follows from the fact that theirprime
divisors pisatisfy
condition(2.19). Applying
the PrimeNumber Theorem to estimate from below the number of
primes
in each oneof the intervals
appearing
in formula(2.19),
weget
whenever p
> c5. We now constructrecursively
a(finite) increasing
se-quence of subsets for m > 1 in the
following
way:Assume that has been constructed and set
bm
:=#Bm.
Assumethat
bm
p - 1(that is, Bm
is not the entire7l.; already).
We then have thefollowing trichotomy:
(i)
Ifbm
>p/2,
we then set :-Bm . Bm,
and notice that =71p
and we can nolonger
continue.(ii)
Ifbm p/2
and there exists an even number apl/4+1/k
such thata/2 0 Bm -
we then set am := a, add a to the list of thex2’s (as
one of the numbers from sublist 2
above),
and we letNotice that
(iii)
Ifp/2
and all even numbers a up to have theproperty
that
a/2
isalready
inBml,
we choose the even number a smallerthan
pl/4+1/k
for which the number ofrepresentations
ofa/2
of theform x .
y-1 with x,
y EBm
is minimal. We then set cm := a, adda to the list of the
xi’s (as
one of the numbers from sublist 2above),
set
and notice that
In
(i)-(iii)
above we have used the set-theoreticnotation, namely
that ifU and V are two subsets of we have denoted
by
U ~ V the set of allelements of
7l.;
of the form u ~ v with u E U and v EV,
andby U-1
the setof all elements of the form
u-1
for u E U.We have to
justify
that(i)-(iii)
above do indeed hold. Notice that(i)
and
(ii)
are obvious. Theonly
detail we have tojustify
is thatinequality
indeed holds in situation
(iii).
But forthis,
weapply
our Theorem 1.2above with u = v =
be, Cl, C2,
... ,Cu
all the residue classes in andDi, D2,
... , 7Du
all the residue classes in We also set S = 0 and T to be thelargest integer
smaller thanp1~4+1~~/2. Clearly,
T >pl/4+l/k /3.
Since we are
discussing
situation(iii) above,
wecertainly
havef (n) >
1 forall
positive integers n
up to T. Let M :=1 ~ ~ ~ T},
and thenam :=
2c,
wheref (c)
= M. Denotebm by
b. Weapply inequality (1.5)
with r := k to
get
Let c6 :=
c6(k)
be the constantimplied
inOk
in(2.25)
above. We show that theinequality
holds for p
large enough. Indeed,
sinceinequality (2.20)),
it follows that in order for(2.26)
tohold,
it suffices thatwhich is
certainly
satisfied when p > C7.Thus, inequalities (2.25)
and(2.26)
show thatinequality
holds,
where the lastinequalities
in(2.28)
follow because bp/2
andp > 13. In
particular,
which proves
inequality (2.24).
The combination of
(2.22), (2.23)
and(2.24)
shows thatholds as
long
asb", p/2.
Now notice that theinequality
will
happen provided
that m >cg log log p,
where one can take Cs :=k + 1
log(4/3)’
forexample,
and for suchlarge
minequality (2.30)
shows thatlog(4/3)
>
p/2.
Inparticular,
situations(ii)
or(iii)
above will not occur formore than
c8loglogp steps
after which we arrive at apoint
where we ap-ply
situation(i)
to construct and we are done.Clearly, (i)-(iii)
andthe above
arguments
prove the existence of a sublist of thexi’s satisfying
conditions
(1)-(3),
which finishes theproof
of Theorem 1.4.References
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Moubariz Z. GARAEV Instituto de Matemdticas
Universidad Nacional Aut6noma de Mexico C.P. 58180, Morelia, Michoacan, Mexico E-mail : garaev0matmor.unam.mx
Florian LUCA
Instituto de Matematicas
Universidad Nacional Aut6noma de Mexico C.P. 58180, Morelia, Michoacdn, Mexico
E-mail : fluca(Omatmor.unam.mx