#A82 INTEGERS 16 (2016) ON A CUBIC MOMENT FOR SUMS OF HECKE EIGENVALUES

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1 #A IEGERS 6 (06) O A CUBIC MOME FOR SUMS OF HECKE EIGEVALUES Khadija Mbarki Department of Mathematics, University of Monastir, Faculty of Sciences, Monastir, unisia mbarkikhadija@yahoo.fr Received: /6/5, Accepted: //6, Published: /3/6 Abstract Let f (n) be the n-th nmalized Fourier coe cient of the Fourier series associated with a holomphic cusp fm f f the full modular group of even weight k and let A f (x) := f (n). During the ELAZ 0 conference in Hildesheim, Germany, napplex K.-L. Kong (University of Hong Kong) presented his result, proved in his Master thesis, that Z (t) ( t) = C( ) 7/ + O " 7/ f some explicit > 0, C( ), where > 0 is fixed and (x) is the err term in the Dirichlet divis problem. A problem posed by Profess Ivić at this conference was to obtain a fmula analogous to the above fmula f the sum A f (x) and especially to discuss the sign of C( ) in the new setting. In this paper, we will solve Ivić s problem and prove that f any " > 0, we have Z A f (t)a f ( t) = C f ( ) 7/ + O," +", f some constant C f ( ) depending on only f, and defined by C f ( ) = / 3 + (i 0,i ){0,} n,m,l= p n+( ) i p p 0 m+( ) i l=0, (nml) 3/, where > 0 is a fixed constant. Our result is new and throws light on the behavi of the classical function A f (x).. Introduction Let k be an even integer and Hk be the set of all primitive cusp fms of weight k f the full modular group SL (Z). If f Hk, then it has the following Fourier expansion at the cusp : f(z) = + n= f (n)n (k )/ e inz (=(z) > 0).

2 IEGERS: 6 (06) By the they of Hecke operats, property: f (n) f (m) = f (n) is real and satisfies the multiplicative d (n,m) f mn d f all integers m and n. Besides, it is also known that f (n) satisfies the deep inequality: f (n) apple d(n) () f all n, where d(n) denotes the number of positive diviss of n (this is the Ramanujan-Petersson conjecture proved by Deligne (see [], [])). he sum of nmalized Fourier coe cients over natural numbers occurs in the study of many imptant problems in number they, such as the number of Hecke eigenvalues of same signs (see [5]). he sequence ( f (n)) is of great arithmetic interest. he generating series L(f, s) = + n= f (n) n s has many analytic properties which o er tools to help achieve this goal. Meover, Hecke (see [3], []) proved that L(f, s) is an entire function that satisfies the following functional equation, which is a special case of the Langlands functiality: where L(f, s) = ( ) k/ (s)l(f, s) if <(s) >, (3) (s) = ( s) Y j= s + applej s + applej and the parameters in the product are apple = k and apple = k+. One of the basic goals of number they is the foundation of asymptotic fmula, as accurate as possible, f the sum A f (x). Rankin (see [7]) showed that f any " > 0 and x, A f (x) f x 3 (log x) +", where = Wu (see [, hereom.]) got a better bound f A f (x), that is where = 0.5. A f (x) f x 3 (log x), In this paper, we shall evaluate the integral R A f (x)a f ( x)dx basing on the truncated Vonoi fmula f the sum A f (x) and we will investigate the sign of the constant C f ( ). he main result of this paper is the following. ()

3 IEGERS: 6 (06) 3 heem. Let f Hk and 0 < Q[p ], where > is a square-free natural number. If there exist integers t, r, s, a 0, b 0, d and d such that t, t odd r odd, t >< s even, gcd(r, s) = b 0 tr + t s a 0 () rs gcd(d, d ) = >: = h d a 0 b0 d tr + t s ± rs p i t, odd r, s odd, gcd(r, s) = >< b 0 tr + t s a 0 rs gcd(d, d ) h= >: = d a 0 b0 d tr + t s ± rs p i, then there exists a constant C f ( ) depending on only and f and defined by (5) C f ( ) = / 3 + (i 0,i ){0,} n,m,l= p n+( ) i p p 0 m+( ) i l=0 (nml) 3/ (6) such that f any " > 0, we have Z A f (t)a f ( t) = C f ( ) 7/ + O," +". Our main tool in the proof of heem is the Vonoi summation fmula. Lau and Wu (see [5, Lemma 3.]) established the truncated Vonoi fmula f A f (x). hey obtained the following result. Lemma. Let f Hk. hen f any A > 0 and " > 0, unifmly f apple M apple xa and x, we have A f (x) = x/ p napplem f (n) n 3/ cos p nx where the implied O-constant depends on A, " and k only. We will use (7) and write x / +O A,",k x " + + x / M A f (x) = B M (x) + O A,",k (x " ) + O A,",k x /+" M /, (), (7)

4 IEGERS: 6 (06) where B M (x) = x/ p napplem ow, we will study the sign of the coe f (n) n 3/ cos p nx. cient C f ( ) in the following theem. heem. Let k be an even integer and let f H k. Set S + := S := so that both S + and S A ± := {a ; f (a)? 0}, + r s l 00 A + + r s l 00 A are positive.! f r s l 00 f (l 00 ), and l 009/! f r s l 00 f (l 00 ), l 009/. Suppose that satisfies (). If d = a 0 b 0 d =, then 9 < C f ( ) > 0 if S + > S = C f ( ) < 0 if S + < S : ;. C f ( ) = 0 if S + = S. Suppose that satisfies (5). If d = and a 0 b 0 d =, then 9 < C f ( ) > 0 if S + > S = C f ( ) < 0 if S + < S : ;. C f ( ) = 0 if S + = S In particular, examples when C f ( ) = 0 are given in the following assertion. 3. If Q + p Q +, then C f ( ) = 0.. Some Lemmas he proof of heem is based on the following lemmas. Lemma. Let be a positive number in Q[ p ], where is a square-free positive integer and let i 0, i {0, }. We put := (i 0, i ) := p n + ( ) i0p m + ( ) i p l, (9)

5 IEGERS: 6 (06) 5 and c (, y) := c ( ) := >: (i 0,i ){0,} (i 0,i ){0,} nappley,mappley lappley =0 + n,m,l= =0 (nml) 3/ (0) (nml) 3/. () hen, we have the following assertions. (a) he equation = 0 is solvable f n, m, l if and only if there exist integers t, r, s, d, d, a 0 and b 0 such that t, t odd r odd, t >< s even, gcd(r, s) = b 0 tr + t s a 0 rs gcd(d, d ) = >< >: = d d a 0 b0 h tr + t s t, odd r, s odd, gcd(r, s) = b 0 tr + t s a 0 rs gcd(d, d ) = = d d a 0 b0 ± rs p i h tr + t s ± rs p (b) On the hypotheses of () and (5), the series c ( ) converges. (c) c (, y) c ( )," y 5/+". Proof. (a) Let = a a + b b p, where a, b Z, a, b and gcd(a, a ) = gcd(b, b ) = and let d i = gcd(a i, b i ). So, we have a i = d i a 0 i and b i = d i b 0 i, i {, }. ote that at most one of the equations (0, ) = 0 and (, ) = 0 is solvable in n, m and l. If the equation (0, ) = 0 is solvable f n, m, l, then by squaring both sides, we obtain a b (n + m) b a l = a b l p a b p nm. () By using the same procedure, we get a b b l p nm = (a b l) + (a b ) nm (a b (n + m) b a l). i.

6 IEGERS: 6 (06) 6 his shows that p nm is necessarily an integer. Also, it is easy to show that p mn is an integer. Hence, Fmula () leads to a b (n + m) b a l = p a b l r mn a b. Since is a square-free integer and p mn is an integer, then we have p is an irrational and p p a b l a b mn is an irrational zero. So, we get a (n + m) = a l and b l = b r mn. (3) his implies that a > 0 and b > 0. Meover, we know that a divides a l and b divides b l. Since gcd(a, a ) = gcd(b, b ) =, it follows that a and b divide l. herefe, there exists a positive integer l 0 such that l = a l 0 and b divides a l 0. hen l 0 = b d l, where l 00 = l a 0 b By considering (3) and (), we get. hus (n + m) = a b 0 l 00 = a 0 d b 0 l 00 and which is equivalent to l = a l 0 = a b d l 00 = a 0 b 0 d l 00. () r mn = b a 0 l 00 = a0 b 0 d l 00, (n + m) = a b 0 l 00 = a 0 d b 0 l 00 and mn = (b0 d a 0 l 00 ). By solving the last two equations f n, we obtain a 0 b 0 ± p (a 0 b0 ) (a 0 b0 ) n = d l 00 Since gcd(a 0, a 0 ) = gcd(b 0, b 0 ) =, we get gcd(a 0 b 0, a 0 b 0 ) =. otice that n is well-defined i.e the equation q (a 0 b0 ) (a 0 b0 ) = e. has solutions in 3 provided we have the following cases: If is odd, there exist integers t, r and s such that >< >: t, t odd r odd, s even, gcd(r, s) = a 0 b 0 = tr + t s a 0 b 0 = rs (5)

7 IEGERS: 6 (06) 7 > < However, if is even, then < : >: t, t odd r, s odd gcd(r, s) = a 0 b 0 = tr + t s a 0 b 0 = rs. t, r odd gcd(r, s) = a 0 b 0 = tr + t s a 0 b 0 = rs. (6) (7) ow, if (5) is valid, then n = d l 00 tr + t s ± (tr t s ). So, we have n = d l 00 tr and m = d l 00 t s, n = d l 00 t s and m = d l 00 tr. herefe b = d b 0 rs = d a and a 0 = d a 0 (tr = d + t s ) b. It follows that 0 = d (tr + t s ) b 0 d a 0 If (6) is valid, then + d rs apple a p 0 d = d b 0 d a 0 tr + b0 t s + rs p. n = d l 00 tr + t s ± tr t s. Assuming that l 00 and d are not both odd, then this implies that n = dl00 tr and that m = dl00 t s, n = dl00 t s and m = dl00 tr rs. hus b = d a 0 and a = d(tr + t s ) b. his leads to 0 = d tr + t s b 0 d a 0 + d rs apple a p 0 d = tr d b 0 d a 0 + t b0 + rs p. If (7) is true, we will obtain the same values of n and m as in (5), and therefe = d apple d a 0 tr + b0 t s + rs p. Subsequently, the numbers n, m and l are well-defined so that the equation (0, ) = 0 is solvable f n, m and l.

8 IEGERS: 6 (06) ow, if the equation (, ) = 0 is solvable then, by following the afementioned steps, we get a (n + m) = a l and b r mn = b l. Let b = b > 0. We replace b by b and we will obtain = d apple d a 0 tr + b0 t s rs p = d apple tr d a 0 + t b0 s rs p. (b) ow, assume that satisfies (); then < n = d l 00 tr m = d : l 00 t s l = a 0 b 0 d l 00 < : n = d l 00 t s m = d l 00 tr l = a 0 b 0 d l 00. () his is valid provided that Fmula () implies that = d d a 0 b0 apple tr + t s + rs p = d apple d a 0 tr + b0 t s rs p. c ( ) + f (d l 00 t s ) f (d l 00 tr ) f (a 0 b 0 d l 00 ) (d s r a 0 b0 d l 003 ) 3/ f (d t s ) f (d tr ) f (a 0 b 0 d ) (d s r a 0 b0 d ) 3/ + f (l 00 ) 3 l 009/. From (), we get f (l 00 ) " l 00" f any " > 0. hus, we have c ( ) " and therefe, the series c ( ) converges. If satisfies (5), then < n = (d tr )l 00 m = : (d t s )l 00 l = a 0 b 0 d l 00 < : n = (d t s )l 00 m = (d tr )l 00 l = a 0 b 0 d l 00, (9)

9 IEGERS: 6 (06) 9 this is valid provided that Fmula (9) implies that = d d a 0 b0 apple tr + t s + rs p = d apple tr d a 0 + t b0 s rs p. c ( ) + + f (d tr )l 00 f (d t s )l 00 f (a 0 b 0 d l 00 ) (d s r a 0 b0 d l 003 ) 3/ f d tr l 00 f d t s l 00 f (a 0 b 0 d (l 00 )) (d s r a 0 b0 d (l) 3 ) 3/ f (d t s ) f (d tr ) f (a 0 b 0 d ) (d s r a 0 b0 d ) 3/ 9/ + f (l 00 ) 3 l 009/ ". Hence, by Fmula (), the series c ( ) converges. (c) Recall that (see (0)) c (, y) = (i 0,i ){0,} nappley,mappley lappley =0 (nml) 3/. ote that if does not satisfy () and (5), then, based on the findings of (a), the equation = 0 is not solvable in n, m and l. hus, we obtain (i 0,i ){0,} + n,m,l= =0 (nml) 3/ = (i 0,i ){0,} nappley,mappley lappley =0 (nml) 3/ = 0, which is equivalent to c ( ) = c (, y) = 0. Suppose now that satisfies () (5).

10 IEGERS: 6 (06) 0 F instance, if satisfies (); then we have c ( ) c (, y) = = (i 0,i ){0,} (i 0,i ){0,} d l 00 tr >y d l 00 t s >y a 0 b0 dl00 >y + n>y m>y l>y =0 + n>y m>y l>y =0 (nml) 3/ (nml) 3/ f (d l 00 tr ) f (d l 00 t s ) f (a 0 b 0 d l 00 ) (d r s a 0 b0 d l 003 ) 3/ f (d tr ) f (d t s ) f (a 0 b 0 d ) (d r s a 0 b0 d ) 3/ d l 00 tr >y d l 00 t s >y a 0 b0 dl00 >y f (l 00 ) 3 l 009/. By using () and since we have d l 00 tr > y d l 00 t s > y a 0 b 0 d l 00 > y, then the last fmula implies that he other case is similar. c ( ) c (, y)," y 5/+". Lemma 3. Let be a positive number in Q[ p ], where > is a square-free integer. Let U i and V i f i =,, 3, be positive real numbers such that U i < V i b and at least two of the U i s are a f some positive real numbers a and b with a apple b. hen we have (a) U <napplev,u <mapplev, U 3<lappleV 3, 6=0 (nml) 3/ Z p t 3/ cos t," 5/+b/+" + 7/ a/+" and

11 IEGERS: 6 (06) (b) napplev,mapplev,lapplev 3 6=0 (nml) 3/ Z p t 3/ cos t," 5/+b/+", where is defined by (9). Proof. (a) ote that (0, 0) ( nml) /6. Also, by using the fact that f any t [, ], we have t 3/ 3/ and that p cos t 0 p, we get Z p t 3/ cos t 5/. (0) hus, U <napplev,u <mapplev, U 3<lappleV 3, 6=0 (nml) 3/," 5/," 5/ Z U <napplev,u <mapplev,u 3<lappleV 3 6=0 U <napplev,u <mapplev,u 3<lappleV 3 6=0 p t 3/ cos t (nml) 3/ (nml) /. ow, since U i < V i b f all i =,, 3, it follows that n, m and l are b and hence, by Fmula (), we obtain: f (n), f (m) and f (l) are " b+". herefe, 5/ U <napplev,u <mapplev,u 3<lappleV 3 6=0 (nml) / 5/+b/+". Suppose now that (i 0, i ) = (0, ) as the other case is similar. here exist integers, 0, M, M 0, L and L 0 such that U < < n apple 0 apple min(v, ), U < M < m apple M 0 apple min(v, M) and U 3 < L < l apple L 0 apple min(v 3, L). Let D = max(, M, L) and d = min(, M, L). F p D, we have {(n, m, l) : n, m M, l L, 0 < < }

12 IEGERS: 6 (06) So, we obtain U <napplev,u <mapplev,u 3<lappleV 3 6=0," D / ML + (nml) 3/ / L D "p ML. () d Z p t 3/ cos t <napple 0,M<mappleM 0, L<lappleL 0, 6=0 0 <napple 0,M<mappleM 0, L<lappleL 0, 0 " (nml) 3/ + <napple 0,M<mappleM 0, L<lappleL 0, / apple < 0 Z + p t 3/ cos t <napple 0,M<mappleM 0, L<lappleL 0,0< < / C A C + C + C 3. " (nml) 3/ Z p t 3/ cos t Since (0, ) < p D and by using Fmulas (0) and () with 0 apple < p D, we get C," 5/+b/+". Meover, we have L max(, M) D and d min(, M), in case (0, ) is between and /0. By using the same equations with / apple < 0 in Fmula (), we obtain C," 5/+b/+" + 7/ a/+". ow, by trivial estimation and by Fmula (), we have C 3," 5/+b/+" + 7/ a/+". Consequently, we get U <napplev,u <mapplev,u 3<lappleV 3 6=0 (nml) 3/ Z," 5/+b/+" + 7/ a/+". p t 3/ cos t

13 IEGERS: 6 (06) 3 (b) We use the same arguments as above f (i 0, i ) = (0, 0), and we get napplev,mapplev,lapplev 3 (0,0)6=0 (nml) 3/ F (i 0, i ) = (0, ), we have napplev,mapplev,lapplev 3 6=0 0 Z t 3/ cos (nml) 3/ <napple 0,M<mappleM 0,L<lappleL 0 0 Z " (nml) 3/ C 0 + C 0. + p t Z," 5 + b +". p t 3/ cos t <napple 0,M<mappleM 0,L<lappleL 0 < 0 p t 3/ cos t By the same token, we get C 0," 5/+b/+". Since using Fmula () with D 3 d / apple, we obtain herefe, we have napplev,mapplev,lapplev 3 6=0 (nml) 3/ < 0 C 0," 5/+b/+". Z t 3/ cos p t C A D 3 d / and by," 5 + b +". 3. Proof of heem In this section, we will use the results of the previous lemmas to prove heem. Let M =. From Fmula (), we have Z Z Z! A f (t)a f ( t) = (B M (t)) B M ( t) + O A,",k " B M (t) A f ( t) = Z! Z! +O A,",k " A f ( t) + O A,",k " A f (t) Z (B M (t)) B M ( t) + O,A,",k 3/+".

14 IEGERS: 6 (06) he last identity is due to Cauchy-Schwarz s inequality and the well-known meansquare results (see [6, heem.]) Z A f (t) 3, where f(t) g(t) means that f(t) g(t) and g(t) f(t). his can be established also by using (). Let B M,y (t) = t/ p and let M 0 = /6. We have hus, we get Z (B M (t)) B M ( t) = M<nappley f (n) n 3/ cos p nx B M (t) = B M0 (t) + B M0,M(t). Z (B M0 (t)) B M0 ( t) + Z Z Z Z Z B M0 (t)b M0,M(t)B M0 ( t) B M0 (t)b M0,M(t)B M0,M( t) (B M0,M(t)) B M0 ( t) (B M0,M(t)) B M0,M( t) = S + S + S 3 + S + S 5 + S 6. (B M0 (t)) B M0,M( t) he fmula cos(a) cos(b) cos(c) = [cos(a + b + c) + cos(a + b c) + cos(a b + c) + cos(a b c)] implies that cos p nt cos = p mt (i 0,i ){0,} cos cos p lt p t, ()

15 IEGERS: 6 (06) 5 where is defined by (9). By using Fmula (), the sum S is written as the following: S = / 3p + / 3p It follows that S / 3p + / 3p (i 0,i ){0,} (i 0,i ){0,} (i 0,i ){0,} napplem 0,mappleM 0 lapplem 0 =0 napplem 0,mappleM 0 lapplem 0 6=0 (i 0,i ){0,} napplem 0 mapplem 0 lapplem 0 6=0 By using Lemma 3, we get (i 0,i ){0,} napplem 0 mapplem 0 lapplem 0 6=0 napplem 0,mappleM 0 lapplem 0 =0 (nml) 3/ (nml) 3/ (nml) 3/ Z Z (nml) 3/ (nml) 3/ Z Z p t 3/ cos t t 3/ cos p t 3/ cos t Z t 3/ cos p t 3/ cos t..," /+". Recall that based on the hypotheses of () and (5), the series c ( ) converges absolutely. hen, we obtain / 3p (i 0,i ){0,} Lemma (c) leads to napplem 0,mappleM 0 lapplem 0 =0 (nml) 3/ Z t 3/ cos = / 3 c (, M 0 ) ( ) 7/ 7/ = / 3 [c ( ) + c (, M 0 ) c ( )] ( ) 7/ 7/ = / 3 c ( ) ( ) 7/ 7/ + O," 7/ c ( ) c (, M 0 ). c ( ) c (, M 0 )," M 5/+" 0," 5/+". Hence, we get / 3 c ( ) ( ) 7/ 7/ + O," 7/ c ( ) c (, M 0 )

16 IEGERS: 6 (06) 6 and therefe, F the sum S, we have = / 3 c ( ) ( ) 7/ 7/ + O," 37/+" S = / 3 c ( ) ( ) 7/ 7/ + O," 37/+". S = / ( p ) 3 Z t 3/ cos Fmula () implies that S = / 3p napplem 0 mapplem 0 M 0<lappleM p nt cos (i 0,i ){0,} napplem 0 mapplem 0 M 0 <lapplem Z (nml) 3/ p mt (nml) 3/ cos p lt p t 3/ cos t. / 3p (i 0,i ){0,} napplem 0 mapplem 0 M 0 <lapplem (nml) 3/ Z p t 3/ cos t / 3p (i 0,i ){0,} napplem 0 mapplem 0 M 0 <lapplem =0 (nml) 3/ Z t 3/ cos + / 3p (i 0,i ){0,} napplem 0 mapplem 0 M 0 <lapplem 6=0 (nml) 3/ Z p t 3/ cos t.

17 IEGERS: 6 (06) 7 By virtue of Lemma (c), we obtain (i 0,i ){0,} napplem 0,mappleM 0 M 0<lappleM =0 (nml) 3/ Z t 3/ cos," 7/ (i 0,i ){0,} + n,m,l= =0 (nml) 3/ (i 0,i ){0,}," 7/ c ( ) c (, M 0 )," 7/ c ( ) c (, M 0 )," 7/ M 5/+" 0," 37/+". napplem 0,mappleM 0,lappleM 0 =0 (nml) 3/ herefe, we get (i 0,i ){0,} napplem 0,mappleM 0 M 0<lappleM =0 (nml) 3/ Z t 3/ cos," 37/+". (3) In addition, we have / 3p = / 3p + / 3p (i 0,i ){0,} napplem 0 mapplem 0 M 0 <lapplem 6=0 (nml) 3/ (i 0,i ){0,} napplem 0,mappleM 0 M 0 <lapple(50m 0 / ) 6=0 (i 0,i ){0,} napplem 0,mappleM 0 (50M 0 / )<lapplem 6=0 Z (nml) 3/ (nml) 3/ p t 3/ cos t Z Z p t 3/ cos t p t 3/ cos t.

18 IEGERS: 6 (06) ow, by applying Lemma 3 (b), we obtain (i 0,i ){0,} napplem 0,mappleM 0 M 0 <lapple(50m 0 / ) 6=0 (nml) 3/," /+". Z p t 3/ cos t If (50M 0 / ) < l, then (i 0,i ){0,} napplem 0,mappleM 0 (50M 0 / )<lapplem 6=0 p l. Hence, by Lemma 3 (a), we get (nml) 3/ Z p t 3/ cos t," /+". It follows that S," /+". By using the same arguments as above, we obtain S 3," /+". F the sum S, we have S / 3p + / 3p (i 0,i ){0,} (i 0,i ){0,} napplem 0,mappleM 0 M 0 <lapplem =0 (nml) 3/ (nml) 3/ napplem 0 mapplem 0 M 0 <lapplem 6=0 Z Z t 3/ cos t 3/ cos p t. Lemma 3 (a) implies that (i 0,i ){0,} napplem 0,mappleM 0 M 0 <lapplem 6=0 (nml) 3/," 3/+" 7/ /+" +," /+". Z p t 3/ cos t hus, by Fmula (3) and the last bound, we get S," /+".

19 IEGERS: 6 (06) 9 By the same arguments applied on the sums S 5 and S 6, we deduce that S 5," /+" and S 6," /+". By summing up S to S 6 above, we obtain Z (B M (t)) B M ( t) = / 3 c ( ) ( ) 7/ 7/ + O," /+". herefe, we get the following fmula Z A f (t)a f ( t) = / 3 c ( ) ( ) 7/ 7/ + O," /+". Put = /, /,... After summing up, we find Let We finally get Z Z A f (t)a f ( t) = / 3 c ( ) 7/ + O," /+". C f ( ) := / 3 c ( ). A f (t)a f ( t) = C f ( ) 7/ + O," /+". Remark. If =, then we find the cubic moment f A f (x) : Z A 3 f (t) = O " /+". We just have to prove that C f () = 0 c () = 0. A simple way to get it is the use of the assertion 3 of heem with = Q +. Generally, if Q + p Q +, we have the same upper bound which means Z A f (t)a f ( t) = O," /+" and this is also a consequence of the assertion 3 of heem.. Sign of the Constant C f ( ) In this section, we are interested in the discussion of the sign of the constant C f ( ). Since C f ( ) = / c 3 ( ), it su ces then to discuss the sign of c ( ). F that, we shall prove heem.

20 IEGERS: 6 (06) 0 Proof.. If satisfies (), then by using the previous steps in the proof of Lemma, f all positive integers n, m and l such that p n ± p m = p l, we get < n = d tr l 00 < : m = d t s l 00 l = a 0 b 0 d l 00 : n = d t s l 00 m = d tr l 00 l = a 0 b 0 d l 00. In this case, we have = d apple tr d a 0 + t b0 s ± rs p. Suppose that d = d a 0 b 0 =. hen, becomes = tr + t s ± rs p. herefe, we obtain c ( ) = = + f (tr l 00 ) f t s l 00 f (l 00 ) (s r l 003 ) 3/ (s r ) 3/ + f (tr l 00 ) f t s l 00 f (l 00 ). l 009/ Accding to (), we have f (tr l 00 ) f t s l 00 = d gcd(tr l 00, t s l 00 ) f r s l 00 d. () But gcd(r, s) = gcd t, t =, since is a square-free integer. It follows that gcd tr, t s = and this implies that gcd tr l 00, t s l 00 = l 00. hus, Fmula () leads to f (tr l 00 ) f t s l 00 = f r s l00 d. (5) d l 00 By applying Fmula (), and since gcd r s l 00, l 00 d l 00 f r s l00 = f r s l 00 f (l 00 ). hus, Fmula () joined on to (5) results in f (tr l 00 ) f t s l 00 d = l 00, one can deduce that f (l 00 ) = f r s l 00 f (l 00 ). (6)

21 IEGERS: 6 (06) By using Fmula (6) and by the expression of c ( ) in our case, we have! + f r s l 00 f (l 00 ) c ( ) = (s r ) 3/ l 009/! + = 6 f r s l 00 f (l 00 ) (s r ) 3/ l 009/ r s l 00 A + = + (s r ) 3/ (S + S ). + r s l 00 A f r s l 00 f (l 00 ) l 009/! It follows that, if S + > S, then c ( ) > 0, and therefe, C f ( ) > 0. Besides, if S + < S, then c ( ) < 0, and therefe, C f ( ) < 0.. If satisfies (5), then we have < n = (d tr l 00 ) m = : d t s l 00 l = a 0 b 0 d l 00 < : n = d t s l 00 m = (d tr l 00 ) l = a 0 b 0 d l 00. ote that n and m are well-defined if l 00 and d are not both odd. Hence, we have = d apple tr d a 0 + t b0 s ± rs p. Suppose that d = and d a 0 b 0 =. So, the expression of becomes = tr + t s ± rs p and thus, c ( ) = (s r ) 3/ + f (tr l 00 ) f t s l 00 f (l 00 ). l 009/ By following the same procedure as in the proof of., we get: c ( ) = (s r ) 3/ (S + S ). We conclude also that, if S + > S, then C f ( ) > 0, and if not, then C f ( ) apple 0.

22 IEGERS: 6 (06) 3. If is a positive rational number and the equation p n + ( ) i 0 p m + ( ) i p l = 0, (i0, i ) {0, } is solvable in n, m and l, then by the result of Lemma (a), we have = d apple tr d a 0 + t b0 s ± rs p = d apple tr d a 0 + t b0 s ± rs p. p Since = a a + b b, then = a a, with gcd(a, a ) = and b = 0. It follows that gcd(b, b ) = gcd(0, b ) = b =. herefe, we obtain d = a > 0 and a 0 = b 0 = d =. hus, we get a = a 0 d =. Hence, we find that = a. So, we have tr + t s = tr + t s = rs = 0 rs = 0. ow, if satisfies (), then tr + t s =. his implies that tr = 0 t s = 0, and r = 0 s = 0. But, we have n = a tr l 00 m = a t s l 00 n = a t s l 00 m = a tr l 00. herefe, n = 0 m = 0. his contradicts the fact that n, m. If satisfies (5), then tr + t s = and therefe, tr = t s =. his leads to n = m = a l 00. Set l 00 = q to be even. hus, we have n = m = a q and l = l 00. he equation p n + ( ) i 0 p m + ( ) i p l = 0, (i0, i ) {0, } leads to p a q + ( ) i0p a q + ( ) i p a q = 0, (i 0, i ) {0, }. his is equivalent to p p i a q h + ( ) i0 + ( ) i = 0, (i 0, i ) {0, }. It is clear that f all (i 0, i ) {0, }, we have + ( ) i0 + ( ) i p 6= 0. Hence, the equation p n + ( ) i 0 p m + ( p ) i l = 0, (i0, i ) {0, }

23 IEGERS: 6 (06) 3 is not solvable in the case of = a = a a, which is a positive rational number, and therefe, c ( ) = 0 = C f ( ). ote that, in this case, we can also consider the fact that rs = 0 which implies that r = 0 s = 0, and since tr = t s =, we then get the contradiction. Suppose now that p Q + and that the equation p n + ( ) i 0 p m + ( ) i p l = 0, (i0, i ) {0, } is solvable in n, m and l. Lemma (a) yields = d d a 0 b0 apple tr + t s ± rs p = d apple tr d a 0 + t b0 s ± rs p. p p Since = a a + b b, then = b b. hus, a = 0, gcd(a, a ) = a =, d = bp > 0 and d = a 0 = b 0 =. Since d = gcd(a, b ) = b =, then = b. Hence, we obtain b tr + t s = 0 b tr + t s = 0 b rs = b b rs = b. In both cases, we have tr + t s = 0, and this is equivalent to tr = t s = 0. If satisfies (), then n = b tr l 00 m = b t s l 00 and if satisfies (5), then n = b tr l 00 m = b t s l 00 n = b t s l 00 m = b tr l 00 n = b t s l 00 m = b tr l 00. herefe, in both cases, n = m = 0. Meover, the equation p n + ( ) i 0 p m + ( ) i p l = 0, (i0, i ) {0, } is solvable, which implies that n = m = l = 0. his contradicts the fact that n, m and l. Hence, we get c ( ) = 0 = C f ( ). ote that we can show the contradiction when satisfies () by using the equality b rs = b, which is equivalent to rs = and this can not happen f positive integers r and s.

24 IEGERS: 6 (06) Acknowledgements. he auth would like to thank the anonymous referee s careful reading and comments which improved the exposition of the paper. hanks are also sent to Profess Alexander Ivić f many suggestions and f his valuable remarks that contributed to the improvement of this wk. References [] P. Deligne, La conjecture de Weil I, Publ. Math. Inst. Hautes Études Sci. (97), [] P. Deligne, La conjecture de Weil II, Publ. Math. Inst. Hautes Études Sci. 5 (9), 33-. [3] E. Hecke, Über modulfunktionen und die Dirichletscher reihen mit Eulercher produktentwicklung I, Math. Ann. () (937), -. [] E. Hecke, Über modulfunktionen und die Dirichletscher reihen mit Eulercher produktentwicklung II, Math. Ann. () (937), [5] Y.-K. Lau and J. Wu, he number of Hecke eigenvalues of same signs, Math. Z. 63 (009), [6] H. Lao, Mean square estimates f coe cients of symmetric power L functions, Acta Appl. Math. 0 (00), [7] R. A. Rankin, Sums of cusp fm coe cients, automphic fms and analytic number they (Montreal, PQ, 99), Univ. Montréal, Montreal, QC (990), 5-. [] J. Wu, Power sums of Hecke eigenvalues and application, Acta Arith. 37 (009),