MODULAR HYPERBOLAS AND THE CONGRUENCE ax 1 x 2 x k + bx k+1 x k+2 x 2k c (mod m)

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1 #A37 INTEGERS 8 (208) MODULAR HYPERBOLAS AND THE CONGRUENCE ax x 2 x k + bx k+ x k+2 x 2k c (od ) Anwar Ayyad Departent of Matheatics, Al Azhar University, Gaza Strip, Palestine anwarayyad@yahoo.co Todd Cochrane Departent of Matheatics, Kansas State University, Manhattan, Kansas cochrane@ath.ksu.edu Sanying Shi School of Matheatics, Hefei University of Technology, Hefei, P.R. China vera23 99@hotail.co Received: 6/2/7, Accepted: 2/20/8, Published: 4/20/8 Abstract For any cube-free integer, and integers a, b, c with (abc, ) =, we obtain the existence of solutions of the congruence x x k c (od ), in any cube of edge length B 4 + p + 2(k+4), and of the congruence ax x 2 x k +bx k+ x k+2 x 2k c (od ), in any cube B of edge length B 4 + 2( p + k+.95). Refineents are given for sall k, and results are also given for arbitrary 2 N.. Introduction A k-diensional odular hyperbola is the set of solutions of a congruence x x 2 x k c (od ), where (c, ) =. Shparlinski [2] has written at length on the properties and applications of odular hyperbolas. Of particular interest is obtaining solutions with coordinates restricted to intervals of short length; see [], [2], [], [2] and [20], in addition to [2]. The first two authors [3] studied the related congruence ax x k + bx k+ x 2k c (od ), with a prie odulus and obtained a nuber of results on the distribution its solutions. In this work we extend the results of [] and [3] to general oduli,

2 INTEGERS: 8 (208) 2 addressing odular hyperbolas in the next three sections and the latter congruence in the reaining sections. The proof of our results on the odular hyperbola is a straightforward generalization and refineent of what was done in [] for the case of prie oduli and in [20] for the case of coposite oduli with intervals of the for [, B]. The ain novelty of this paper is the ethod of proof provided for our results on the second congruence. For the case of a prie odulus, the authors in [3] ade use of additive cobinatorics, in particular a result of Hart and Iosevich [3] on when we have a su-product relation A B +A 2 B 2 Z p. Here, no appeal is ade to additive cobinatorics, but rather a ore delicate evaluation of character sus is ade in order to obtain results of the sae strength as the od p results of [3] for a general odulus, for boxes in general position. 2. The Congruence x x k c (od ) For k, 2 N, and integers c with (c, ) =, we consider the congruence x x 2 x k c (od ). () with variables restricted to a general box B with sides of length B i, B = {(x,..., x k ) 2 Z k : h i + apple x i apple h i + B i, apple i apple k}; (2) for convenience we take h i, B i 2 Z, apple i apple k and assue apple B i <. If all of the B i are equal, say B i = B, apple i apple k, then we call B a cube with edge length B. Let Z = Z/ and Z be the group of units in Z. By identifying Z with an appropriate set of integer representatives we ay view B as a box of points in Z k. Let I i be the interval in Z given by and I i = [h i +, h i + B i ] \ Z, (3) B = I I 2 I k = B \ Z k. If all B i = B we will continue calling B a cube with edge length B, although it ay be the case that the cardinalities I i of the edges are not equal. Generalizing the work of the first author [] for prie oduli, and Shparlinski [20, Theore 9] for coposite oduli, we obtain the following result. Theore. Suppose that k 4, r 2 N. Let c be an integer with (c, ) =. Then the nuber n c of solutions of the congruence x x k c (od ), (4)

3 INTEGERS: 8 (208) 3 with x i 2 I i, apple i apple k, is given by n c = B + O B r 2 k + 4 r+ kr 4r 2 (k 4)+ 3r 4r2 (k 4), where = Q p e e 3 p e. The ter ay be reoved for r =, 2 or 3. Shparlinski proved the special case where each interval is of the for [, B i ] \ Z and r =, 2 or 3. The theore yields an asyptotic estiate for n c provided that B k 4 + k(k 4+2r 2 ) 4r(k 4+2r) + (3r )k(k 4) 4r(k+2r 4), (5) where the ter ay be dropped for r =, 2, or 3. In particular, using r = 2, 3 we see that n c > 0 for any and any cube B with edge length ( k +, for k = 4, 5, 6; B 3 + k+2 +, for k 7. Thus for a general odulus we can only get down to a threshold of 3 + for k su ciently large. In particular, this is the best we can do for the case where =. At the other extree, where =, we can reduce the exponent further to 4 + for k su ciently large. To be precise, if = then choosing the optial value of r (as shown in []), we obtain for k 4 that n c > 0 for any cube with For a general box B, the sae holds for B 4 + p + 2(k+4). (6) B k 4 + p k + 2(k+4). Next, let us exaine the cases where k = 2, 3 or 4. For k = 2 it is well known that n c > 0 for any and any cube B of edge length B, provided that B ; see for exaple [23], [5] or [2]. The proof akes use of the Kloosteran su estiate. For k = 4, the ter in (5) goes away altogether, and we get n c > 0 for any and box B with B 2+. (7) We deduce a result for k = 3 fro our k = 4 result by applying it to a box with I 4 = {}. In this case the inequality in (7) yields a solution in any box with

4 INTEGERS: 8 (208) 4 I I 2 I Thus, any cube of side length B contains a solution of () provided that ( 2 3 +, if k = 3; B 2 +, if k = 4. For k = 3 the sae estiate was given in [, Theore ] for prie oduli. weaker result for k = 3 was given for coposite in [20, Theore 8]. A 3. Estiating the Cardinality of an Interval in Z Before proceeding with the proof of Theore, let us reind the reader of a well known estiate for the cardinality of an interval I = [h +, h + B] \ Z, (8) in Z. We prove a nueric lower bound for our purposes here. Lea. For > 30 and any interval I of edge length B in Z, I > B 3 log log.96/ log log. In particular, if B > and is su ciently large, then I > 4 follows fro the next two leas. B log log. The result Lea 2. For any B, d 2 N, h 2 Z, the nuber of ultiples of d in the interval [h +, h + B] is B d + r d, for soe r 2 Z with r apple d. Proof. Say h + = q d + r, h + B = q 2 d + r 2, with q, q 2 2 Z, 0 apple r, r 2 < d. If r > 0, there are q 2 q ultiples of d in [h +, h + B], and we have q 2 q = B r 2 + r d = B d + r r 2, d with r r 2 apple d. If r = 0, then there are q 2 q + ultiples of d in the interval, and we have with d r 2 apple d. q 2 q + = B r 2 + r + d d = B d + d r 2, d Lea 3. For any 2 N, and interval I in Z, we have I = B + 2!, for soe apple, where! is the nuber of distinct prie divisors of.

5 INTEGERS: 8 (208) 5 Proof. h+applexappleh+b (x,)= = h+applexappleh+b d (x,) = d µ(d) h+applexappleh+b d x µ(d). By the preceding lea, the su over x is just B d + (d) for soe real nuber (d) with (d) apple. Thus = B µ(d) d + d d h+applexappleh+b (x,)= = B d µ(d) d + d µ(d) (d) = B + d µ(d) (d). The latter su is bounded by the nuber of square-free divisors of, 2!. Proof of Lea. By the work of Robin [8] we have! apple.384 log / log log for 3, whence, 2! <.96/ log log for 3. Also, by the work of Rosser and Shoenfeld [9] we have / < 3 log log for > 30. Thus for > 30, it follows fro the preceding lea that for any interval I in Z, as desired. I > B 3 log log.96/ log log, 4. Proof of Theore The theore is an easy consequence of the following lea. We let 0 denote the principal character (od ) and write P to indicate a su over all ultiplicative characters (od ) with. Lea 4. For any interval of points in Z as in (8), we have x2i 4 I 2. Proof. Let B denote the length of the interval I. If B /4, then by Lea, B I log log. Using the ean value estiate x=a+ 4 8! (log ) 3 (log log ) 7 B 2, (9)

6 INTEGERS: 8 (208) 6 of Cochrane and Shi [8], where is the nuber of divisors of and! is the nuber of distinct prie divisors of, we obtain x2i 4 B 2 I 2. If B < /4, the trivial bound iplies x2i 4 B 4 apple. Lea 5. We have for any nonprincipal character where = Q p e k e 3 p e. x2i I r r+ 4r 2 + 3r 4r 2, (od ), Proof. The proof is siilar to the preceding lea. If B, the upper bound of Burgess [4], [5] for a general (see for exaple [4, equation (2.56)] or [7, Theore.6]), x=a+ B r r+ 4r 2 + 3r 4r 2 yields the result. If B <, the trivial bound gives the result., (0) Lea 6. For any positive integers k,, r, with k 4, and intervals I i as above, i= (x i ),k B r 2 k + 4 r+ kr 4r 2 (k 4)+ 3r 4r2 (k 4). Proof. By the preceding two leas, we have for any interval I i, x2i i k apple ax x2i i k 4 I i 2 ax x2i i,k I i k( r ) 2+ 4 r r+ k 4 x2i i 4r 2 (k 4)+ 3r 4r2 (k 4) 4. ()

7 INTEGERS: 8 (208) 7 By Hölder s inequality, i= (x i ) apple i= (x i ) k A/k. Inserting the upper bound in () copletes the proof. Proof of Theore. The nuber of solutions of (4) with x i 2 I i, apple i apple k, is given by n c = (c x x k ) (2) Q k i= = I i + The result follows fro the preceding lea. (c ) i= (x i ). (3) 5. The Congruence ax x k + by y k c (od ) We turn now to the congruence Let I i, J i be intervals in Z given by ax x 2 x k + by y 2 y k c (od ). (4) I i = [h i +, h i + B i ] \ Z, J i = [h k+i +, h k+i + B k+i ] \ Z, (5) for soe h i, B i 2 Z, with apple B i apple, apple i apple k. Theore 2. Suppose that k,, r 2 N with k 4, r 2, and that a, b, c 2 Z with (abc, ) =. The nuber N of solutions of (4) with x i 2 I i, y i 2 J i, apple i apple k, is given by Q k N i= = I i J i Y p 2 p + O p ( + O + O Q k i= I i Q k i= J i p i= i= i= J i ( r ) 2 k + 4 kr r+ I i ( r ) 2 k + 4 kr r+ I i J i ) ( r ) 2 k + 4 kr r+ 4r 2 (k 4)+ 3 4r (k 4) 4r 2 (k 4)+ 3 4r (k 4) For r = 2 or 3, the ter ay be dropped fro the error ters.!!. 2r 2 (k 4)+ 3 2r (k 4)!

8 INTEGERS: 8 (208) 8 We note that if 2 the ain ter of the theore vanishes. Indeed, in this case it is plain that N = 0, since for any odd integers x i, the left-hand side of (4) is even while the right-hand side is odd. Aside fro this case, the theore yields an asyptotic forula for N provided that the following three inequalities hold, and i= I i J i k 2 + k(r 2 +k 4) 2r(k I i i= J i i= 4+2r) + 3k(k 4) 2(k 4+2r) k(k 4)(r+) 4r(k 4+2r) + 3k(k 4) 4(k 4+2r) k(k 4)(r+) 4r(k 4+2r) + 3k(k 4) 4(k 4+2r), (6), (7). (8) The ter ay be dropped if r = 2 or 3. The result obtained here generalizes the result of [3] for prie oduli. Using r = 3 we obtain for k 4 and any positive integer, that N > 0 provided that i= I i J i 2k 3 + k 2(k+2) +, I i k 3 i= 2k k+2 +, J i k 3 i= 2k k+2 +. (9) (There is no advantage in using r = 2 for any value of k.) For a general odulus this is the best we can do. For a cube, it is easy to verify that the condition in (6) iplies the conditions in (7) and (8). In particular, taking r = 3, we see that for k 4 and arbitrary, any cube with edge length B 3 + 4(k+2) +, (20) contains a solution of (4). Suppose now that =, k choice of r is an integer satisfying 5. Then the optial r 2 + k 4 2r 2 + rk 4r < 2 p, k +.95 as shown in [3, Lea 4.2]. For this choice of r we see that any cube with edge length B p + k+3.9, contains a solution of (4). Although Theore 2 requires k 4, we can deduce a result for k = 3 by applying it with k = 4 and I 4 = J 4 = {}. In this anner we obtain fro (9) that N > 0 for k = 3 and any cube with B 2 +.

9 INTEGERS: 8 (208) 9 6. Using Multiplicative Characters to Estiate N We ay assue that a = and write (4), x x k c by y k (od ). Let I i, J i be intervals as in (5). For any A 2 Z, let n A denote the nuber of solutions of y y k A (od ), with y i 2 J i, apple i apple k, and n c ba the nuber of solutions of x x k c ba (od ), with x i 2 I i, apple i apple k. Using the forula in (2) for n A and n c N = n c ba n A ba, we have = (c ba,)= (c ba,)= = 2 I i J i i= ((c ba) x x k ) (c ba,)= say, where the three error ters are given by say. E : = 2 E 2 : = 2 E 3 : = 2 (c ba,)= (c ba,)= (A y y k ) + E + E 2 + E 3, (2) ((c ba) ) (A ) (c ba,)= (x x k ) 6.. Estiation of the Main Ter To estiate the ain ter we need,!! (x x k ), (22) (y y k ), (23) ((c ba) ) (A )! (24) (y y k ), (25)

10 INTEGERS: 8 (208) 0 Lea 7. For any integers b, c with (bc, ) =, = Y p (c ba,)= p 2 p. (26) Proof. This is actually a special case of Lea 0 below applied to the principal character, but let s give a quick proof here. The su is plainly ultiplicative in, and for a prie power = p e, the su just counts the nuber of A 2 Z with A 6 0, cb (od p), which equals p e (p 2) = (p e ) p 2 p. Thus the ain ter in (2) is just Q i= I i J i Y p 2 p. (27) p 6.2. Estiation of E and E 2 Let us first recall a couple of notions about ultiplicative characters. For any ultiplicative character (od ) and divisor d of, we say that is induced by a character (od d) (or siply is a character (od d)) if whenever x y (od d), then = (y). Viewed as a character (od d) there is a slight di erence in the definition of on values of x with (x, d) = but (x, ) 6=. As a character (od ), = 0, but as a character (od d),. This distinction is not a concern in what follows, for we will always restrict our attention to values of x with (x, ) =. There is a unique inial divisor d such that is a character (od d), called the conductor of, written cond( ). For this d, is a priitive character (od d). For the principal character 0 we have cond( 0 ) =. Lea 8. If d and is a character (od ) that is not a character (od d), then there exists u (od d), with (u, ) = and (u) 6=. Proof. Say has prie power factorization = p e pe2 2 pe` `, for distinct pries p i and exponents e i, apple i apple `. Then can be expressed = 2 ` for soe characters i (od p ei ). Say d = p f pf2 2 pf` ` with f i apple e i, apple i apple `. Since is not a character (od d), there exists an i apple ` such that i is not a character (od p fi i ). Let a i be a priitive root (od p ei i ), and let j i be the unique integer with 0 < j i apple p ei i (p i ) and i(a i ) = e 2 ij i (p e i i ). Since i is not a character (od p fi i ), then pei fi i have u 0 (od p fi i ) and - j i. Setting u 0 = a (pfi i i ), we i(u 0 ) = i (a (pf i i ) i 2 ij i ) = e (p f i i ) (p e i 2 ij i f i i ) p = e e i i 6=,

11 INTEGERS: 8 (208) since p ei i fi - j i. By the Chinese Reainder Theore, there exists an integer u with ( u 0 (od p ei i u ); (od p ej j ), for j 6= i. Then (u, ) =, u (od d) and (u) 6=. Lea 9. For any character of we have (od ), integer c with (c, ) = and divisor d d t=0 (c + td) = ( (c) (d), if is a character (od d); 0, if is not a character (od d). Proof. If is a character (od d) the clai is iediate, since there are / (d) choices for t such that (c + td, ) =. If is not a character (od d), then there exists u (od d), with (u, ) = and (u) 6= by the preceding lea. Then d t=0 (c + td) = d t=0 (u(c + td)) = (u) d t=0 (c + td), and so the su ust be zero. Lea 0. For any ultiplicative character c with (c, ) =, we have Proof. Now, (c A,)= (c A,)= (A) = (A) = (c A,)= (c) µ = d µ(d) = d µ(d) (A) (od ) of conductor e and integer d (c Y p e p-e A,) A c (od d) d t=0 p 2 p. µ(d) (c + td). Thus letting e denote the conductor of, we get fro the preceding lea, (A) = (c) µ(d) (d) d e d = (c) f e µ(ef) (ef). (A)

12 INTEGERS: 8 (208) 2 Now, the only contribution to the su over f coes fro square-free values of ef. Thus if (e, f) > there is no contribution, and so we ay assue (e, f) =, whence µ(ef) = µµ(f) and (ef) = (f). Thus (c A,)= (A) = (c) µ f e (f,e)= µ(f) (f) = (c)µ Y p e p-e p 2 p. Next we have to obtain character su bounds over intervals with restricted variables. Again let I be an interval of points in Z, I = [a +, a + B] \ Z. First we obtain the following Burgess-type estiate. Lea. For any e, positive integers B, r and non-principal character (od e), we have x=a+ (x,)= I r e r+ 4r 2 e 3 4r, where e is the product of the prie-power divisors of e of ultiplicity at least 3. Proof. We have x=a+ (x,)= = x=a+ = µ( ) x=a+ x = µ( ) ( ) (x,) µ( ) (a+)/ appletapple()/ (t). (28) Then by the Burgess bound in (0), x=a+ (x,)= apple (t) (a+)/ appletapple()/ (B/ + ) r+ r e 4r 2 + e 3 4r B r e r+ 4r 2 + e 3 4r. Replacing B with I in the stateent of the lea now follows as in the proof of Lea 4.

13 INTEGERS: 8 (208) 3 Generalizing the result of Cochrane and Shi [8], we have Lea 2. For any e, integer a and positive integer B we have (od e) x=a+ (x,)= 4 I 2. Proof. By (28), Hölder s inequality and then eploying the upper bound in (4), we get (od e) x=a+ (x,)= 4 apple (od e) apple 3 (a+)/ appletapple()/ (od e) (t) (a+)/ appletapple()/ 4 (t) 4 apple 3 8! (log e) 3 (log log e) 7 (B/ + ) 2 4 8! (log ) 3 (log log ) 7 B 2 B 2. Lea 3. Suppose that k we have 4. For any e, integer a and positive integers r, B (od e) x=a+ (x,)= k I k( r ) 2+ 4 r e r+ 4r 2 (k 4) e 3 4r (k 4). (29) Proof. Fro the preceding two leas we have (od e) x=a+ (x,)= k apple ax x=a+ (x,)= k 4 x=a+ (x,)= 4 I k( r ) 2+ 4 r e r+ 4r 2 (k 4) e 3 4r (k 4). Again, replacing B with I in the stateent of the lea follows as before. Turning to E we have by Lea 0, letting e denote the conductor of,

14 INTEGERS: 8 (208) 4 cond( ), E : = 2 = 2 0 Q k i= = J i (c ba,)= (c ) µ(e ) (e ) e e> µ Y p /e p-e ((c ba) C ) A (x x k ) p 2 p Y p e p-e (od e) cond( )=e p 2 p Thus, by Holder s inequality and the preceding lea, Q k i= E J i I i r Q k i= J i e e> i= I i r i= (x x k ) (c ) (x x k ). 2 k + 4 r+ kr e 4r 2 (k 4) e 3 4r (k 4) 2 k + 4 r+ kr 4r 2 (k 4)+ 3 4r (k 4). In a siilar anner we obtain the following upper bound for E 2, Q k i= E 2 I i J i r i= 2 k + 4 r+ kr 4r 2 (k 4)+ 3 4r (k 4). (30) 6.3. Estiation of E 3 Lea 4. For any ultiplicative characters (c, ) = we have,, (od ) and integers c, A with (c A,)= (A) ((c A) ) apple p [cond( ), cond( )], where [cond( ), cond( )] denotes the least coon ultiple of the conductors of and. Proof. We first consider the case of a prie power = p e. Let a be a priitive root (od p e ) and be the generator of the character group defined by (a j ) = e 2 ij (p e ).

15 INTEGERS: 8 (208) 5 For any rational function q = q = f/g with integer coe the character su S p e(q) = p e x= (fg,p)= (q), cients, we define where it is understood that /g eans the ultiplicative inverse of g (od p e ). Following the ethod of critical points developed in [9], [0] and [6], we define t to be the axiu power of p dividing all of the coe cients of gf 0 g 0 f, and the critical point congruence to be the congruence p t (gf 0 g 0 f) 0 (od p). (3) The critical points are the solutions x of the congruence (3) with p - fg. By [6, Theore.], if e t + 2 and (3) has a unique critical point of ultiplicity then S p e(q) = p e+t 2. If e t + 2 and there is no critical point, then S p e(q) = 0. We clai that one of these two options always occurs for the case at hand. Let, be characters (od p e ) and say = u, = v, for soe positive integers u, v apple (p e ). Then x ((c x) u ) = (c x) v. With q = xu (c x) we have v gf 0 g 0 f = ux u (c x) v + vx u (c x) v = x u (c x) v (uc + (v u)x), and so p t k(u, v), and so there is either no critical point or a single critical point of ultiplicity one. Thus, if t apple e 2 then p e x= p-x(c x) ((c x) ) = S p e(q) apple p e+t 2. Otherwise, t = e. In this case, both and are characters (od p) and we have by the Weil bound [22] for character sus over finite fields (see eg. [7]) that p e x= p-x(c x) ((c x) ) = p e p x= p-x(c x) x u (c x) v apple p e p p = p e+t 2. Suppose that p t ku, p t2 kv. Then cond( ) = p e t while cond( ) = p e t2, and we see that [cond( ), cond( )] = p e in(t,t2) = p e t,

16 INTEGERS: 8 (208) 6 and p e+t 2 = pe p p e t = p e p [cond( ), cond( )]. For general we write = p e pe2 2 pe` ` with the p i distinct pries, with i, i characters (od p ei i ), and = 2 `, = 2 `, Ỳ S (q) = S e (A p i iq), i i= for appropriate integers A i with (A i, p i ) =. Then Ỳ S (q) apple i= p ei i p [cond( i ), cond( i )] = p [cond( ), cond( )]. Fro the preceding lea we have, Thus, E 3 : = 2 (c ba,)= (x x k ) ((c ba) ) (A ) (y y k ).! E 3 apple 2 apple apple d d> 2 d d> e e> e e> (d) 2 d d> e e> (d) p [e, d] (d) p [e, d] cond( )=d (d) p [e, d] (od d) cond( )=d cond( )=e (x x k ) (x x k ) (x x k ) cond( )=e (od e) (y y k ) (y y k ) (y y k )

17 INTEGERS: 8 (208) 7 Letting L = [d, e] and applying Lea 3 we obtain E 3 2 (Y I i J i ) ( r ) 2 k + 4 kr p i= L L p ( Y i= I i J i ) ( r ) 2 k + 4 kr r+ d L (d)d r+ 2r 2 (k 4)+ 3 2r (k 4). 4r 2 (k 4) d 3 4r (k 4)! Proof of Theore 2 Fro equation (2), the value for the ain ter in (27) and the estiates for the error ters E, E 2 and E 3 in the preceding sections we see that Q N i= = Ii Ji Y p 2 p + O p ( Y! I i J i ) ( r ) 2 k + 4 r+ kr 2r2 (k 4)+ 3 (k 4) 2r p i= Q! k i= + O Ii J i ( r ) 2 k + 4 r+ kr 4r 2 (k 4)+ 4r 3 (k 4) i= Q! k i= + O Ji I i ( r ) 2 k + kr 4 r+ 4r 2 (k 4)+ 4r 3 (k 4). i= References [] A. Ayyad, The distribution of solutions of the congruence x x 2 x n c (od p), Proc. Aer. Math. Soc. 27 (999), no. 4, [2] A. Ayyad and T. Cochrane, Lattices in Z 2 and the congruence xy + uv c (od ), Acta Arith. 32 (2008), no. 2, [3] A. Ayyad and T. Cochrane, The congruence ax x 2 x k + bx k+ x k+2 x 2k c (od p), Proc. Aer. Math. Soc. 45 (207), no. 2, [4] D. A. Burgess, On character sus and L-series I, Proc. London Math. Soc. 2 (962), no. 2, [5] D. A. Burgess, On character sus and L-series II, Proc. London Math. Soc. 3 (963), no. 3, [6] T. Cochrane, Exponential sus odulo prie powers, Acta Arith. 0 (2002), no. 2, [7] T. Cochrane and C. Pinner, Using Stepanov s ethod for exponential sus involving rational functions, J. Nuber Theory 6 (2006), no. 2, [8] T. Cochrane and S. Shi, The congruence x x 2 x 3 x 4 (od ) and ean values of character sus, J. Nuber Theory, 30 no. 3 (200), [9] T. Cochrane and Z. Zheng, Pure and ixed exponential sus, Acta Arith. 9 (999), no. 3,

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