#A29 INTEGERS 17 (2017) EQUALITY OF DEDEKIND SUMS MODULO 24Z

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1 #A29 INTEGERS 17 (2017) EQUALITY OF DEDEKIND SUMS MODULO 24Z Kurt Girstmir Institut für Mthemtik, Universität Innsruck, Innsruck, Austri Received: 10/4/16, Accepted: 7/3/17, Pulished: 7/17/17 Astrct Let S(, ) = 12s(, ), where s(, ) denotes the clssicl Dedekind sum. In recent note E. Tsukermn gve necessry nd su cient condition for S( 1, ) S( 2, ) 2 8Z. In the present pper we show tht this condition is equivlent to S( 1, ) S( 2, ) 2 24Z, provided tht 9 -. Tsukermn lso otined congruence mod 8 for T (, ), where T (, ) is the lternting sum of the prtil quotients of the continued frction expnsion of /. We show tht the respective congruence holds mod 24 if 3 - nd mod 72 if Introduction nd Results Let e n integer, nturl numer, nd (, ) = 1. The clssicl Dedekind sum s(, ) is defined y Here ((x)) = s(, ) = X ((k/))((k/)). k=1 ( x xc 1/2 if x 2 R r Z; 0 if x 2 Z (see [2, p. 1]). It is often more convenient to work with S(, ) = 12s(, ) insted (see, for instnce, formul (7) elow). Since S( +, ) = S(, ), we otin ll Dedekind sums if is restricted to the rnge 0 pple pple 1, (, ) = 1. In the recent note [4], E. Tsukermn gve necessry nd su cient condition for the equlity of S( 1, ) nd S( 2, ) modulo 8Z. This condition involves the function µ, which is defined, for, s ove, s follows: ( 2 2 µ(, ) =, if is odd; ( 1)( + 1), if is even.

2 INTEGERS: 17 (2017) 2 Here is the Jcoi symol. Tsukermn s condition is phrsed y mens of the residue clss ( 2 µ(, 1 ) 1 µ(, 2 )) mod 8. We oserve, however, tht this residue clss depends only of the residue clsses of µ(, 1 ) nd µ(, 2 ) modulo 8, not of the vlues of µ(, 1 ) nd µ(, 2 ) themselves. Therefore, we my replce the function µ y the following simpler function, which we henceforth lso cll µ. 8 >< 2 2, if is odd; µ(, ) = 4, if 0 mod 4 nd 3 mod 4; >: 0, otherwise. In this pper we show Theorem 1. Let 1, 2 2 N e reltively prime to 2 N. Suppose, further, tht 9 -. Then S( 1, ) S( 2, ) 2 24Z if, nd only if, ( 2 µ(, 1 ) 1 µ(, 2 )) ( 1 2 )( 1)( ) mod 8. (1) This equivlence cnnot e extended to the cse 9 in n ovious wy, s we show in Section 3. Tsukermn showed tht (1) is equivlent to S( 1, ) S( 2, ) 2 8Z for ritrry nturl numers, i.e., he needed not ssume 9 -. For 2 Z nd 2 N, let = [ 0, 1,..., n ] denote the regulr continued frction expnsion of /. The prtil quotients 1,..., n re nturl numers. We do not ssume n 2, ut require n to e odd, insted. Define nx T (, ) = ( 1) k 1 k. (2) k=0 In the sid pper, Tsukermn showed, for, 2 N, (, ) = 1, T (, ) µ(, ) mod 8, (3) with 2 {1,..., 1}, 1 mod. Our new definition of µ suggests more explicit form of (3), which we use in the following Theorem. To this end we define " 2 {±1} y the congruence " mod 3 (4) for ech 2 Z, 3 -.

3 INTEGERS: 17 (2017) 3 Theorem 2. Let 2 Z e reltively prime to 2 N. () Let e odd. If 3 -, then T (, ) mod 24. If 3, then T (, ) " mod 72. () Let 2 mod 4 or let oth 0 mod 4 nd 3 mod 4 hold. If 3 -, then T (, ) 6 mod 24. If 3, then T (, ) 54 16" mod 72. (c) Let 0 mod 4 nd 1 mod 4. If 3 -, then T (, ) 18 mod 24. If 3, then T (, ) 18 16" mod 72. In Section 3 we exhiit mny exmples tht illustrte oth Theorem 1 nd the fct tht this theorem does not hold if Proofs Our min tools re two congruences modulo 3 for Dedekind sums. First we oserve tht S(, ) is n integer; moreover, if 3 does not divide, then S(, ) 0 mod 3. (5) These ssertions follow from [2, p. 27, Th. 2]). On the other hnd, if 3, where " is defined s in (4) (see [3, formul (70)]). S(, ) 2" mod 9, (6) Proof of Theorem 1. Suppose, first, tht 3 -. Becuse of (5), we my write S( 1, ) = 3k 1, S( 2, ) = 3k 2 with integers k 1, k 2. By [4, Th. 3.1], the congruence (1) is equivlent to S( 1, ) S( 2, ) 2 8Z. Accordingly, (1) is lso equivlent to 3(k 1 k 2 ) = 8r, r 2 Z.

4 INTEGERS: 17 (2017) 4 However, 3 -, nd so this mens 3 r. This proves Theorem 1 in the cse 3 -. Suppose now tht 3. Then the congruence (1) implies ( 1 2 )( 1 2 1) 0mod3. Hence we otin, from (6) S( 1, ) = 2" + 9k 1, S( 2, ) = 2" + 9k 2 with common vlue " 1 2 mod 3 nd k 1, k 2 2 Z. Accordingly, (1) is equivlent to 9(k 1 k 2 ) = 8r, r 2 Z. If 9, this simply mens S( 1, ) S( 2, ) 2 8Z, so this is just Tsukermn s result. However, if 9 -, we otin 3 r, which yields the theorem in the cse 3, 9 -. Proof of Theorem 2. The Brkn-Hickerson-Knuth formul sys S(, ) = T (, ) (7) (see, for instnce, [1]). Note tht this formul is often enuncited only for the cse 0 pple <, ut it is, in fct, vlid for ritrry integers reltively prime to, provided tht T (, ) is defined s in (2). Hence we otin, y (5) nd (7), if 3 -. In the cse 3, (6) nd (7) give T (, ) mod 3 (8) T (, ) 2" mod 9 (9) insted. Further, Tsukermn s congruence (3) is lso vlid for ritrry integers reltively prime to, s we esily check. We comine (3) with the congruences (8) nd (9) y mens of the Chinese reminder theorem. This redily gives Theorem A Proposition Yielding Exmples Our exmples rise from the following proposition. Proposition 1. Let c, d e odd nturl numers, d 3. Put = cd 2 nd = cd+1. Then S(1, ) S(, ) = c(d 2 1).

5 INTEGERS: 17 (2017) 5 Proof. We pply the reciprocity lw for Dedekind sums (see [2, p. 5]), which gives S(, ) = S(, ) Now d mod, hence the reciprocity lw sys S(, ) = S( d, ) = S(, d) However, 1 mod d, nd so S(, d) = S(1, d) = d = cd 2 nd = cd + 1 gives d S(, ) = c d 1 d /d. Inserting the vlues Since S(1, ) = 3 + 2/, we otin the desired result. In the setting of the proposition, let 3 - d. Then d mod 24, so the proposition yields mny exmples with S(1, ) S(, ) > 0 nd S(1, ) S(, ) 0 mod 24. On the other hnd, if 3 d, then d mod 8, ut d mod 24. If, therefore, 3 - c, we otin mny exmples with S(1, ) S(, ) 0 mod 8, ut S(1, ) S(, ) 6 0 mod 24. References [1] D. Hickerson, Continued frctions nd density results for Dedekind sums, J. Reine Angew. Mth. 290 (1977), [2] H. Rdemcher, E. Grosswld, Dedekind Sums, Mthemticl Assocition of Americ, [3] H. Slié, Zum Wertevorrt der Dedekindschen Summen, Mth. Z. 72 (1959), [4] E. Tsukermn, Equlity of Dedekind sums modulo 8Z, Act Arith. 170 (2015),