Annales Univ. Sci. Budapest., Sect. Comp. 36 (2012) 355–373
ON REAL VALUED ADDITIVE FUNCTIONS MODULO 1 Kalyan Chakraborty (Allahabad, India) Imre K´ atai and Bui Minh Phong (Budapest, Hungary) Communicated by Antal J´arai (Received December 23, 2011; accepted January 10, 2012)
Abstract. We determine class of five completely additive real valued functions satisfying particular relations.
1.
1.1.
Introduction
Notations
Let G be an additive commutative semigroup with identity element 0. Let AG and A∗G denote the set of G valued additive and completely additive functions respectively. In case G = R, then we simply write A (respectively A∗ ) and when H = = C, then we write M (respectively M∗ ). The domain of f ∈ AG (A∗G ) can be extended to Z by defining f (−1) = f (0) = 0. Then f (n) = f (|n|), and f (nm) = f (n) + f (m) remain valid in n, m ∈ Z∗ := Z \ {0}. Similarly, for g ∈ MH , defining g(−1) = g(0) = 1 and g(−n) = g(n), we can extend g over Z by g(n) = g(|n|). Then g(nm) = g(n)g(m) holds, if (n, m) = 1 and m, n ∈ Z∗ . Key words and phrases: Additive functions. 2010 Mathematics Subject Classification: Primary: 11A25. The Research of the last two authors is supported by the European Union and co-financed by ´ the European Social Fund (grant agreement no. TAMOP 4.2.1./B-09/1/KMR-2010-0003).
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K. Chakraborty, I. K´ atai and B.M. Phong
1.2.
Regular behaviour of additive and multiplicative functions
P. Erd˝ os [2] proved that if f ∈ A be such that f (n + 1) − f (n) → 0 as n → ∞, then f (n) is a constant multiple of log n. Since then this beautiful and simple assertion saw a plenty of generalizations. It is natural to determine all g ∈ M for which g(n + 1) − g(n) → 0 as n → ∞. It clearly holds if g(n) → 0 (n → ∞), or if g(n) = ns (n ∈ N) and 12 for which ai (n0 ) 6≡ 0 (mod 1). Then n0 should be a prime p ≥ 13. Let a2 (p) ≡ ξ 6≡ 0
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On real valued additive functions modulo 1
(mod 1). Using H(p − 2) ≡ 0 (mod 1) we have that a0 (p + 2) ≡ −ξ (mod 1) and p + 2 ∈ P. Thus p≡2
(2.4)
(mod 3).
Using (2.4) and p ≥ 13, we have 2|p + 3, 3|p + 4, 2|p + 5, consequently we infer from H(p + 2) ≡ 0 (mod 1) that a0 (p + 6) ≡ −ξ (mod 1) and p + 6 ∈ P. Since H(p + 6) = a0 (p + 6) + a1 (p + 7) + a2 (p + 8) + a1 (p + 9) +a0 (p + 10) ≡ 0
(mod 1)
and 2|p + 7, 2|p + 9, 3|p + 10, therefore a2 (p + 8) ≡ −ξ (mod 1) and p + 8 ∈ P. Thus we have proved that p, p + 2, p + 6, p + 8 ∈ P, which implies that p≡1
(2.5)
(mod 5).
Next, we prove the following assertion: (2.6)
if
p ∈ P, q < 2p − 3, then
a1 (q) ≡ 0
(mod 1).
This clearly holds if q < p. Let p ≤ q < 2p − 3. Then either 3|q − 2 or 3|q + 2. Since H(q − 1) = a0 (q − 1) + a1 (q) + a2 (q + 1) + a1 (q + 2) + a0 (q + 3) ≡ 0
(mod 1)
and H(q − 3) = a0 (q − 3) + a1 (q − 2) + a2 (q − 1) + a1 (q) + a0 (q + 1) ≡ 0 and 2|q + `,
q+` 2
(mod 1)
< p if ` = −3, −1, 1, 3. Thus
a1 (q) + a1 (q + 2) ≡ 0
(mod 1)
and
a1 (q − 2) + a1 (q) ≡ 0
(mod 1).
Since either a1 (q − 2) ≡ 0 (mod 1) or a1 (q + 2) ≡ 0 (mod 1), consequently a1 (q) ≡ 0 (mod 1). Hence (2.6) is proved. From H(2p + 1) = a0 (2p + 1) + a1 (2p + 2) + a2 (2p + 3) + a1 (2p + 4) +a0 (2p + 5) ≡ 0
(mod 1),
observing from (2.4), (2.5) and (2.6) that 4|2p + 2, 5|2p + 3, 3|2p + 5, and that a1 (2p + 4) ≡ a1 (p + 2) ≡ 0 (mod 1), we deduce that a0 (2p + 1) ≡ 0 (mod 1). Therefore, H(4p − 2) ≡ 0 (mod 1) implies that a0 (4p − 2) + a1 (4p − 1) + a2 (4p) + a1 (4p + 1) + a0 (4p + 2) ≡ 0
(mod 1).
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K. Chakraborty, I. K´ atai and B.M. Phong
Since 6|4p − 2, 5|4p + 1 and a0 (2p + 1) ≡ 0 (mod 1), therefore a2 (p) + a1 (4p − 1) ≡ 0
(mod 1).
Thus, a1 (4p − 1) ≡ −ξ
(2.7)
(mod 1), and 4p − 1 ∈ P.
Since H(2p − 3) ≡ 0 (mod 1), 4|2p − 2, 3|2p − 1 and a0 (2p + 1) ≡ a1 (p) ≡ ≡ 0 (mod 1), therefore a0 (2p − 3) ≡ 0 (mod 1). From H(4p − 6) ≡ 0 (mod 1) we deduce that a0 (4p − 6) + a1 (4p − 5) + a2 (4p − 4) + a1 (4p − 3) + a0 (4p − 2) ≡ 0
(mod 1).
It is obvious that 6|4p − 2 implies a0 (4p − 2) ≡ 0 (mod 1), 3|4p − 5. Thus is not a prime. In both cases either 4p − 5 = 3q, q ∈ P, q < 2p − 3, or 4p−5 5 we deduce from (2.6) that a1 (4p − 5) ≡ 0 (mod 1). Thus we derive, a0 (2p − 3) + a1 (4p − 3) ≡ 0
(mod 1).
Consequently, a1 (4p − 3) ≡ 0
(2.8)
(mod 1).
Finally, from H(4p − 4) ≡ 0 (mod 1) we have a0 (4p − 4) + a1 (4p − 3) + a2 (4p − 2) + a1 (4p − 1) + a0 (4p) ≡ 0
(mod 1).
Since a0 (p) ≡ 0 (mod 1), 8|4p − 4, 6|4p − 2, we get from (2.8) that a1 (4p − 1) ≡ ≡ 0 (mod 1). This contradicts (2.7). Lemma 3. Let a0 , a1 , a2 ∈ A∗ and H(n) = a0 (n) + a1 (n + 1) + a2 (n + 2) + a1 (n + 3) + a0 (n + 4). If (2.9)
H(n) ≡ 0
(mod 1)
for all
n ∈ N,
(mod 1)
for
then (2.2) is true, i.e. a0 (n) ≡ a1 (n) ≡ a2 (n) ≡ 0
n ≤ 12.
Proof. Let B be the subgroup of Q3+ generated by the sequences Ln = n(n + 4), (n + 1)(n + 3), n + 2 (n ∈ N).
363
On real valued additive functions modulo 1
It is easy to see ( by (2.9)) that, (2.10)
a0 (a) + a1 (b) + a2 (c) ≡ 0
(mod 1)
for all
(a, b, c) ∈ B.
We use the following notations for a prime p: ap = (p, 1, 1), bp = (1, p, 1)
and cp = (1, 1, p).
We show that ap , bp , and cp ∈ B for all primes p ≤ 11. This assertion along with (2.10) would imply (2.2). Using a simple Maple program, for, n ∈ {1, 2, 3, 4, 5, 8, 12, 7, 11, 14, 48, 9, 13, 16, 22, 23, 19, 15, 25, 26, 28, 31}, we can give ap , bq , cr for primes p, q ≤ 31, and r ≤ 17 in terms of Ln and a2 , a3 , b2 , b3 , c2 , c3 and c5 . Table 1 n 1
Ln (5, 23 , 3)
ap , bq , cr a5 = bL3 c13
2
(22 .3, 3.5, 22 )
3
(3.7, 23 .3, 5)
4
(25 , 5.7, 2.3)
5
(32 .5, 24 .3, 7)
8
(25 .3, 32 .11, 2.5)
12
(26 .3, 3.5.13, 2.7)
7
(7.11, 24 .5, 32 )
11
(3.5.11, 23 .3.7, 13)
14
(22 .32 .7, 3.5.17, 24 )
48
(26 .3.13, 3.72 .17, 2.52 )
9
(32 .13, 23 .3.5, 11)
13
(13.17, 25 .7, 3.5)
16
(26 .5, 17.19, 2.32 )
22
(22 .11.13, 52 .23, 23 .3)
23
(33 .23, 24 .3.13, 52 )
19
(19.23, 23 .5.11, 3.7)
15
(3.5.19, 25 .32 , 17)
L2 , a22 a3 b3 c22 a7 = a3 bL3 b33 c5 2 b7 = a5 bL5 c42 c3 = LL4 2aa3 3b3c3c2 2 2 c7 = a2 aL5 5b4 b3 = L1La52cb32 b3 3 2 3 b11 = a5 a3Lb28c2 c5 = a5 a3Lb28c2 c5 2 3 2 3 L1 L12 a23 b3 c2 b2 = b13 = a6 a3 bL312 b5 c2 c7 L2 L5 a42 c3 2 L7 a23 b23 c5 a22 c22 L7 a11 = a7 b4 b5 c2 = L3 b2 L2 c2 2 3 3 L2 L L b c4 a c13 = a3 a5 aL1111b3 b3 b7 = L12L43L711a4 b24 c35 c23 2 3 3 2 L b3 b c b17 = a2 a2 aL714b3 b5 c4 = L142 L23 c32 5 2 3 2 2 L3 L L c23 a13 = a6 a3 b3Lb248b17 c2 c2 = L2 L142 a33b348 3 4 2 7 5 4 3 2 b3 c2 c5 L24 L9 L14 a23 b43 c32 c35 a22 L9 c11 = a2 a13 b3 b3 b5 = L3 L42 L48 c23 3 2 L4 L13 L14 a23 b33 c25 a32 L13 a17 = a13 b5 b7 c3 c5 = L2 L3 L48 c2 b2 2 2 3 2 L2 L3 L16 c2 16 b19 = a6 a5Lb17 2 = L L a6 c b c c c 2 1 14 3 3 5 2 3 2 L24 L14 L22 a33 b42 b43 c25 L22 b23 = a2 a11 a13 b2 c3 c3 = L42 L7 L48 c3 2 5 2 L2 L5 L23 a42 c3 L23 a23 = a3 b4 b3 b13 c2 = L1 L12 a5 b5 b2 c2 c2 3 2 5 3 2 3 5 L21 L12 L19 a93 b32 b63 a32 c42 c35 a19 = a23 b3 bL519 = 2 L2 L L c3 b c c L 11 3 7 2 2 5 8 23 3 L22 L25 L8 L15 L23 c43 c17 = a3 a5La15 = 5 2 5 8 4 3 3 L1 L12 L19 a32 a10 19 b2 b3 3 b2 b3 c2 c5
2
b5 =
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K. Chakraborty, I. K´ atai and B.M. Phong
n
Ln 2
3
3
25
(5 .29, 2 .7.13, 3 )
26
(22 .3.5.13, 33 .29, 22 .7)
28
(27 .7, 29.31, 2.3.5)
31
(5.7.31, 26 .17, 3.11)
ap , bq , cr L2 L L b22 c3 a72 L25 a29 = a2 b3 b7 b13 c3 = L23 L54 L25 3 2 2 12 a3 b3 c2 5 2 3 1 L24 L14 L26 a43 b72 b23 c35 L26 b29 = a2 a3 a5 a13 b3 c2 c7 = L3 L3 L5 L48 a2 c2 c2 2 3 2 2 2 3 L3 L5 L28 L48 c3 b31 = a7 a7 bL2928c2 c3 c5 = L2 L142 L26a 5 a3 b4 b c3 2 4 2 3 2 3 5 L52 L3 L31 L48 c23 L31 a31 = a5 a7 b6 b17 c3 c11 = L1 L2 L9 L2 b3 a3 b4 c3 c2 a2 2 4 14 2 3 5 2
Now, by using the above relations for n = 6, 10, 18, 24, 32, 30, 54 and n = 62, we will get the following 8 equations.
(2.11)
E1 :=
L2 L6 a2 b3 c4 = 3 33 22 ∈ B, L1 L4 a2 b2 c3
(2.12)
E2 :=
L2 L5 L10 c2 = 7 5 22 2 ∈ B, 2 L1 L3 L8 L12 a2 b2 b3 c3 c5
(2.13)
E3 :=
L1 L2 L14 L18 a5 b3 c6 c5 = 3 5 3 2 4 ∈ B, L4 L7 L16 a2 b2 c3
(2.14)
E4 :=
L1 L4 L7 L24 a22 c43 ∈ B, = L42 L23 L11 a63 b22 b43 c62 c25
(2.15)
E5 :=
L31 L12 L19 L32 2 L2 L4 L25 L28 L15 L23
=
c33 6 9 a2 a3 b52 b93 c52 c45
∈ B,
(2.16)
E6 :=
b3 c 4 L3 L4 L26 L30 = 5 92 2 2 ∈ B, L1 L2 L5 L8 L13 L28 a2 a3 b2 c3 c5
(2.17)
E7 :=
L51 L4 L12 L14 L54 b2 c3 = 4 4 6 2 ∈B L42 L3 L25 L8 L16 L25 a2 a3 b3 c2 c5
and (2.18)
E8 :=
a43 b3 c72 L4 L5 L9 L214 L62 = ∈ B. L32 L7 L12 L31 L48 a72 b32 c23 c25
365
On real valued additive functions modulo 1
This system has solutions in a2 , a3 , b2 , b3 , c2 , c3 , c5 , which are given in terms of E1 , · · · , E8 . Thus a2 , a3 , b2 , b3 , c2 , c3 , c5 are elements of B. The solutions of the above equations (2.11)–(2.18) can be obtained now as follows: we express a2 from the expression of E1 in (2.11), similarly c5 from (2.13). After taking these expressions of a2 and c5 into equations (2.11)–(2.18), E 17 E2 E6 E4 and E411 ,respectively. we get c3 , b3 , c2 and a3 from the expressions of E 2 , E ,E 7 5 4 7 Finally the solution b2 can be gotten from (2.14) and (2.18) in the expression E 15 of E832 . The solutions are: 4
E1905 E2151 E372 E4109 E5228 , E6528 E777 E875 E 135 E225 E326 E438 E521 , b2 = 1 E677 E815 E 266 E 20 E 45 c2 = 461 6 82 762 892 94 , E1 E2 E3 E4 E5
a2 =
and c5 =
E111 E22 E56 E34 E45 E67 E74 E 18 E 25 E 43 E 23 b3 = 3 694 8 6 37 7 , E1 E2 E5 E 969 E 109 E 150 c3 = 1670 6287 7176 8263 383 , E1 E2 E3 E4 E5 a3 =
E1898 E2138 E321 E433 E5274 . E6531 E7118 E860
Finally, it is obvious from Table 1 that a5 , b5 , a7 , b7 , a11 , b11
and c11
are elements of B. This completes the proof of Lemma 3.
Proof of Theorem 2. Assume that f0 , f1 , f2 , f3 , f4 ∈ A∗ satisfy the condition Af (n) := f0 (n) + f1 (n + 1) + f2 (n + 2) + f3 (n + 3) + f4 (n + 4) ≡ 0
(mod 1)
for all n ∈ Z. Then Af (−n−4) = f4 (n)+f3 (n+1)+f2 (n+2)+f1 (n+3)+f0 (n+4) ≡ 0
(mod 1).
Let ϕ0 (n) = f0 (n) − f4 (n)
and ϕ1 (n) = f1 (n) − f3 (n)
for all
n ∈ Z.
Thus, we deduce from the above relations that, ϕ0 (n) + ϕ1 (n + 1) − ϕ1 (n + 3) − ϕ0 (n + 4) ≡ 0
(mod 1)
for all
n ∈ Z.
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K. Chakraborty, I. K´ atai and B.M. Phong
From Lemma 1, we have that ϕ0 (n) ≡ ϕ1 (n) ≡ 0 (mod 1), consequently f0 (n) ≡ f4 (n) (mod 1) and f1 (n) ≡ f3 (n) (mod 1) for all n ∈ Z. Hence Af (n) ≡ f0 (n) + f1 (n + 1) + f2 (n + 2) + f1 (n + 3) + f0 (n + 4) ≡ 0
(mod 1)
is true for all n ∈ Z. The conditions of Lemma 2 and Lemma 3 are satisfied by taking aj (n) = fj (n) (j = 0, 1, 2) and H(n) = f0 (n) + f1 (n + 1) + f2 (n + 2) + f1 (n + 3) + f0 (n + 4). Thus f0 (n) ≡ f1 (n) ≡ f2 (n) ≡ f3 (n) ≡ f4 (n) ≡ 0
(mod 1)
holds for all n ∈ Z. This completes the proof.
We can deduce an interesting result from Lemma 3. Theorem 5. If B denotes the subgroup of Q3+ generated by the sequences Ln = n(n + 4), (n + 1)(n + 3), n + 2 (n ∈ N), then we have B = Q3+ .
3.
Proof of Theorem 3
We follow similar strategy as in the case of Theorem 2 and prove a couple of lemmas before completing the proof of the theorem. Lemma 4. Let b0 , b1 , b2 ∈ A∗ . Assume that S(n) := b0 (n) + b1 (n + 2) + b2 (n + 3) + b1 (n + 4) + b0 (n + 6) ≡ 0
(mod 1)
for all n ∈ N. If b0 (n) ≡ b1 (n) ≡ b2 (n) ≡ 0
(mod 1)
for
n ≤ 10,
then b0 (n) ≡ b1 (n) ≡ b2 (n) ≡ 0
(mod 1)
for all
n ∈ N.
Proof. Let n0 be the minimal positive integer for which bj (n0 ) 6≡ 6≡ 0 (mod 1) holds for some j ∈ {0, 1, 2}. It is clear that n0 should be a prime P, P ≥ 11. S(P − 6) ≡ (mod 1) implies that b0 (P ) ≡ 0 (mod 1),
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On real valued additive functions modulo 1
S(P − 3) ≡ 0 (mod 1) similarly that b2 (P ) ≡ 0 (mod 1). It remains to consider the case when b1 (P ) ≡ ξ (mod 1). Then S(P − 4) ≡ 0 (mod 1) implies that b0 (P + 2) ≡ −ξ (mod 1), P + 2 is a prime, thus P ≡ 2 (mod 3). From S(P − 2) ≡ 0 (mod 1) we obtain that b0 (P − 2) + b1 (P ) + b2 (P + 1) + b1 (P + 2) + b0 (P + 4) ≡ 0
(mod 1),
which, by 3|P + 4, 2|P + 1 implies that b1 (P ) + b1 (P + 2) ≡ 0
(mod 1), i.e b1 (P + 2) ≡ −ξ
(mod 1).
Finally, we infer from 4|2P + 2, 3|2P + 5, 4|2P + 6, 6|2P + 8 and S(2P + 2) ≡ ≡ 0 (mod 1) that b0 (2P + 2) + b1 (2P + 4) + b2 (2P + 5) + b1 (2P + 6) + b0 (2P + 8) ≡ 0
(mod 1),
and so b1 (P + 2) ≡ 0
(mod 1).
This contradicts to the fact that b1 (P + 2) ≡ −ξ (mod 1) and consequently the Lemma 4 is proved. Lemma 5. Let b0 , b1 , b2 ∈ A∗ . If b0 (n) + b1 (n + 2) + b2 (n + 3) + b1 (n + 4) + b0 (n + 6) ≡ 0
(mod 1),
for all n ∈ N, then b0 (n) ≡ b1 (n) ≡ b2 (n) ≡ 0
(mod 1)
for
n ≤ 10.
Proof. The proof is similar to the proof of Lemma 3. Let D be the subgroup of Q3+ generated by the sequences Dn := n(n + 6), (n + 2)(n + 4), n + 3 (n ∈ N). From (3.4) we obtain that, b0 (a) + b1 (b) + b2 (c) ≡ 0
(mod 1)
for all
(a, b, c) ∈ D.
We shall use the following notations (p is prime): Ap := (p, 1, 1) ∈ D, Bp := (1, p, 1 ∈ D)
and Cp := (1, 1, p) ∈ D.
We shall prove that Ap , Bp ∈ D and Cp ∈ B for all primes p ≤ 7. This will prove Lemma 5.
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K. Chakraborty, I. K´ atai and B.M. Phong
First, by using a simple Maple program, we shall give Ap , Bq , Cπ for primes p ≤ 23, q ≤ 23, π ≤ 23 in terms of Ln and A2 , A3 , B2 , B3 , C2 , C3 and A5 . n 2
Dn (24 , 23 .3, 5)
Ap , Bq , Cπ 2 C5 = A4 D B 3 B3
4
(23 .5, 24 .3, 7)
6
(23 .32 , 24 .5, 32 )
1
(7, 3.5, 22 )
3
(33 , 5.7, 2.3)
18
(24 .33 , 23 .5.11, 3.7)
5
(5.11, 32 .7, 23 )
7
(7.13, 32 .11, 2.5)
8
(24 .7, 23 .3.5, 11)
9
(33 .5, 11.13, 22 .3)
10
(25 .5, 23 .3.7, 13)
14
(23 .5.7, 25 .32 , 17)
11
(11.17, 3.5.13, 2.7)
21
(34 .7, 52 .23, 23 .3)
16
(25 .11, 23 .32 .5, 19)
20
(23 .5.13, 24 .3.11, 23)
30
(23 .33 .5, 26 .17, 3.11)
13
(13.19, 3.5.17, 24 )
15
(32 .5.7, 17.19, 2.32 )
17
(17.23, 3.7.19, 22 .5)
D4 A32 A5 B24 B3 6 B5 = A3 AD 2 4 2 2 3 B2 C3 D A3 A2 B 4 C 2 1 A7 = B 3 D = 1D26 B33 C22 3 B5 C22 2 D A32 B24 C3 B7 = A3 BD5 C3 2 C3 = A3 3 D 6 C2 3 D18 A22 A5 B25 B3 C3 B11 = A4 A3 BD318 = D4 D6 A3 2 3 2 B5 C3 C7 D5 D6 A3 5 A11 = A5 BD 2 B C 3 = D A3 A B 4 B 2 C 2 C 3 2 5 2 3 2 3 3 7 2 D4 D62 D7 C2 D7 = A13 = A7 B 2 B 6 3 C C D D D 11 2 5 1 2 18 A2 A3 A5 B2 B3 C3 3 D8 C22 D8 C11 = A4 A7 B 3 B3 B5 = D1 A4 B 3 2 2 2 2 B13 = A3 A5 BD119 C 2 C3 = D18 A2 AD24AD26BD5 9B3 C 2 C 2 3 2 2 3 5 2 2 3 A3 C2 = D3DA68DA10 C13 = A5 A5DB10 3B B 7B C B 3 7 5 3 3 2 2 2 2 D6 D14 C22 = C17 = A3 A5DA14 5 2 6 2 D1 A2 A3 A5 B3 B29 C32 7 B2 B3 2 A17 = A11 B3 BD511 B13 C2 C7 = 3 4 17 3 3 5 D3 D11 D18 A11 A A 2 3 5 B2 B3 C 2 C 3 = D42 D5 D63 D9 D A32 B24 B3 C3 B23 = A4 A7DB21 = 21 2 3 D1 D6 A23 C2 3 5 C2 C3 D3 D16 A2 A3 A5 B25 C22 C33 D16 C19 = A5 A11 B 3 B 2 B5 = D5 D62 2 2 3 D1 D2 D20 A23 C32 D20 C23 = A3 A5 A13 B 4 B3 B11 = D6 D7 A4 A5 B 3 B3 C2 2 2 2 2 A2 30 = D8 AD31AD530 B17 = A3 A3 AD 6C C 3C2C B B 5 3 11 2 2 3 3 2 2 3 D2 D8 D13 D18 A32 A63 A25 B213 C36 D13 A19 = A13 B3 B5 B17 C 4 = D4 D63 D7 D30 C23 2 D6 D8 D15 B3 C23 D15 B19 = A2 A5 A7 B17 C2 C 2 = D2 D 4 3 3 3 1 30 A2 A3 B2 C3 D17 A23 = A17 B3 B7 B19 C 2 C5 = 2 D12 D42 D5 D63 D9 D17 D30 = D2 D2 D8 D11 6 D18 D15 A2 A3 A45 B217 B34 C27 C33 3
2
2
C7 =
Table 2 Now, by using the above relations for n = 12, 19, 22, 24, 32, 42, 46 and n = 48, we will get 8 equations. For n = 12, we have D12 = A32 A33 B25 B7 C3 C5 = quently (3.1)
F1 :=
D2 D3 A22 A23 B26 C32 . D6 C2 B3
D6 D12 A2 A2 B 6 C 2 = 2 3 2 3 ∈ D. D2 D3 B3 C 2
Conse-
369
On real valued additive functions modulo 1
For n = 19, we infer from A19 , B7 , B23 , C11 and D19 = A25 A19 B3 B7 B23 C2 C11 =
D2 D3 D82 D13 D18 D2 1A52 A33 A45 B218 B32 C38 D12 D4 D65 D7 D30 C22
that (3.2)
F2 :=
D19 D12 D4 D65 D7 D30 A5 A3 A4 B 18 B 2 C 8 = 2 3 5 22 3 3 ∈ D. 2 D2 D3 D8 D13 D18 D21 C2
As, D22 = A32 A7 A11 B24 B3 B13 C52 =
D1 D22 D4 D5 D6 D9 A3 , D3 D18 A72 A35 B27 B35 C3 C26
we have, (3.3)
F3 :=
D3 D18 D22 A3 ∈ D. = 7 3 7 5 6 D1 D22 D4 D5 D6 D9 A2 A5 B2 B3 C2 C3
Similarly, we get from B7 and B13 that D24 = A42 A23 A5 B23 B7 B13 C33 =
D3 D4 D9 A52 B22 C32 . D18 A3 A5 B3 C23
This implies that, (3.4)
F4 :=
A52 B22 C32 D18 D24 = ∈ D. D3 D4 D9 A3 A5 B3 C23
For n = 32, we get from A19 , B17 , C5 , C7 that D32 = A62 A19 B23 B32 B17 C5 C7 =
D1 D22 D13 D18 A32 A33 B26 C35 , D63 D7 C25
which gives (3.5)
F5 :=
D63 D7 D32 A3 A3 B 6 C 5 = 2 3 5 2 3 ∈ D. 2 D13 D18 D2 D1 C2
For n = 42, 46, and 48, we get the following equations:
(3.6)
(3.7)
F6 :=
F7 :=
D4 D63 D42 A9 A3 A5 B213 C36 = 2 ∈ D, D2 D18 D21 C23
2 D46 1 D22 D11 D32 D15 D8 D18 = 16 6 7 34 6 6 10 ∈ D, 7 5 D7 D1 D6 D4 D17 D5 D9 D30 A2 A3 A5 B2 B3 C2 C3
370
K. Chakraborty, I. K´ atai and B.M. Phong
(3.8)
F8 :=
D1 D18 D48 1 = 9 4 3 19 2 7 ∈ D. 4 D4 D6 D9 D14 A2 A3 A5 B 2 B 3 C 3
The solutions of the above equations (3.1)-(3.8) can be obtained now as follows: we express B3 from (3.1), similarly A5 from (3.4), A3 from (3.6). Therefore, we get C2 from (3.3) and (3.6), C3 from (3.3) and (3.5), B2 from (3.2) and (3.5). Finally the solution A2 can be gotten from (3.3) and (3.7). The solutions are: A2 =
F111 F27 F417 F719 F1838 F2503 F41174 F71178 F31816 F51812 F64536 F86342 , A = , A = , 3 5 F310 F511 F625 F839 F3672 F5673 F61679 F82357 F12274 F21362 F43176 F73173 B2 =
F3125 F5127 F6282 F8452 F11526 F2919 F42158 F72090 , B = , 3 F1142 F283 F4185 F7228 F31203 F51199 F63055 F84186
C2 =
F1431 F2250 F4554 F7684 F1181 F2103 F4222 F7304 , C = . 3 F3165 F5167 F6351 F8598 F3377 F5380 F6845 F81352
They are elements of D and so Lemma 5 is proved.
Proof of Theorem 3. Let f0 , f1 , f2 , f3 , f4 ∈ A∗ and, Bf (n) = f0 (n) + f1 (n + 2) + f2 (n + 3) + f3 (n + 4) + f4 (n + 6) ≡ 0
(mod 1)
for all n ∈ Z. Then Bf (−n−6) = f4 (n)+f3 (n+2)+f2 (n+3)+f1 (n+4)+f0 (n+6) ≡ 0 Let ψ0 (n) := f0 (n) − f4 (n) Thus, we have,
and ψ1 (n) := f1 (n) − f3 (n)
ψ0 (n) + ψ1 (n + 2) − ψ1 (n + 4) − ψ0 (n + 6) ≡ 0
(mod 1)
(mod 1). n ∈ Z.
for all
n ∈ Z.
for all
From Lemma 1 we have that ψ0 (n) ≡ ψ1 (n) ≡ 0 (mod 1), consequently f0 (n) ≡ ≡ f4 (n) (mod 1) and f1 (n) ≡ f3 (n) (mod 1) for all n ∈ Z. Hence, Bf (n) ≡ f0 (n) + f1 (n + 2) + f2 (n + 3) + f1 (n + 4) + f0 (n + 6) ≡ 0
(mod 1)
is true for all n ∈ Z. The conditions of Lemma 4 and Lemma 5 are satisfied by taking bj (n) = fj (n) (j = 0, 1, 2) and S(n) = b0 (n) + b1 (n + 2) + b2 (n + 3) + b1 (n + 4) + b0 (n + 6). Thus f0 (n) ≡ f1 (n) ≡ f2 (n) ≡ f3 (n) ≡ f4 (n) ≡ 0 holds for all n ∈ Z and this completes the proof.
(mod 1)
371
On real valued additive functions modulo 1
From Lemma 4 and Lemma 5 we obtain Theorem 6. If D denotes the subgroup of Q3+ generated by the sequences Dn = n(n + 6), (n + 2)(n + 4), n + 3 (n ∈ N), then we have D = Q3+ .
4.
Proof of Theorem 4
Lemma 6. Let c0 , c1 , c2 ∈ A∗ . If (4.1)
c0 (n) + c1 (n + 1) + c2 (n + 3) + c1 (n + 5) + c0 (n + 6) ≡ 0
(mod 1)
for all n ∈ N, then (4.2)
c0 (n) ≡ c1 (n) ≡ c2 (n) ≡ 0
(mod 1)
for
n ∈ N.
Proof. In order to prove Lemma 6, we shall use the following fact: (4.3)
If (4.1) holds for all n ∈ N, then (4.2) holds for n ≤ 11.
This can be shown in the same way as we proved Lemma 3 and lemma 5. Let, T (n) = c0 (n) + c1 (n + 1) + c2 (n + 3) + c1 (n + 5) + c0 (n + 6) ≡ 0
(mod 1).
Let n0 be the smallest positive integer n for which cj (n) 6≡ 0 (mod 1) for at least one j. Then n0 is a prime p and p > 11. It is easily seen that T (p − 5) ≡ 0 (mod 1) and T (p − 6) ≡ 0 (mod 1) imply that c0 (p) ≡ c1 (p) ≡ ≡ 0 (mod 1). Let c2 (p) ≡ ν 6≡ 0 (mod 1), then T (p − 3) ≡ 0 (mod 1) implies that c1 (p + 2) ≡ −ν (mod 1). From T (p + 1) ≡ 0 (mod 1) we have that c1 (p + 6) ≡ ν (mod 1) and from T (p + 5) ≡ 0 (mod 1) we have c2 (p + 8) ≡ −ν (mod 1). As p ≡ 2 (mod 3), and < p. It is obvious from p, p + 2, p + 6, p + 8 ∈ P so 3|p + 10, 2|p + 11 and p+11 2 that p ≡ 1 (mod 5). We have 0 ≡ T (2p − 3) ≡ c0 (2p − 3) + c2 (p) (mod 1), thus c0 (2p − 3) ≡ −ν (mod 1). Let us consider now 0 ≡ T (2p − 6j − 3) ≡ 0
(mod 1)
372
K. Chakraborty, I. K´ atai and B.M. Phong
for j = 1, 2, 3, 4, 5. Since 2|2p − 6j − 2, 2|2p − 6j, 2|2p − 6j + 2, we have c1 (2p − 6j − 2) + c2 (2p − 6j) + c1 (2p − 6j + 2) + c0 (2p − 6j + 3) ≡ 0
(mod 1),
and so c1 (2p − 6j − 3) + c1 (2p − 6j − 3) ≡ 0
(mod 1) (j = 1, 2, 3, 4, 5).
Hence c0 (2p − 9) ≡ ν (mod 1), c0 (2p − 15) ≡ −ν (mod 1), c0 (2p − 21) ≡ ≡ −ν (mod 1), c0 (2p − 27) ≡ ν (mod 1), which with 5|2p − 27 implies that ν = 0. Proof of Theorem 4. Let f0 , f1 , f2 , f3 , f4 ∈ A∗ and, Cf (n) = f0 (n) + f1 (n + 1) + f2 (n + 3) + f3 (n + 5) + f4 (n + 6) ≡ 0
(mod 1)
for all n ∈ Z. Then Bf (−n−6) = f4 (n)+f3 (n+1)+f2 (n+3)+f1 (n+5)+f0 (n+6) ≡ 0
(mod 1).
Let κ0 (n) = f0 (n) − f4 (n)
and κ1 (n) := f1 (n) − f3 (n)
for all
n ∈ Z.
Thus, we deduce from the above relations that κ0 (n) + κ1 (n + 1) − κ1 (n + 5) − κ0 (n + 6) ≡ 0
(mod 1)
for all
n ∈ Z.
From Lemma 1 we have κ0 (n) ≡ κ1 (n) ≡ 0 (mod 1), and so f0 (n) ≡ f4 (n) (mod 1) and f1 (n) ≡ f3 (n) (mod 1) for all n ∈ Z. Hence, Cf (n) ≡ f0 (n) + f1 (n + 1) + f2 (n + 3) + f1 (n + 5) + f0 (n + 6) ≡ 0
(mod 1)
is true for all n ∈ Z. Thus the conditions of Lemma 6 are satisfied by taking cj (n) = fj (n) (j = 0, 1, 2) and T (n) = c0 (n) + c1 (n + 1) + c2 (n + 3) + c1 (n + 5) + c0 (n + 6). Thus f0 (n) ≡ f1 (n) ≡ f2 (n) ≡ f3 (n) ≡ f4 (n) ≡ 0 holds for all n ∈ Z.
(mod 1)
Thus we obtain (from the last lemma), Theorem 7. If T denotes the subgroup of Q3+ generated by the sequences Tn = n(n + 6), (n + 1)(n + 5), n + 3 (n ∈ N), then we have T = Q3+ .
373
On real valued additive functions modulo 1
References
[1] Elliott, P.D.T.A., Arithmetic Functions and Integer Products, SpringerVerlag, New York, (1985). [2] Erd˝ os, P., On the distribution function of additive functions, Ann. Math., 47 (1946), 1–20. [3] K´ atai, I., On arithmetical functions with regularity properties, Acta Sci. Math., 45 (1983), 253–260. [4] K´ atai, I., On additive functions satisfying a congruence, Acta Sci. Math., 47 (1984), 85–92. [5] K´ atai, I., A remark on additive functions satisfying a relation, Ann. Univ. Sci. Budapest., Sect. Math., 27 (1985), 229–233. [6] K´ atai, I., Arithmetical functions satisfying some relations, Acta Sci. Math., 55 (1991), 249–268. [7] K´ atai,I. and M. van Rossum-Wijsmuller, Additive functions satisfying congruences, Acta Sci. Math., 56 (1992), 63–72. [8] Kov´ acs, A. and B.M. Phong, On completely additive functions satisfying a congruence, Ann. Univ. Sci. Budapest., Sect. Comp., 19 (2000), 237–249. [9] van Rossum-Wijsmuller, M., Additive functions on the Gaussian integers, Publ. Math. Debrecen, 38 (1991), 255–262. [10] Styer, R., A problem of K´ atai on sums of additive functions, Acta Sci. Math., 55 (1991), 269–286. [11] Wirsing, E., Tang Yuansheng and Shao Pintsung, On a conjecture of K´ atai for additive functions, J. Number Theory, 56 (1996), 391–395.
K. Chakraborty Department of Mathematics Harish-Chandra Research Institute Allahabad 211 019 Chhatnag Road, Jhusi, India
[email protected]
I. K´ atai and B.M. Phong Department of Computer Algebra E¨otv¨os Lor´and University H-1117 Budapest P´azm´any P´eter S´et´any 1/C Hungary
[email protected] [email protected] 
