A characterization of the identity function

Transcription

1 Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, ) pp. 3 9 A characterization of the identity function BUI MINH PHONG Abstract. We prove that if a multiplicative function f satisfies the euation fn +m +3)=fn +1)+fm +) for all positive integers n m, then either fn) is the identity function or fn +m +3)=fn +1)=fm +)=0 for all positive integers. Throughout this paper N denotes the set of positive integers let M be the set of complex valued multiplicative functions f such that f1) = 1. In 199, C. Spiro [3] showed that if f M is a function such that fp + ) = fp) + f) for all primes p, then fn) = n for all n N. Recently, in the paper [] written jointly with J. M. de Koninck I. Kátai we proved that if f M, fp+n ) = fp)+fn ) holds for all primes p n N, then fn) is the identity function. It follows from results of [1] that a completely multiplicative function f satisfies the euation fn + m ) = fn ) + fm ) for all n,m N if only if f) =, fp) = p for all primes p 1 mod 4) f) = or f) = for all primes p 3 mod 4). The purpose of this note is to prove the following Theorem. Assume that f M satisfies the condition 1) f n + m + 3 ) = fn + 1) + fm + ) for all n, m N. Then either ) fn + 1) = fm + ) = fn + m + 3) = 0 for all n,m N, or fn) = n for all n N. Corollary. If f M satisfies the condition 1) fn 0 + 1) 0 for some n 0 N, then fn) is the identity function. First we prove the following lemma. Lemma. Assume that the conditions of Theorem 1 are satisfied. Then either ) is satisfied for all n N or the conditions 3) fn + 1) = n + 1, fn + m + 3) = n + m + 3 fm + ) = m + Research partially) supported by the Hungarian National Research Science Foundation, Operating Grant Number OTKA 153 T 0095.

2 4 Bui Minh Phong simultaneously hold for all n, m N. Proof. From 1), we have for all n,m N, so fn + 1) + fm + ) = fm + 1) + fn + ) 4) fn + ) fn + 1) = f3) f):= D for all n N. Thus, the last relation together with 1) implies that 5) fn + m + 3) = fn + 1) + fm + 1) + D holds for all n,m N. Let S j := fj + 1). It follows from 5) that if k,l,u v N satisfy the condition k + l = u + v, then fk + 1) + fl + 1) + D = fu + 1) + fv + 1) + D, which shows that 6) k + l = u + v implies S k + S l = S u + S v. We shall prove that 7) S n+1 = S n+9 + S n+8 + S n+7 S n+5 S n+4 S n+3 + S n holds for all n N. Since j + 1) + j ) = j 1) + j + ) j + 1) + j 7) = j 5) + j + 5), we get from 6) that 8) S j+ S j = S j S j+ S j+ S j 7 = S j 5 + S j+5.

3 These with 8) imply that A characterization of the identity function 5 S j+5 S j+ + S j S j 7 = S j 1 S j 5 = S j+1 S j 3 + S j 3 S j 5 = S j+1 S j 3 + S j S j 4, which proves 7) with n = j 7. By 8), we have S 7 = S 3+1 = S 5 S 1, S 9 = S 4+1 = S 7 + S 6 S = S 6 + S 5 S S 1 S 11 = S 5+1 = S 9 + S 7 S 3 = S 6 + 4S 5 S 3 S S 1. Finally, by using 6) the facts = 7 + 4, = = 9 + 8, we have S 8 = S 7 + S 4 S 1 = S 5 + S 4 S 1, S 10 = S 1 S S 5 = S 6 + 3S 5 S 3 S 1 S 1 = S 9 + S 8 S 1 = S 6 + 4S 5 + S 4 S 4S 1. Thus, to complete the proof of the lemma, by using 1), 4), 5) 7), it is enough to prove that either S 1 = S = S 3 = S 4 = S 5 = S 6 = 0 or 9) S j = j + 1 for j = 1,,3,4,5,6. Repeated use of 1), using the multiplicativity of f, gives S 1 = f1 + 1) = f), 10) S = f + 1) = f5) = f ) = f) + f3), 11) S 3 = f3 + 1) = f10) = f)f5) = f) + f)f3). thus f11) = f + + 3) = f5) + f6) = f) + f3) + f)f3).

4 6 Bui Minh Phong On the other h, it follows from 4) that f11) = f3 + ) = f10) + D = f)f5) + D = f) + f)f3) + f3) f), which, together with the last relation, implies 1) f) = f), f13) = f ) = f) + f11) = f) + f)f3) + f3). Finally, the relation 10) together with the fact f8) = f ) = f) + f6) = f5) + f3) show that 13) f)f3) = f3). Moreover 14) S 5 = f5 + 1) = f6) = f)f13) = 4f) + 6f3), 15) S 6 = f6 + 1) = f37) = f ) = f11) + f6) = 5f) + 9f3), 16) f17) = f ) D = f35) D = f5)f7) D, 17) f3)f7) = f1) = f ) = f10) + D = 3f) + 5f3). The euation 1) shows that either f) = 0 or f) =. Assume that f) = 0. Then 13) implies that f3) = 0 so, by using 10) 17) we have S 1 = S = S 3 = S 4 = S 5 = S 6 = 0, from which follows that ) is true.

5 A characterization of the identity function 7 Assume now that f) =. In this case we have f5) = + f3), f8) = + f3). We shall prove that f3) = 3. It follows from 15) using the fact that f37) = f ) f3) = f5)f8) f3) = f3) + 5f3) ) f3) 4f3) 6 = 0. On the other h, from 4) we infer that conseuently f6)f11) f3)f13) = f66) f65) = f3) f), 3f3) 7f3) 6 = 0. This together with 17) proves that f3) = 3, so 10) 17) imply that S j = j + 1 j = 1,,3,4,5,6). This completes the proof of 9) so the lemma is proved. Proof of the theorem In the proof of the theorem, using the lemma, we can assume that 3) is satisfied, that is 18) fn + 1) = n + 1, fn + m + 3) = n + m + 3. fm + ) = m + It is clear from 18) that fn) = n for all n 7. Assume that fn) = n for all n < T, where T > 7. We shall prove that ft) = T. It is clear that T must be a prime power, that is T = α with α N some prime. It is easily seen that if α = 1, then > 7 there are positive integers n,m 1 such that n + m + 3 = N,,N) = 1 N <. Thus, we have f) =. Assume now that α > 3. We consider the congruence n + m mod α ). Let { A 3):= 1 m 1: m ) 3 } = 1.

6 8 Bui Minh Phong Then we have A 3) = = 1 1 m=1 m +3,)=1 1 m=0 m m )) 3 = m 3 ) 1 m=0 )) 1 3 )). 1 )) 3 m )) 3 By our assumption, the last relation implies that A 3) 1. Thus, there are integers m {1,..., 1}, 1 n 1 α 1, n 1,) = 1 1 n := α n 1 α 1 such that n i + m + 3 = α N i i = 1,). It follows from the above relations that that is α N N 1 ) = α n 1 ) n 1 = α α n 1, N N 1 = α n 1. Since n 1,) = 1, we obtain that at least one of N 1 or N is coprime to. Let n {n 1,n } N {N 1,N } such that n + m + 3 = α N, N,) = 1. Then α implies that Thus, N 1 [ ] α α 1) + 1) + 3 < α. Nf α ) = fn)f α ) = fn α ) = fn + m + 3) = n + m + 3 = N α, which shows that f α ) = α as we wanted to establish. To complete the proof of the theorem, it remains to consider the cases = = 3. Let = T = α, where α 3. Since 7 1 mod 8), we have 7 is a uadratic residue modulo α therefore there exists n α [0, α 1 1] such that n α +7 = n α mod α ), conseuently, [n α + α 1 ] mod α ). Define N 1,

7 A characterization of the identity function 9 N by n α + 7 = α N 1 [n α + α 1 ] + 7 = α N. We easily deduce from these two euation the fact 7 < T = α that N 1 < α, N < α N N 1 = n α + α. It follows from the last relation the fact does not divide n α that one of N 1 or N is odd, so f α ) = α. Finally, let = 3 T = 3 α, where α > 1. We consider the congruence n + 0 mod 3 α ). Similarly as above, one can deduce that there are positive integers n,n N such that n + = 3 α N, N,3) = 1 N < 3 α. Thus these together with 18) implies that f3 α ) = 3 α. The theorem is proved. References [1] P. V. Chung, Multiplicative functions satisfying the euation fm + n ) = fm ) + fn ), Math. Slovaca, ), No. 3, [] J.-M. De Koninck, I. Kátai B. M. Phong, A new charateristic of the identity function, J. Number Theory to appear). [3] C. Spiro, Additive uniueness set for arithmetic functions, J. Number Theory 4 199), Bui Minh Phong Department of Computer Algebra Eötvös Loránd University Múzeum krt. 6-8 H-1088 Budapest, Hungary