Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2001. Sectio Mathematicae (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 28)
Bui MINH PHONG, Multiplicative functions satisfying a congruence property IV
40 Bui Minli Phong and so by (18) we have (19) /(1 + tp) - /(1) = 0 (mod p) for all t 6 N and for every prime p > poLet q,r be distinct primes and let a(q ) > a(r). Then there is a prime p such that p > max(p 0,9 a(3 )" a(r )) and qr s - 1 = 0 (mod p) for some positive integer s. Using (19), we have f(qr s) = /(1) = 1 (mod p) and f(qr s) = q aWr s a^ = q<i)-<r) (mod p), which implies a(p) = ci(q ) = a . Hence, f(n) = n a for all n £ N. This completes the proof of Theorem 2. 3. Proof of Theorem 1 Let f £ M* and f(n) f 0 for all n £ N. We denote by I f the set of all polinomials P with rational coefficients for which there exists a suitable non-zero integer Ap such that ApP(E)f{n + rn) = A PP(E)f(n ) (mod m) holds for all n, m £ N. By our assumption (6), we have If ^ 0. It is clear to check that (i) cP(x) £ Ij for every P £ If , c £ Q (ii) P(x) + P'{x) £ I f for every P, P' £ I f (iii) xP(x) £ If for every P £ If. Thus, (i)-(iii) show that If is an ideal in Q[x]. Let 5(«) = c 0 + c 1x + h c kx k (c k = 1) be a polynomial of minimum degree in If. If k — 0, then Theorem 1 follows from Theorem 2. In the following we assume that k > 1. Let From the fundamental theorem of symmetric polynomials it follows that for a fixed integer s > 1 the polynomial RRFLZL Ü j: - o.
