Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1995-1996. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 23)

CHUNG, P. v., Multiplicative functions satisfying the equation f(m2+n2) = (f(m))2 + (f(n))2

28 Pham Van Chung (a-3) f(q

2

k)= 1 for all primes q = 3 (mod 4) and all integers k > 1. In the present paper we consider a similar question, and we give solutions of multiplicative functions / satisfying the equation (B) f(m

2 + n

2) = (f(n))

2 + (f(n))

2 for all positive integers m and n. Investigating this question we have same result by using the method of [1]. Our main result is to prove the following. Theorem. Let / / 0 be a multiplicative function. The / fulfills the condition (C) f(m

2+n

2) = (f(m))

2 +(f(n))

2 for all positive integers m and n if and only if (C-l) /(2

f

c) = 2

f

c for all integers k > 0, (C-2) f(p

k) = p

k for all primes p ~ 1 (mod 4) and all integers k > 1, and (C-3) f(q

2

k) = q

2

k for all primes q = 3 (mod 4) and all integers k > 1. We shall prove our theorem in the next two sections. First in Section 2, we verify some auxiliary lemmas. Finally, in Section 3, we give the proof of the theorem. 2. Lemmas Lemma 1. If / satisfies the hypotheses of the theorem, then we have /(2) - 2 and /(4) = 4. Proof. Since / ^ 0 multiplicative, we have /(1) = 1. Thus, (C) yields /(2) = /(l

2 + l

2) - (/(l))

2 + (/(l))

2 — 1 + 1 = 2. Therefore, by using the equation 5 = 2

2 + l

2, (C) implies that /(5) = (/(2))

2 + (/(I))

2 = 5. So, we conclude from (C) and the multiplicativity of / that (/(3))

2 = /(10) — (/(

l))

2 = 9. The fact that 20 = 4

2 + 2

2, coupled with /(2) = 2, /(5) = 5 and (C), forces (1) (/(4))

2 = /(20) - (/(2))

2 = 5/(4) - 4.