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

30 Pham Vari Chung for all positive integers m. Proof. Prom the multiplicativity of / and Lemma 2. we easy reduce that /(2m) = 2f(m) for all positive integers m. Thus, by (C) we have 2/(m

2) = /(2m

2) = /(m

2 +m

2) = 2(/(m))

2

, from which /(m

2) = (/(m))

2 follows. Lemma 4. If / fulfills the hypotheses of the theorem, then we have (/(m))

2 = m

2 for all positive integers m. We note that from (C

) and Lemma 3 condition (A) follows. So we have, using Lemma 2, that (AI) - (A3) are satisfied. From this we have /(m

2) = m

2, which proves the lemma. But here we give another proof. Proof. We argue by induction on m. In the proof of Lemma 1. we have shown that (/(m))

2 = m

2 for m = 1,2 and 3. Suppose that n is an integer with n > 3, and (/(m))

2 = m

2 for all positive integers m < n. We will show that (/(n + 1)) = (n + 1) . If n + 1 is even, then n + 1 = 2

kh where h < n and h is odd. Thus, by the multiplicativity of /, Lemma 2. and the hypotheses of the induction, we obtain (f(n + l))

2 = {f(2

k))\f(h))

2 = 2

2

kh

2 = (n + l)

2. 2 2» To show that (/(n -f 1)) = (n + 1) , when n +

1 is odd, we use the equation (n - l)

2 + (n + l)

2 = 2(n

2 +1), where (2, n

2 + 1) = 1. From this and (C) we have (/(n + l))

2 = / [2(n

2 + 1)] -(/(n - l))

2 = /(2) [(/(n))

2 + (f(l))

2 (/(n — I))

2 - Since /(2) = 2, and the hypotheses of the induction, we reduce that (/(n + I))

2 = 2(n

2 + 1) - (n - l)

2 = (n + l)

2, which proves the lemma. Now we return to prove the theorem. 3. The proof of the theorem First we verify the necessity of the condition. Assume that / fulfills the condition of the theorem. Then, by Lemma 2., we have shown that (C

—1) is satisfied. Moreover, by using Lemmas 3. and 4., we obtain Hi

2") = (/(9

f

c))

2 = (I*)' =