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

Multiplicative functions satisfying 29 On the other hand, from (C), the multiplicativity of /, and the equation 52 = 4

2 + 6

2, the equation (2) (/(4))

2 = /(52) - (/(6))

2 = /(13)/(4)~ (/(2)) (/(3)) = /(13)/(4) — 2

23

2 follows. But, by using (C) and the fact 13 = 3

2 +2

2, we have /(13) = (/(3))

2 + (/(2))

2 = 9 + 4 = 13. So, from (2), it follows that (3) (/(

4))

2 = 13/(4) - 36. Thus, by (1) and (3), we have 5/(4) - 4 = 13/(4) - 36. Consequently, /(4) = 4. So, the lemma is proved. Lemma 2. If / fulfills the hypotheses of the theorem, then we have f(2

k) = 2

k for all integers k > 0. Proof. By Lemma 1, we have /(2

k) = 2

k for A: = 1,2, and it is clear for cases k — 0 and k = 3. Assume that n is an integer with n > 3, and that we have f(2

k) = 2

k for all integers 1 < k < n. We will show that /(2

n+

1) = 2

n+

1. If n + 1 is even then n + 1 = 2k, where k + 1 < n. Thus, (C), the multiplicativity of /, /(5) = 5, and the hypotheses of the induction yield 5/(2

n+

1) = /(5 • 2

n+

1) = /(2

2k+

2 + 2

2

k) = (/(

2

k+

1))

2 + (^^))

2 _ 2

22

2

k — 5 2

2

k which gives /(

2

n+

1) = 2

n+

1

. So, it remains to show that f(2

n+

1) = 2

n+

1

, when n + 1 is odd. If n + 1 is odd, then n + 1 = 2k + 1 where k < n. By using (C) and the hypotheses of induction, we obtain /(

2

n+

1) = /(2

2

k + 2

2A

:) = 2(/(2

f

c))

2 = 2-2

2f

c = 2

n+

1. Thus, the lemma is proved. Lemma 3. If / satisfies the hypotheses of the theorem, then we have /(m

2) = (/(m))

2