if f and g are two multiplicative function that are not identically zero and satisfying f(pk)=g(pk) for each prime p and k>=1 prove f is congruence to g
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Thus \(f(1) = 1\) unless \(f\) is the zero function, and a multiplicative function is completely determined by its behaviour on the prime powers. Examples : We have seen that the Euler totient function \(\phi\) is mulitplicative but not totally multiplicative (this is one reason it is convenient to have \(\phi(1) = 1\)).
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