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If an additive function $a(x)$ with the property that $a(xy)=a(x)+a(y)$, when $\gcd(x,y)=1$, then is $a(1)=0$, I think it should be, otherwise I would have for example $a(1)=a(1)+a(1)..., a(1)=c\cdot a(1)$, which implies $1$ is equal to everything?

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To rephrase your argument: we have $a(1) = a(1\cdot 1) = a(1) + a(1)$. Subtracting $a(1)$ from both sides, we get $a(1) = 0$.

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    In fact, concluding $a(1) = 0$ also depends on $a(x)$ being a valid definition of a function, although in this case $a(x)=0$ is an obviously valid example of such.2012-11-27