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 $a(1) = 0$.
Right, you're not allowed to divide by $a(1)$ because it is zero. – Ben Nov 25 '12 at 2:57
@boby: It does imply $a(1)$ must be zero for the exact reason Ben gave above: $$a(1) = 2a(1) \iff 2 a(1) - a(1) = 0 \iff a(1) = 0.$$ In general it is a bit sloppy to divide by an unknown, derive a contradiction, and then conclude the unknown must have been zero. It is far easier (and less error-prone) to subtract one side from another and factor. – JavaMan Nov 25 '12 at 2:59
In the argument JavaMan and I are trying to make, we don't make an assumption that $a(1)$ is nonzero and derive a contradiction; we prove directly that $a(1)$ is zero. We see that the definition of additive implies that $a(1) = 2a(1)$, and subtracting $a(1)$ (which is always legal!) gives us that $a(1) = 0$. – Ben Nov 25 '12 at 3:09
In this case $a(1)$ does happen to be zero (which we show without any contradictions), so there is no need to divide by it. It is acceptable (although unusual, as JavaMan remarks) to say that we have $2a(1) = a(1)$, assume $a(1)$ is nonzero, and divide by it to get a contradiction. The reason why the contradiction must come from assuming $a(1) \neq 0$ is that we haven't made any other assumptions. – Ben Nov 25 '12 at 3:25