# $\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)$ Double Summing a (Not Completely) Multiplicative Function

Let $$f(n)$$ be a multiplicative function that is not completely multiplicative, i.e $$f(m)\cdot f(n)= f(m\cdot n)$$ only if $$gcd(m,n)=1$$. Let $$S(x)$$ be the double sum over $$f$$, that is:

$$S(x)=\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)$$

It is not difficult to see that if $$f(n)$$ were completely multiplicative, then $$S(x)$$ could be simplified:

$$S(x)=\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)= \sum_{i=1}^xf(i)\sum_{j=1}^xf(j)= \biggl(\sum_{k=1}^xf(k)\biggr)^2$$

But since $$f(n)$$ is not completely multiplicative, this simplification is not completely true, and it fails in every combination where $$gcd(i,j)\neq1$$. Hence, $$S(x)$$ can be written this way provided we add some additional error term, let’s call it $$E$$:

$$S(x)=\sum_{i=1}^x\sum_{j=1}^xf(i\cdot j)= \biggl(\sum_{k=1}^xf(k)\biggr)^2+E$$

$$E$$ is either negative or positive, I’m not sure. Obviously, $$E$$ is comprised of all the small errors generated by the initial sum term, when $$gcd(i,j)\neq1$$. I am mainly interested in the cases where $$f(n)$$ takes the form of:

1. Euler totient function: $$S_{\varphi}(x)=\sum_{i=1}^x\sum_{j=1}^x\varphi(i\cdot j)$$
2. Sum of divisors function: $$S_{\sigma_1}(x)=\sum_{i=1}^x\sum_{j=1}^x\sigma_1(i\cdot j)$$
3. Moebius function: $$S_{\mu}(x)=\sum_{i=1}^x\sum_{j=1}^x\mu(i\cdot j)$$

My question is, what is this error term $$E$$ exactly? how can I calculate it? How can I properly sum all those small errors to get a correct evaluation of $$S(x)$$? For clarification, I am concerned with evaluating $$S(x)$$, but I think I must evaluate $$E$$ first in order to do it. I am taking this approach because I can compute $$\biggl(\sum_{k=1}^xf(k)\biggr)^2$$ very efficiently, and so, finding the error term $$E$$ will solve my question.