Show that function $E$ is non-multiplicative. Definition 1:
A function $f$ is said to be non-multiplicative if
$$f(ab)\ne f(a)f(b)$$
for all coprime integers $a,b>1$.
Definition 2:
We define the function $E$ as
$$E(n)= n+1-\tau (n)- \phi(n)$$
$$=\sum_{(n,d)\notin\{1,d\} \\ \ \ \ \ 1<d<n}1.$$
Here, $(n,d)$ denotes gcd$(n,d)$
The values of $E(n)$ A045763

Question:
Show that the function $E$ is non-multiplicative.

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Example:
Let $a=10$, $b=7$, $ab=n=70$.
We have $E(70)=39$, $E(10)=3$, $E(7)=0$, and
$E(70)\ne E(10)E(7)$.

First recall that $n = \sum_{d | n} \phi(d)$, so using this we have 
$$E(n) = \sum_{d | n} \phi(d) - \sum_{d | n} 1 - (\phi(n) - 1) = \sum_{d | n,\, d < n} (\phi(d) - 1)$$
Now note that (i) for coprime $a, b$, and $d | ab$, we have a unique decomposition $d = d_1 d_2$ for $d_1 | a$ and $d_2 | b$, and that (ii) for coprime $d_1, d_2$ we have 
$$\phi(d_1d_2) - 1 = \phi(d_1)\phi(d_2) - 1 \geq (\phi(d_1) - 1)(\phi(d_2) - 1),$$ 
since $\phi(d_1), \phi(d_2) \geq 1$. Using these facts, it follows that for coprime $a, b > 1$, we have
\begin{align*}
E(ab) 
&= \sum_{d | ab,\, d < ab} (\phi(d) - 1) \\
&= \sum_{\substack{d_1 | a,\, d_2 | b \\ (d_1, d_2) \neq (a, b)}}(\phi(d_1d_2) - 1) \\
&> \sum_{\substack{d_1 | a,\, d_2 | b \\ d_1 < a,\, d_2 < b}}(\phi(d_1d_2) - 1) \\
&\geq \sum_{\substack{d_1 | a,\, d_2 | b \\ d_1 < a,\, d_2 < b}}(\phi(d_1) - 1)(\phi(d_2) - 1) \\
&= E(a)E(b)
\end{align*}
where the first inequality is strict because the difference is 
$$\sum_{d_1 | a,\,d_1 < a} (\phi(d_1b) - 1) + \sum_{d_2 | b,\,d_2 < b} (\phi(ad_2) - 1) \geq \phi(b) + \phi(a) - 2 > 0$$
since at least one of $a, b$ is divisible by a prime $\geq 3$. Thus $E(ab) > E(a)E(b)$ for coprime $a, b > 1$, so $E$ is non-multiplicative.
