Given that $\gcd(m, n) = 2$, prove that $\phi (mn) = 2 \phi (m) \phi (n)$.
We're given $\gcd(m,n)=2$. This means there exist coprime integers $a$ and $b$ such that $m=2a$ and $n=2b$. Because $a$ and $b$ are coprime, they share no prime factors.
Let's write the prime decomposition of $a$ and $b$,
$$ \begin{align} a & = p_1^{j_1} \times p_2^{j_2} \times \ldots \times p_r^{j_r} \\ \\ b & = q_1^{k_1} \times q_2^{k_2} \times \ldots \times q_s^{k_s} \end{align} $$
Since $a$ and $b$ don't share any prime factors, we have $p_x \ne q_y$.
Without loss of generality, let's set $p_1=2$. This means there is no prime factor 2 in $b$.
$$ \begin{align} a & = 2^{j_1} \times p_2^{j_2} \times \ldots \times p_r^{j_r} \\ \\ b & = q_1^{k_1} \times q_2^{k_2} \times \ldots \times q_s^{k_s} \end{align} $$
And so
$$ \begin{align} m = 2a & = 2^{j_1+1} \times p_2^{j_2} \times \ldots \times p_r^{j_r} \\ \\ n = 2b & = 2 \times q_1^{k_1} \times q_2^{k_2} \times \ldots \times q_s^{k_s} \end{align} $$
We now consider $\phi$,
$$ \begin{align} \phi(mn) & = \phi \bigl( (2a)(2b) \bigr) \\ \\ & = \phi(2^2ab) \\ \\ & = \phi(2^{j_1+2} \times \underbrace{ p_2^{j_2} \times \ldots \times p_r^{j_r}}_{a/2^{j_1}} \times \underbrace{q_1^{k_1} \times q_2^{k_2} \times \ldots \times q_s^{k_s}}_{b}) \\ \\ & = \phi(2^{j_1+2}) \times \phi(a/2^{j_i}) \times \phi(b) \tag{i} \\ \\ & = 2^{j_1 + 1} \times \phi(a/2^{j_1}) \times \phi(b) \\ \\ &=2^{j_1 + 1} \times \frac{1}{2^{j_1}} \phi(2^{j_1+1} \times a/2^{j_1}) \times \frac{1}{1} \phi(2b) \\ \\ & = 2 \times \phi(2a)\phi(2b) \\ \\ & = 2\phi(m)\phi(n) \end{align} $$
Step (i) is justified because $(2^{j_1+2})$, $(p_2^{j_2} \times \ldots \times p_r^{j_r})$ and $(q_1^{k_1} \times q_2^{k_2} \times \ldots \times q_s^{k_s})$ are all co-prime to each other, allowing us to use the multiplicity of $\phi$.
We have shown that $\gcd(m, n) = 2$ implies $\phi (mn) = 2 \phi (m) \phi (n)$.