Problem: For a positive integer $n$, let $\phi (n)$ be the number of positive integers which is coprime to $n$. If $n$ is an even integer prove that $\phi (2n) = 2 \phi (n)$.
If $\gcd(a,b)=1$, then $\phi(ab) = \phi(a)\cdot \phi(b)$.
For any prime $p$, $\phi(p^k) = p^{k-1}(p-1)$.
Solution: Since $n$ is even, let us write $n = 2^k m$, where $m$ is an odd number. Then the number of positive integers which are coprime to $n$ is given by
\begin{align*}
\phi \left( 2^k m \right) & = \phi(2^k) \cdot \phi(m) \\
& = 2^{k-1} \cdot \phi(m).
\end{align*}
Also,
\begin{align*}
\phi (2n) & = \phi \left( 2 \cdot 2^k m \right) = \phi \left( 2^{k+1} m \right) \\
& = 2^{k} \cdot \phi(m) = 2 \cdot 2^{k-1} \cdot \phi(m) \\
& = 2 \phi (n).
\end{align*}