For what integers $n$ does $\phi(2n) = \phi(n)$? For what integers $n$ does $\phi(2n) = \phi(n)$?
Could anyone help me start this problem off? I'm new to elementary number theory and such, and I can't really get a grasp of the totient function.
I know that $$\phi(n) = n\left(1-\frac1{p_1}\right)\left(1-\frac1{p_2}\right)\cdots\left(1-\dfrac1{p_k}\right)$$ but I don't know how to apply this to the problem. I also know that $$\phi(n) = (p_1^{a_1} - p_1^{a_1-1})(p_2^{a_2} - p_2^{a_2 - 1})\cdots$$
Help

Hint If $n$ is odd, then gcd$(n,2)=1$ thus 
$$\phi(2n)=\phi(2) \phi(n) \,.$$
If $n$ is even, write $n=2^km$ with $m$ odd and $k \geq 1$.
$$\phi(n)=\phi(2^k) \phi(m) \,.$$
$$\phi(2n)=\phi(2^{k+1}) \phi(m) \,.$$
