Find all the positive integers $n$ such that $\phi(4n) = 2\phi(n)$.

Question

Find all the positive integers $n$ such that $\phi(4n) = 2\phi(n)$.

I know that when $n$ is odd you have that

$\phi(4n) = \phi(4)\phi(n) = \phi(2^ 2 )\phi(n) = 2\phi(n)$

I'm not sure how to show it for if $n$ is even to show that it wont have a solution if $n$ is even. n

Answer 1

You can use the formula $\phi(n)$ = $n(1-\frac{1}{p_1}) \cdots (1-\frac{1}{p_k})$ where $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$. Now note that if $n$ is even then $4n$ and $n$ has same set of prime divisiors. Hence $\phi(4n)$ = $4n(1-\frac{1}{p_1}) \cdots (1-\frac{1}{p_k}) = 4 \phi(n)$. So no solution.

Answer 2

Hint: if $n$ is even, then we can write it as $2^a \cdot b$, where $b$ is odd. Now you can compute both $\phi(n)$ and $\phi(4n)$ in terms of $\phi(b)$.