Equality of the totient function of two multiples of $x$ I'm looking to solve for $x \in \mathbb{N}$ in the equation $\phi(4 x) = \phi(5 x)$. I know the totient function $\phi(y)$ just gives the number of integers less than or equal to $y$ that are coprime to $y$. I tried approaching it like a normal equation and expanding out $\phi(4 x) - \phi(5 x) = 0$ into its prime number decomposition, but I didn't get anywhere. Any ideas? Graphing it, I noticed the equation seems to hold only when n is even, but I can't figure out why it fails at certain even values (like $n=10$, for instance).
 A: We use the fact that the totient function is multiplicative. Let:
$$x=2^a5^by$$
where $\gcd(y,10)=1$. Then:
$$\phi(4x)=\phi(5x) \implies \phi(2^{a+2}5^bx)=\phi(2^a5^{b+1}x)$$
Using the fact that the totient function is multiplicative, we yield:
$$\phi(2^{a+2})\phi(5^b)\phi(x)=\phi(2^a)\phi(5^{b+1})\phi(x)$$
Cancelling $\phi(x)$, we have:
$$\phi(2^{a+2})\phi(5^b)=\phi(2^a)\phi(5^{b+1})$$
We know that $\phi(2^{a+2})=2^{a+1}$ and $\phi(5^{b+1})=4\cdot 5^b$. If $b>0$, then $\phi(5^b)=4 \cdot 5^{b-1}$. However, this would be a contradiction as the LHS as one less factor of $5$ than required. Thus, $b=0$. Substituting:
$$\phi(2^{a+2})=4\phi(2^a)$$
which holds for all $a \geqslant 1$. Thus, $x=2^ay$ where $a>0$. This means that $x$ can be any even number not divisible by $5$.
A: Note the following implications:
\begin{align*}
5 \mid x &\implies \varphi(5x) = 5\varphi(x) \\
5 \not\mid x &\implies \varphi(5x) = 4\varphi(x) \\
2 \mid x &\implies \varphi(4x) = 4\varphi(x) \\
2 \not\mid x &\implies \varphi(4x) = 2\varphi(x)
\end{align*}
As $\varphi(x) \neq 0$, the only possibility of equality is if $5 \not\mid x$, but $2 \mid x$. That is $x$ is even, but not a multiple of $10$.
