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).

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 $10$.
