Let $d,n\in\mathbb{Z}^+$. The question is to prove that $d\mid n$ $\implies$ $\varphi(d)\mid\varphi(n)$.
Here, $\varphi:\mathbb{Z}^+\to\mathbb{Z}^+$ is the Euler totient function.
For $n=1$, we must have $d=1$. Then, $\varphi(d)=\varphi(n)=1$ and thus $\varphi(d)\mid\varphi(n)$.
For $n>1$ and $d=1$, $\varphi(d)\mid\varphi(n)$ since $1\mid z$ for all $z\in\mathbb{Z}^+$.
For $n,d>1$, since $d\mid n$, we can express $\varphi(d)$ and $\varphi(n)$ as follows:
$$\varphi(d)=p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)...p_s^{\alpha_s-1}(p_s-1)$$
$$\varphi(n)=p_1^{\beta_1-1}(p_1-1)p_2^{\beta_2-1}(p_2-1)...p_s^{\beta_s-1}(p_s-1)q_1^{\gamma_1-1}(q_1-1)q_2^{\gamma_2-1}(q_2-1)...q_t^{\gamma_t-1}(q_t-1)$$
where
- $p_i,q_i$ are distinct primes;
- $s,\alpha_i,\beta_i\in\mathbb{Z}^+$ with $\alpha_i\le\beta_i$;
- $q_i$ account for those prime factor(s) of $n$ which do not divide $d$. If such factors exist, then $t,\gamma_i\in\mathbb{Z}^+$, otherwise we don't have those terms in $\varphi(n)$.
Now, let $\omega=p_1^{\beta_1-\alpha_1}p_2^{\beta_2-\alpha_2}...p_s^{\beta_s-\alpha_s}q_1^{\gamma_1-1}(q_1-1)q_2^{\gamma_2-1}(q_2-1)...q_t^{\gamma_t-1}(q_t-1)$. Clearly $\omega\in\mathbb{Z}$.
We then have $\varphi(d)\omega=\varphi(n)$, that is, $\varphi(d)\mid\varphi(n)$.
May I ask if my proof is valid?
Many thanks!
