Regarding square-free integers and Euler's function

Question

Let $\phi$ be Euler's totient function. Prove that if $\text{gcd}(\phi(n), n) = 1$, then $n$ is a square free integer.

Answer 1

Hey and welcome to MSE.

I guess you might be aware of the multiplicative properties of φ(n).

Start by writing $n=p_1^{r_1}...p_m^{r_m}$ It follows that $φ(n)=φ(p_1^{r_1})φ(p_2^{r_2})...φ(p_m^{r_m})$, and we have that $φ(p_i^{r_i})=p_i^{r_i-1}(p_i-1)$.

This means that if the $gcd(n, φ(n))=1$, then it means that $r_i=1$ for all i, otherwise $p_i$ would be a common factor.