I'm having trouble approaching the following exercise:

Let $n$ and $k$ be positive integers. Prove that $\phi(n^k) = n^{k-1}\phi(n)$.

I've tried examining the prime factorization $n^k = (p_1^{a_1} \cdot\cdot\cdot p_r^{a_r})^k$ and plugging it into the Euler phi-function but I'm not really sure where to go from there.

Any hints?


The method you have suggested should work fine as long as you know the formula $$\phi(n)=(p_1-1)p_1^{a_1-1}\cdots(p_r-1)p_r^{a_r-1}\ .$$ Then we get $$n^k=p_1^{ka_1}\cdots p_r^{ka_r}\quad\Rightarrow\quad \phi(n^k)=(p_1-1)p_1^{ka_1-1}\cdots(p_r-1)p_r^{ka_r-1}\ ,$$ and I think I can leave the rest up to you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.