Euler's phi function application

Question

Please advise on how to solve the following textbook question below:

Let phi(n)/n = a/b. Where n = positive odd integer and a and b are relatively prime.

Prove that the largest prime factor of n = largest prime factor of b

So far I have thought of using prime factorization φ(n)/n = {(p1-1)(p2-1)...(pr 1)}/p1*p2*...*pr

I'm not sure where to go from here. Should I conduct proof by cases?

Thank you!

Answer 1

Let $n=p^kQ$ where $Q$ is a product of primes less than $p$.

Then $\phi(n)=p^{k-1}(p-1)\phi(Q)$, where $p$ and $\phi(Q)$ are coprime.

Therefore $\frac{\phi(n)}{n}=\frac{(p-1)\phi(Q)}{pQ}$ and the factor of $p$ in the denominator cannot cancel with any factor of the numerator.