Find integers such that $\varphi(n)^3\le n^2$ (Mathematical Reflections)

Question

For which integers $n$ is the relation $\varphi(n)^3\le n^2$ true where $\varphi$ is the totient fuction of Euler.

I used the property of $\varphi$ function ($\varphi(n)=n\prod_{p|n}(1-\frac1 p)$ ) to obtain $n\prod_{p|n}(p-1)^3\le \prod_{p|n}p^3$ Now, how do we proceed further. Thanks beforehand.

This is problem U391 from the Problem column of Mathematical Reflections - Issue 6 2016.

Answer 1

We know that $\varphi(n)$ is multiplicative (if $n,m$ are coprime then $\varphi(nm)=\varphi(n)\varphi(m)$), therefore so is $f(n)=\varphi(n)^3/n^2$. We're looking for numbers with $f(n)\leq 1$, and so if we write $n$ as a product of distinct prime powers we can just multiply the $f$ values together. They key is that there are very few prime powers with $f(p^k)\leq 1$.

If $k>1$ then $f(p^k)=p^{k-1}f(p)$, and it is easy to check that $p=2$ and $p=3$ are the only primes with $f(p)\leq 1$. It follows that $2$, $4$, $8$ and $3$ are the only prime powers with this property. Then you can enumerate a few prime powers which give the smallest $f(p^k)>1$ -- the $f$ values get too large to be relevant very quickly -- and complete the classification.