# $\phi(n)^{\sigma(n)^{\tau(n)}}=n^2$ find all natural numbers $n$ such that the equality is true

I found this problem in a old number theory test about arithmetic functions. The problem says that a number $n \in N$ is "perfectly crazy" if $$\phi(n)^{\sigma(n)^{\tau(n)}}=n^2,$$ and, as an example, 1 is "perfectly crazy" and I need to find all "perfectly crazy" numbers. I believe there aren't any prime numbers perfectly crazy after some tests and I don't know how to prove that there aren't any at all.

$\phi(n)$ being Euler totient function

$\sigma(n)$ is the sum of positive divisors of $n$

$\tau(n)$ is the number of positive divisors of $n$

• what do you mean exactly? $\left(\phi(n)\right)^{\sigma(n)^{\tau(n)}}$ or $\left(\phi(n)^{\sigma(n)} \right)^{\tau(n)}$? – Serser Jun 10 '18 at 22:13
• @Serser I had trouble formatting, but it is the first one – Xandin Jun 10 '18 at 22:21
• @Dzoooks I found this as well. What should a do now? if I were a student taking this test I would said that 1 is the only perfectly crazy number and moved to the next question. – Xandin Jun 10 '18 at 22:23

## 1 Answer

The answer is $n=1$. For $n \geq 2$, we have $\tau(n) \geq 2$ and $\sigma(n) \geq n+1$ and $\phi(n) \geq 2$. Hence $$n^2=\phi(n)^{\sigma(n)^{\tau(n)}} \geq 2^{(n+1)^2}$$ implies $$n \geq 2^{n+1},$$ which is impossible.