Which $n$ will satisfy $\phi(n)=2n$?

Question

The full question says: Perfect numbers satisfy $\sigma(n)=2n$. Which $n$ satisfy $\phi(n)=2n$?

My response is that for $n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$, an alternate expression of the $\phi$-function gives

$$\phi(n)=n\left(1-\frac {1} {p_1}\right)\left(1-\frac {1} {p_2}\right)\dots\left(1-\frac {1} {p_k}\right)$$

And this expression suggests that $\phi(n)$ must be less than or equal to $n$, so that there could be NO $n$ (other than one) that would satisfy the condition.

Is my argument sound?

Answer 1

Mostly, yes.

More precisely, your conclusion is that the formulation of $\phi$ given shows if $n$ has any prime divisors, then $\phi(n) < n$.

But that doesn't mean that, for your final candidate of concern, $\phi(1)=2$. In fact, $\phi(1) = 1$.

So no positive integers whatsoever satisfy $\phi(n)=2n$.