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

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?

• yes, $\phi(n) < n$ unless $n=1$ Commented Jul 20 at 1:15
• You don't need the formula. More simply by definition $\,\phi(n)\,$ counts which of the $\,\color{#c00}n\,$ integers $\,1,2,3,\ldots,n\,$ are coprime to $\,n,\,$ so $\,\phi(n)\le \color{#c00}n < 2n.\ \$ Commented Jul 20 at 1:38
• For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. Commented Jul 20 at 1:39
• @BillDubuque, thank you for your comments. Could you please point me to some place where I can review the purposes of the site so as not to abuse the privilege of using the site? I sought to find a similar question so as not to duplicate an existing question. I am studying by myself, so I have no one that I can ask about these things. So any suggestions are appreciated. Commented Jul 20 at 2:22
• It would be more interesting to consider the equation $\phi(2n)=n$ instead. As it turns out, the solutions to that equation are exactly the powers of $2$. Commented Jul 20 at 3:27

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 $$\phi(n)=2n$$.