Find all $x \in \mathbb{N}$ that satisfy $pφ(x)=x$, where $p$ is a prime.

Find all $x \in \mathbb{N}$ that satisfy $pφ(x)=x$, where $p$ is a prime.

This is a generalization of Solve the equation $2φ(x)=x $ for $x\in\mathbb N^+.$, where this is solved for $p=2$.

My partial solution:

I can show that, if $x = p^kh$ where $h$ and $p$ are relatively prime, then $h$ can have only prime factors less than $p$ and that many values of $p$ have no solutions.

In particular, the only solutions for a prime of the form $p=2^a3^b+1$ are $p=3$ with $x = 2^m3^n$.

More generally, it depends on finding a set of primes $Q$ such that $p-1 =\prod_{q \in Q} \frac{q}{q-1} $. I can find all possible primes in $Q$ for any particular $p$, but not for all $p$.

Here's what I have got:

I assume that $p$ is odd, so $p = 2^ab+1$ where $a, b \in \mathbb{N}$ and $b$ is odd.

Let $x=p^kh$ with $k\in\mathbb Z_{\ge 0}$, $h\in\mathbb Z_{\ge 1}$, and $(p, h) = 1$.

$p\phi(x)=x$ implies $k\ge 1$, so $\phi(p^kh)=p^{k-1}h$.

Now use the multiplicativity of the $\phi$ function and the fact that $\phi(p^k) =p^{k-1}(p-1) $:

$p^{k-1}h =\phi(p^{k})\phi(h) =p^{k-1}(p-1)\phi(h) $, or $h =(p-1)\phi(h) $, so that $p-1 =\frac{h}{\phi(h)} =\prod_{q | h} \frac{q}{q-1} $.

Let $q$ be the largest prime factor of $h$. Then $q | (p-1)$ (because it can not be cancelled out by any of the smaller primes). Since $p = 2^ab+1$ where $b$ is odd and $a \ge 1$, then $q | 2^ab$. In particular, $q \le \max(2, b)$, so there are only a finite number of primes (those $\le \max(2, b)$) that can divide $h$.

If $q = 2$, then $h = 2^m$ for some $m$, so $p-1 = \frac{h}{\phi(h)} = \frac{2^m}{2^{m-1}} =2 $ or $p = 3$.

To check, if $x = 2^m 3^j$, $\phi(x) =\phi(2^m)\phi(3^j) =2^{m-1}2\,3^{j-1} =2^m3^{j-1} =\frac{x}{3} $.

If $q = 3$, then the possible primes dividing $h$ are $2$ and $3$, so $x = p^k2^u$, $x = p^k3^u$, or $x = p^k2^u3^v$.

If $x = p^k2^u$, $\phi(x) =(p-1)p^{k-1}2^{u-1} =\frac{p-1}{2}p^{k-1}2^{u} =\frac{p-1}{2}\frac{x}{p} $, which only works if $p=3$ as shown previously.

If $x = p^k3^u$, $\phi(x) =(p-1)p^{k-1}2\,3^{u-1} =\frac{2(p-1)}{3}p^{k-1}3^{u} =\frac{2(p-1)}{3}\frac{x}{p} $, which never works since it needs $2(p-1) = 3$.

If $x = p^k2^u3^v$, $\phi(x) =(p-1)p^{k-1}2^{u-1}2\,3^{v-1} =\frac{2(p-1)}{6}p^{k-1}2^{u}3^v =\frac{p-1}{3}\frac{x}{p} $, which never works.

Solution 1:

$\varphi(x)|x$ if and only if $x=2^a3^b$ with $a>0$. From here it is clear if $p\varphi(x)=x$ then $p=2$ or $p=3$, to be precise, if $b>0$ we have $p=3$ and otherwise $p=2$.

For a proof of the previous, clearly $x$ must be even, if $x$ has two or more odd prime factors we get $v_2(\varphi(x))>v_2(x)$. So $x=2^ap^b$, from here $\varphi(x)=2^{a-1}(p-1)p^b$, which gives $(p-1)|2$. So $p=3$.