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.
 A: $\varphi(x)|x$ 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 $p=3$.
A: In fact, we can solve $n\phi(x)=x$ with $n,x\in\mathbb Z^+$.
$n\phi(x)=x$ implies $\phi(x)\mid x$. Now see this question, in particular my detailed answer there.
Therefore either $x=1$ or $x=2^a$ or $x=2^b3^c$ for some $a,b,c\in\mathbb Z^+$.
If $x=1$, then $n\phi(x)=x$ gives $n=1$.
If $x=2^a$ for some $a\in\mathbb Z^+$, then $n\phi(x)=x$ gives $n2^{a-1}=2^a$, i.e. $n=2$.
If $x=2^b3^c$ for some $b,c\in\mathbb Z^+$, then $n\phi(x)=x$ gives
$n\left(2^{b-1}\right)\left(3^{c-1}(3-1)\right)=2^b3^c$, i.e. $n=3$.
Answer: All the solutions are $(n,x)=(1,1),\left(2,2^a\right),\left(3,2^b3^c\right)$ for any $a,b,c\in\mathbb Z^+$.
A: Let $x=\displaystyle\prod_{q\mid x} q^\alpha$ where $q$'s are all primes. Then $\varphi(x)=\displaystyle\prod_{q\mid x} q^{\alpha-1}(q-1)$. Now, $$p\varphi(x)=x\implies p\displaystyle\prod_{q\mid x} q^{\alpha-1}(q-1)=\displaystyle\prod_{q\mid x} q^\alpha\implies p=\displaystyle\prod_{q\mid x}\dfrac{q}{q-1}=\dfrac{\displaystyle\prod_{q\mid x}q}{\displaystyle\prod_{q\mid x}(q-1)}$$ Now, the exponent of $2$ in the product term the product term $\displaystyle\prod_{q\mid x}q$ (if it divides $x$ at all) is $1$ whereas that of the term $\displaystyle\prod_{q\mid x}(q-1)$ is at least $0$. If it is greater than $1$ then $p$ can't be an integer (why?). So it must be equal $1$ or $0$. 

Case 1.

If it is $0$ then $q=2$ is the only prime dividing $x$ and we have $x=2^\alpha$ and $p=2$.

Case 2.

If it is $1$ then $q=3$ is the only odd prime dividing $x$ and we have $x=2^\alpha3^\beta$ and $p=3$.
