find all $m$,$n$ such that $\varphi(m) + \varphi(n) = \varphi(mn)$.

Question

We have to find all $m,n$ belongs to natural numbers such that $\varphi(m) + \varphi(n) = \varphi(mn)$.

I know if $(m,n)=1$, then $ \varphi(m)\varphi(n) = \varphi(mn)$. but, I am confused how I can use this fact here.

What have you tried? An obvious place to start is to search for examples. — lulu, Sep 30 '19 at 16:17
It's also obvious that no pair of prime numbers can be a solution except $(2,2)$ — The Demonix _ Hermit, Sep 30 '19 at 16:23
@ajotatxe $(2,2) $ is a valid solution as $\phi(2) = 1$ and $1+1 = 2$ — The Demonix _ Hermit, Sep 30 '19 at 16:31
If $(m,n)=1$ then we have $\varphi(m)\varphi(n)-\varphi(m)-\varphi(n)=0$. This implies that $(\varphi(m)-1)(\varphi(n)-1)=1$. From here it follows that $\varphi(m)=\varphi(n)=2$ — , Sep 30 '19 at 16:33
Using a computer , the only solutions I found till now are $(2,2) , (3,4) and (4,3)$ — The Demonix _ Hermit, Sep 30 '19 at 16:35
$\varphi(m)\mid \varphi(mn)$ .Hence $\varphi(m)\mid\varphi(n)$. Similarly $\varphi(n)\mid \varphi(m)$. Therefore, $\varphi(m)=\varphi(n)$. — , Sep 30 '19 at 17:05
https://math.stackexchange.com/questions/301837/is-the-euler-phi-function-bounded-below — , Sep 30 '19 at 17:14

lhf · Answer 1 · 2019-09-30T18:22:42.750

Let $d=\gcd(m,n)$, $D=\varphi(d)$, $M=\varphi(m)$, $N=\varphi(n)$.

Then $D(M+N)=MNd$, because $\varphi(d) \varphi(mn) = \varphi(m) \varphi(n) d$.

Also, $\varphi(n)$ divides $\varphi(kn)$. Therefore, $M=AD$ and $N=BD$.

Thus, $D^2(A+B)=ABD^2d$ and so $A+B=ABd$. This implies $A \mid B \mid A$ and so $A=B$. Then, $2A=A^2d$ implies $d=1$ or $d=2$.

If $d=1$, then $M+N=MN$, which implies $M=N=2$. Therefore, $(m,n)=(3,4)$ or $(m,n)=(4,3)$, because $\varphi(x)=2$ iff $x\in\{3,4,6\}$.

If $d=2$, then $M+N=2MN$, which implies $M=N=1$. Therefore, $(m,n)=(2,2)$.

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