About the solutions of $x^{\varphi(yz)}+y^{\varphi(xz)}=z^{\varphi(xy)}$, being $\varphi(n)$ the Euler's totient

Question

In this post we denote the Euler's totient function as $\varphi(n)$ for integers $n\geq 1$. I wondered about the solutions of the equation $$x^{\varphi(yz)}+y^{\varphi(xz)}=z^{\varphi(xy)}\tag{1}$$ for integers $x\geq 1$, $y\geq 1$ and $z\geq 1$.

Question. I think that $(1)$ has a finite number of such solutions $(x,y,z)$. Am I right? Many thanks.

I am waiting about a rigorous proof or a remarkable reasoning that can invoke heuristics or conjectures.

Computational fact. For integers $1\leq x,y,z\leq 100$ we can find (using a Pari-GP program; these can also be found by inspection) the solutions $(x,y,z)=(1,1,1),(1,3,2)$ and $(x,y,z)=(3,1,2)$.

Answer 1

The only solutions are those you have found. If we are careful, we can avoid using Fermat's last theorem or Catalan's conjecture:

Claim 1: The only solutions with one of $x,y,z$ equal to $1$ are those you found.

Proof. Clearly $z>1$. Wlog $x=1$ and we are left with $1 + y^{\phi(z)} = z^{\phi(y)}$. We can observe that $\phi(n)$ is even for $n>2$, and that there are no nonzero consecutive squares. Hence either:

• $\phi(z)=1$ so $z=2$, and $1+y = 2^{\phi(y)}$. One shows that for $y$ odd the RHS is strictly larger unless $y=1$ or $y=3$, in which case there is equality. The details are at the bottom.
• $\phi(y)=1$ so either $y=1$ which gives $z=2$, or $y=2$ which gives $1+2^{\phi(z)}=z$. The LHS is now strictly larger, by the same inequalities as above.

We will suppose $x, y, z > 1$ from now on.

Claim 2: Every prime divisor $p$ of $z$ divides $x$ and $y$.

Proof. For $p>2$ this follows by Fermat's little theorem. Let $p=2$. Suppose $x,y \geq 3$ are odd. We have $xy, yz, xy > 2$ so all exponents are even. Odd squares are $1$ mod $4$, hence the LHS is $2$ mod $4$, while the RHS is $0$.

Claim 3: $z$ is a power of $2$.

Proof. With FLT: If $p>2$ is a prime divisor of $z$, then $p$ is a divisor of $\phi(xy),\phi(yz),\phi(xy)$, which contradicts Fermat's last theorem. Without FLT, we can proceed as follows: If $z$ has two prime divisors, then $4$ divides $\phi(xy),\phi(yz),\phi(xy)$ contradicting FLT for exponent $4$. Hence $z=p^k$ is a prime power. Also by FLT(4), $p=2$ or $p \equiv 3 \pmod 4$. But $p \equiv 3 \pmod 4$ is a contradiction with the fact that $p^{k\phi(xy)}$ is not a sum of two nonzero squares. Thus $p=2$.

Claim 4: There are no other solutions.

Proof. We have that $x,y,z$ are even. By FLT for exponent $4$, we cannote have $4 \mid z$. So $z=2$. By FLT for exponent $4$, not both $x$ and $y$ can have an odd prime divisor or be divisible by $4$. Wlog, $y=2$. But for $y=z$ the equation is $$x^{\phi(yz)} + y^{\phi(yx)} = y^{\phi(yx)}$$ which would imply $x=0$, contradiction.

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On the comparison of $2^{\phi(y)}$ and $y$. We use the following inequalities:

Fact 1: $2^n \geq n+2$ for $n \geq 2$ with equality only for $n=2$.

Fact 2: $ \prod b_i \geq \sum b_i$ for $b_i \geq 2$ with equality only for $b_1 \geq b_1$ and $2\cdot2 \geq 2+2$.

(You can show both by induction.)

Back to the problem. Let $y$ be odd and $y = \prod p_i^{a_i}$. By fact 2, we have $\phi(y) \geq \sum a_i(p_i-1)$ with equality only for $y=1$ or $y$ prime. By fact 1 applied to $n=p_i-1$ and mulitplying the inequalities: $$2^{\phi(y)} \geq 2^{\sum a_i(p_i-1)} > \prod (p_i+1)^{a_i} \geq y+1$$ with equality only if $y=1$ or if $y=p$ is prime and $p=3$.

Answer 2

Hint: Let $p$ be a prime factor of $z$ not dividing $x$ or $y$ and compute $x^{\varphi(yz)}+y^{\varphi(zx)}$ modulo $p$.

Answer 3

Observation 1: The greatest common divisor of $x$, $y$ and $z$ divides $2$.

Let $d:=\gcd(x,y,z)$, so that there exist integers $a,b,c\geq1$ such that $$\varphi(yz)=a\varphi(d^2),\qquad \varphi(xz)=b\varphi(d^2),\qquad \varphi(xy)=c\varphi(d^2).$$ If $d\geq3$ we get a contradiction with Fermat's last theorem because then $$(x^a)^{\varphi(d^2)}+(y^b)^{\varphi(d^2)}=(z^c)^{\varphi(d^2)},$$ where $x^a,y^b,z^c\geq1$ and $\varphi(d^2)>2$. Hence $d\leq2$.

Observation 2: $z$ is a power of $2$.

Let $p>2$ be a prime dividing $z$. Then by the above $p$ does not divide $x$ or $y$, and there are integers $a,b\geq1$ such that $\varphi(xz)=a\varphi(p)$ and $\varphi(yz)=b\varphi(p)$, so by Fermat's little theorem we have $$z^{\varphi(xy)}=x^{\varphi(yz)}+y^{\varphi(xz)}=(x^b)^{\varphi(p)}+(y^a)^{\varphi(p)}\equiv2\pmod{p},$$ contradicting the fact that $p$ divides $z$ and $p>2$. Hence no such $p$ exists, so $z$ is a power of $2$.

Observation 3: $z=2$.

Clearly $z=1$ is impossible, and if $4\mid z$ then $2$ divides both $\varphi(xz)$ and $\varphi(yz)$, so the congruence $$x^{\varphi(yz)}+y^{\varphi(xz)}=z^{\varphi(xy)}\equiv0\pmod{4}$$ implies that $4$ divides both $x$ and $y$. But then $4$ divides both $\varphi(xz)$ and $\varphi(yz)$, and $2$ divides $\varphi(xy)$, yielding $u,v,w\geq1$ such that $$u^4+v^4=w^2,$$ a contradiction. Hence $z=2$.