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

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 $(x,y,z)=(3,1,2)$.

• What led you to asking this question? – Mastrem May 17 '18 at 9:22
• Many thanks for your attention @Mastrem . I was playing with variants of an equation from Guy's book (Unsolved Problems in Number Theory). After that I ran the Pari-GP program I would like to know why previous equation seems to have a finite number of solutions. It is not clear to me if only by using Euler-Fermat theorem can it be solved. – user243301 May 17 '18 at 9:25
• 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 as 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 $d\leq2$. – Servaes May 17 '18 at 10:10
• I think that your comment has the merit to be written as an answer. Thus many thanks and feel free to edit it as an answer @Servaes – user243301 May 17 '18 at 10:12
• I think it is helpful, but it does not answer the question. I will give the problem some more thought and hopefully expand this into an answer soon. – Servaes May 17 '18 at 10:13

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.

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 $p=3$.

• Many thanks I am going to study your answer and previous answers. You and the other users answering this question are true mathematicians. – user243301 May 17 '18 at 11:23
• I am going to need some days to undestand your proof and the proof below, many thanks to you and all users that did contributions in answers and comments. Many thanks for your great answer and share the calculations in this site, you're generous. – user243301 May 17 '18 at 13:57

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

• ... which leaves the case where there is no such prime factor... – punctured dusk May 17 '18 at 9:31
• From your hint I see it. $x^{\varphi(yz)}+y^{\varphi(zx)}\equiv 2\text{ mod }p$, that is zero if $p=2$. – user243301 May 17 '18 at 9:32
• @barto yes, you're right. For a second, I thought that would mean that $z$ divided $x$ and $y$, which is obviously ffalse for anything divisible by a square. – Mastrem May 17 '18 at 9:32
• @user243301 no, I'm saying that any prime factor $p$ of $z$ must divide both $x$ and $y$ except if it's equal to $2$. Sadly, that doesn't solve the problem. As barto pointed out, my hint is a good start, but doesn't solve the problem. – Mastrem May 17 '18 at 9:36
• @servaes corrected. For about the 918,752nd time I would have preferred to be able to edit my own comments instead of having to copy, delete and paste. – Oscar Lanzi May 17 '18 at 11:31

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 $z=2$.

• Many thanks for your answer. – user243301 May 17 '18 at 10:57
• Many thanks for your great answer, I am going to accept the answer that currently can avoid the use of Fermat Last Theorem. – user243301 May 17 '18 at 13:56