# Find all $x,y,z\in\mathbb N$, $x,y,z>1$ such that satisfy $x\mid yz+1$, $y\mid xz+1$, and $z\mid xy+1$

Find all $x,y,z\in\mathbb N$, $x,y,z>1$ such that satisfy $$\begin{cases}x\mid yz+1\\y\mid xz+1\\z\mid xy+1\end{cases}$$

I've found out easily that $$\begin{cases}x\nmid yz\\y\nmid xz\\z\nmid xy\end{cases}\Rightarrow \begin{cases}x\nmid y\\x\nmid z\\y\nmid x\\y\nmid z\\z\nmid x\\z\nmid y\end{cases}$$ I can't find a way to continue from here, so I'd like to get some ideas. Thanks.

• Much stronger than what you have, $\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1$. Jan 17, 2014 at 21:48
• If prime $p$ were to divide both $x$ and $y$, then it would also divide $yz$. Hence, $p|yz$ and $p|yz+1$, so $\gcd(yz,yz+1)\neq 1$. This is a contradiction. Jan 17, 2014 at 22:23

The only solutions are permutations of $(2,3,7)$:

Assume $x \mid yz+1$, $y \mid xz+1$, $z \mid xy+1$. We have $(x,y)= (y,z) = (z,x) = 1$. Let $x< y< z$. Then as $x$,$y$ and $z$ divide $xy+xz+yz+1$ we have $$xyz \mid xy+xz+yz+1 < 3yz.$$ By $1<x$ we conclude $x= 2$. But then $y \mid (2z+1)$ and $z \mid (2y+1)$ leading to $$yz \mid 2(y+z)+1 <4z$$ from which we conclude $y=3$ using $x<y$. Finally $z \mid xy+1 = 7$ and indeed $(2,3,7)$ is a solution.

• I'm sorry but how do we know that $xy+xz+yz+1 < 3yz$? Jan 18, 2014 at 12:28
• @mathh By assumption $x<y<z$. Therefore $xz< yz$ and $xy<yz$. This implies $xz \leq yz-1$. All in all $xy+xz+yz+1 \leq xy+yz+yz < yz+yz+yz = 3yz$.
– benh
Jan 18, 2014 at 14:52
• Oh, sorry, forgot about the assumption. Jan 18, 2014 at 14:55
• How does $1<x$ show that $x=2$? Nevermind, I understand it now. Jan 18, 2014 at 14:59
• @janmarqz I am sorry, I don't know about the $(2,3,7)$-triangle group. I suspect it's the law of small numbers at work.
– benh
Jan 19, 2014 at 20:01