we have $\gcd(x,y)=1$ (easy to prove) $\Rightarrow lcm(x^z,y^z)=x^zy^z$ $\Rightarrow 1+x^z+y^z=x^zy^z \Rightarrow (x^z-1)(y^z-1)=2$(easy)

thuanz123 wrote: we have $\gcd(x,y)=1$ (easy to prove) $\Rightarrow lcm(x^z,y^z)=x^zy^z$ $\Rightarrow 1+x^z+y^z=x^zy^z \Rightarrow (x^z-1)(y^z-1)=2$(easy) Of course. After all it's just the first problem of a Macedonian National Olympiad

thuanz123 wrote: we have $\gcd(x,y)=1$ (easy to prove) $\Rightarrow lcm(x^z,y^z)=x^zy^z$ $\Rightarrow 1+x^z+y^z=x^zy^z \Rightarrow (x^z-1)(y^z-1)=2$(easy) How do you prove that $\gcd(x,y)=1$ ?

Let it be $d$. Since $d \mid x$ and $d \mid y$ $\Rightarrow d \mid x^z+y^z$. But $d \mid \text{lcm}(x^z,y^z)=1+x^z+y^z$, so $d \mid 1$, hence $d=1$

Let $d = gcd(x^z, y^z)$ We have: $d | RHS$ $=>$ $d | 1 + x^z + y^z$ => $d | 1$, so $d = 1$. From this, $1 + x^z + y^z = x^zy^z$ (From the known equation if d = 1: $lcm(n, m) * gcd(n, m) = nm$) This is equiavalent to: $x^zy^z - x^z - y^z - 1 = 0$, adding 2 to both sides gives us $(x^z -1)(y^z - 1) = 2$. Since $x, y, z$ are natural numbers, we have two answers: ($x^z = 3$, $y^z = 2$) and the permutation. It is easy to realise that the final answer is: (x, y, z) $\in$ (3, 2, 1), (2, 3, 1)

x,y,z natural integers 1+x^z+y^z=LCM(x^z,y^z) claim)gcd(x,y)=1 . Assume,d=gcd(x,y)>1 d doesnt divide LHS but divide RHS -->contradiction. so,d=1-->LCM(x^z,y^z)=(xy)^z x^z+1=y^z(x^z-1) 1+2/(x^z-1)=y^z x^z=2 we get---> (x,y,z)=(2,2,1) x^z=3 we get--->(x,y,z)=(3,2,1)

Claim: $\gcd(x^z,y^z)=1$ Proof: Let it be $d$ As $d\mid x$ and $d\mid y$, $d\mid x^z+y^z=1-\text{lcm}(x^z,y^z) \implies d \mid 1 \implies d=1$ From that, we have $$1+x^z+y^z=x^zy^z\implies (x^z-1)(y^z-1)=2$$From that, $\boxed{(x,y,z)=(3,2,1),(2,3,1)}$

how can we prove that (x, y, z) $\in$ (3, 2, 1), (2, 3, 1) is the only solution.