Posted here. And a nice solution by pco: pco wrote: It's rather easy to check that $(x,y,z)=(\frac{p^nu+p^a\frac{p^a+1}u}t,p^{n-a}u,\frac{p^a+t}u)$ is a solution whenever these are positive integers. For $n>1$ ========= let us look at the following solutions : A first group of S1 of $4n-4$ solutions : Setting $t=1$, $a\in[1,n-1]$ and $u\in\{1,2,\frac{p^a+1}2,p^a+1\}$ gives solutions $(x,y,z)=(p^nu+p^a\frac{p^a+1}u,p^{n-a}u,\frac{p^a+1}u)$ In order to show that all these solutions are distinct, consider $(x,y,z)$ such a solution : Two of the three integers are divisible by $p$ and the third is not. The littlest of the two element divisible by $p$ need to be $p^{n-a}u$. Since $u$ is not divisible by $p$, $a$ is uniquely determined as $n-v_p(\text{ this number})$. Then $u$ is uniquely determined. A second group of $n-1$ solutions : Setting $t=p$, $a\in[1,n-1]$ and $u=1$ gives solutions $(x,y,z)=(p^{n-1}+p^{a-1}(p^a+1),p^{n-a},p^a+p)$ In order to show that all these solutions are distinct, consider $(x,y,z)$ such a solution : Only one of the three numbers is a perfect power of $p$, and this uniquely gives $a$. In order to show that all these solutions are distinct from S1 : If $a>1$, the three numbers are divisible by $p$ and so cant be in S1. If $a=1$, the solution is $(p^{n-1}+p+1,p^{n-1},2p)$ and both element divible by $p$ are $<p^n$ while in S1, the greatest is $>p^n$ Four other solutions : S3 : $t=1,u=1,a=0$ $\implies$ solution $(x,y,z)=(p^n+2,p^n,2)$ S4 : $t=1,u=2,a=0$ $\implies$ solution $(x,y,z)=(2p^n+1,2p^n,1)$ S5 : $t=1,u=1,a=n$ $\implies$ solution $(x,y,z)=(p^{2n}+2p^n,p^n,p^n+1)$ S6 : $t=1,u=2,a=n$ $\implies$ solution $(x,y,z)=(2p^{n}+\frac{p^n(p^n+1)}2,2,\frac{p^n+1}2)$ It's immediate to see that these 4 solutions are distinct And they are also distinct from S1 and S2 since they contain exactly one element divisible by p while group S1 always contains exactly 2 and group S2 contains 3 or 2 (in case of $a=1$) So we got $5n-1$ solutions and since $5n-1\ge 3n+3$ $\forall n>1$, we got the result (in fact a stronger result) For $n=1$, ========= We need to find 6 solutions. Choose for example : $a=0$ and $t=1$ and $u=1$ $\implies$ solution $(x,y,z)=(p+2,p,2)$ $a=0$ and $t=1$ and $u=2$ $\implies$ solution $(x,y,z)=(2p+1,2p,1)$ $a=0$ and $t=2$ and $u=1$ $\implies$ solution $(x,y,z)=(\frac{p+3}2,p,3)$ $a=0$ and $t=2$ and $u=3$ $\implies$ solution $(x,y,z)=(\frac{3p+1}2,3p,1)$ $a=1$ and $t=1$ and $u=1$ $\implies$ solution $(x,y,z)=(p^2+2p,1,p+1)$ $a=1$ and $t=1$ and $u=2$ $\implies$ solution $(x,y,z)=(2p+\frac{p(p+1)}2,2,\frac{p+1}2)$ All these 6 solutions are clearly distinct (order their element).

This is also IMO Short List 1986 - P4.