the equation :
sorry, it's not in N but in Z
there isn't any difference ^^
First, we obviously have the trivial solution
Now suppose there's another solution
If and are not multiples of 3 then a contradiction.
Again if only one of them ( x or z) is multiple of 3 we get a contradiction ( try it).
So both, and are multiples of 3. And thus is multiple of 9, so it must be that is multiple of 3. Now set ; ;
We get: (dividing by 9)
And it follows by infinite descent that there can be no other solution.