Solve for diophantine equation $x^n + y^n + z^n =1$

Question

Solve for diophantine equation

1. $x^n + y^n + z^n =1$
2. $x^n+y^n+z^n=2$

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Is this equation solve-able ?

Answer 1

I didn't find any for 5th or higher odd powers (looked for $-100 \le x,y,z \le 100$). [Note I don't count $(a,-a,1)$ or permutations; looking for three distinct absolute values.]

But for cubes, with $x,y,z \in [-200,200]$ there are three cases of $x^3+y^3+z^3=2$: $$(-161,-54,163),\ (-47,-24,49),\ (-6,-5,7).$$ There are also several cases of $x^3+y^3+z^3=1$, but since 1 is a cube this is a special case of four cubes adding to zero, which there is literature on. The six cases found, again with $x,y,z \in [-200,200]$: $$(-150,73,144),\ (-138,-135,172),\ (-138,-71,144),$$ $$(-103,64,94),\ (-12,9,10),\ (-8,-6,9).$$ My guess is that the equation with 2 on the right might not have much literature on it. And there may be known methods to rule out 1 on the right for odd powers greater than 3.

Answer 2

There are the known identities,

$$(1-9t^3)^3 + (9t^4)^3 + (3t-9t^4)^3 = 1$$

$$(1+6x^3)^3 + (1-6x^3)^3 + (-6x^2)^3 = 2$$