I think it's natural to look at quadratic extensions and $n=3$, which I believe eliminates obvious solutions. I used $(5-\sqrt 6)^3 + (3\sqrt 6)^3 = (5+\sqrt 6)^3$ for this problem.

This paper in Integers has some useful references, and a new proof that the case $n=3$ has no nontrivial solutions over the Gaussian integers $\mathbb Z[i]$.

"In L.E. Dickson’s History of the Theory of Numbers, one can find that this question has already been answered by R. Feuter within the frame of algebraic number theory. Namely, he has proven that if $a^3+b^3+c^3 = 0$ is solvable by numbers $\ne 0$ of an imaginary quadratic domain $k(\sqrt m)$, where $m \lt 0$, $m = 2 (\mod 3)$, then the class number of $k$ is divisible by 3. Now $k = \mathbb Z$, $m = −1$ and $\mathbb Z$ is a Principal Ideal Domain having class number 1 not divisible by 3."

I think it's natural to look at quadratic extensions and $n=3$, which I believe eliminates obvious solutions. I used $(5-\sqrt 6)^3 + (3\sqrt 6)^3 = (5+\sqrt 6)^3$ for this problem.

This paper in Integers has some useful references, and a new proof that the case $n=3$ has no nontrivial solutions over the Gaussian integers $\mathbb Z[i]$.

"In L.E. Dickson’s History of the Theory of Numbers, one can find that this question has already been answered by R. Feuter within the frame of algebraic number theory. Namely, he has proven that if $a^3+b^3+c^3 = 0$ is solvable by numbers $\ne 0$ of an imaginary quadratic domain $k(\sqrt m)$, where $m \lt 0$, $m = 2 (\mod 3)$, then the class number of $k$ is divisible by 3. Now $k = \mathbb Z$, $m = −1$ and $\mathbb Z$ is a Principal Ideal Domain having class number 1 not divisible by 3."

I think it's natural to look at quadratic extensions first. I found $(5-\sqrt 6)^3 + (3\sqrt 6)^3 = (5+\sqrt 6)^3$ for this problem. It was the second example I tried, so I think there are a lot of counterexamples.

I think it's natural to look at quadratic extensions and $n=3$, which I believe eliminates obvious solutions. I used $(5-\sqrt 6)^3 + (3\sqrt 6)^3 = (5+\sqrt 6)^3$ for this problem.

This paper in Integers has some useful references, and a new proof that the case $n=3$ has no nontrivial solutions over the Gaussian integers $\mathbb Z[i]$.

"In L.E. Dickson’s History of the Theory of Numbers, one can find that this question has already been answered by R. Feuter within the frame of algebraic number theory. Namely, he has proven that if $a^3+b^3+c^3 = 0$ is solvable by numbers $\ne 0$ of an imaginary quadratic domain $k(\sqrt m)$, where $m \lt 0$, $m = 2 (\mod 3)$, then the class number of $k$ is divisible by 3. Now $k = \mathbb Z$, $m = −1$ and $\mathbb Z$ is a Principal Ideal Domain having class number 1 not divisible by 3."