But $11^{2018}$ is also a cube

So, by FLT, there exist no solutions.

RagvaloD wrote: Are there natural solution of $$a^3+b^3=11^{2018}$$? Just observe that $11^{2018}$ is $4 \mod 9$.

Note how since we have $\phi (9)=6$, $11^{2018}\equiv 2^{2018} \equiv 2^{6\cdot 336+2}\equiv 1^{336}\cdot 2^2\equiv 4\pmod{9}$; however, since $a^3, b^3\in \{-1,0,1\} \pmod{9}$, it's clear that $a^3+b^3\equiv 4\pmod{9}$ yelds no solutions.

jrc1729 wrote: So, by FLT, there exist no solutions. Perhaps the proof doesn't end there. I think it would be better if someone proofs Fermat's Last Theorem for $n=3$ in the test itself.

ythomashu wrote: But $11^{2018}$ is also a cube Wait no, what, it's not a cube.

oh, oops

Math.Is.Beautiful wrote: jrc1729 wrote: So, by FLT, there exist no solutions. Perhaps the proof doesn't end there. I think it would be better if someone proofs Fermat's Last Theorem for $n=3$ in the test itself. Hmm, you're right. It's not too tough.