# How do I prove the following result in number theory? [closed]

There exist no $(n, m) ∈ \mathbb{N}$ so that $n + 3m$ and $n ^2 + 3m^2$ both are perfect cubes.
• Also, Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. May 18, 2016 at 16:23
• May 18, 2016 at 17:46
• May 18, 2016 at 18:29

Suppose they are. Then $(n+3m)(n^2+3m^2)$ is a perfect cube.
$(n+3m)(n^2+3m^2) = (n+m)^3 + (2m)^3$
• @MXYMXY Citing Fermat's Last Theorem for the relatively simple case of $a^3+b^3=c^3$ is overkill. See fermatslasttheorem.blogspot.lt/2005/05/… for an elementary proof. May 18, 2016 at 16:29