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There exist no $(n, m) ∈ \mathbb{N}$ so that $n + 3m$ and $n ^2 + 3m^2$ both are perfect cubes.
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Suppose they are. Then $(n+3m)(n^2+3m^2)$ is a perfect cube.

$(n+3m)(n^2+3m^2) = (n+m)^3 + (2m)^3$

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    $\begingroup$ Great! +1 Don't add anything and hopefully @MXYMXY will delete his comment, so that people can deduce that by themseleves $\endgroup$
    – DonAntonio
    May 18, 2016 at 16:26
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    $\begingroup$ A neat solution! $\endgroup$
    – almagest
    May 18, 2016 at 16:26
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    $\begingroup$ @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. $\endgroup$
    – user236182
    May 18, 2016 at 16:29
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    $\begingroup$ @user236182 Now you need to delete your comment so that people can deduce either case. $\endgroup$
    – DonAntonio
    May 18, 2016 at 16:31
  • $\begingroup$ @Joanpemo It is not a require to make people deduce the case, though I do think that it would be advisable in this case. $\endgroup$
    – S.C.B.
    May 18, 2016 at 16:32

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