$\begingroup$
There exist no $(n, m) ∈ \mathbb{N}$ so that $n + 3m$ and $n
^2 + 3m^2$ both are
perfect cubes.
$\endgroup$
3
-
3$\begingroup$ 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. $\endgroup$– S.C.B.May 18, 2016 at 16:23
-
$\begingroup$ Compare with Show that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes $\endgroup$– Dietrich BurdeMay 18, 2016 at 17:46
-
$\begingroup$ Also math.stackexchange.com/questions/1595262/… $\endgroup$– user236182May 18, 2016 at 18:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
8
Suppose they are. Then $(n+3m)(n^2+3m^2)$ is a perfect cube.
$(n+3m)(n^2+3m^2) = (n+m)^3 + (2m)^3$
-
2$\begingroup$ Great! +1 Don't add anything and hopefully @MXYMXY will delete his comment, so that people can deduce that by themseleves $\endgroup$ May 18, 2016 at 16:26
-
1
-
1$\begingroup$ @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. $\endgroup$ May 18, 2016 at 16:29
-
2$\begingroup$ @user236182 Now you need to delete your comment so that people can deduce either case. $\endgroup$ 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