Strangely enough, it seems to me that if $n$ is even we get not only $4|n$, but also $8|n, 16|n,...$ and so on, so n =0.

Can you guys check what is wrong with the following; We expand and simplify the LHS to get $3n(n^2+2) = m^3$ Then let $m = nk$. $n = 0$ works so suppose $n \ne 0$. Then $3(n^2+2) = n^2k^3$. Suppose $3|n^2$. Then take $\pmod 9$ which gives a contradiction. So, $3|k^3$ or $k = 3p$. But then we rearrange to get $n^2(9p^3-1) = 2$. But this has no integer solutions. So the only answer is $n = 0$ ??

ssilwa wrote: Can you guys check what is wrong with the following; We expand and simplify the LHS to get $3n(n^2+2) = m^3$ Then let $m = nk$. $n = 0$ works so suppose $n \ne 0$. Then $3(n^2+2) = n^2k^3$. Suppose $3|n^2$. Then take $\pmod 9$ which gives a contradiction. So, $3|k^3$ or $k = 3p$. But then we rearrange to get $n^2(9p^3-1) = 2$. But this has no integer solutions. So the only answer is $n = 0$ ?? No another example is when $n=4$

$4^3+3^3+5^3=216=6^3$

ssilwa wrote: Can you guys check what is wrong with the following; We expand and simplify the LHS to get $3n(n^2+2) = m^3$ Then let $m = nk$. $n = 0$ works so suppose $n \ne 0$. Then $3(n^2+2) = n^2k^3$. Suppose $3|n^2$. Then take $\pmod 9$ which gives a contradiction. So, $3|k^3$ or $k = 3p$. But then we rearrange to get $n^2(9p^3-1) = 2$. But this has no integer solutions. So the only answer is $n = 0$ ?? I think you are wrong on the part you said $m=nk$ because this is not necesarily true, one example is 4 doesn't divide 2, but 4 divides 2^3

Oh ok, It should be $m^3 \equiv 0 \pmod n$ which doesnt necessarily imply that $n = nk$. Thanks!

ssilwa wrote: Can you guys check what is wrong with the following; We expand and simplify the LHS to get $3n(n^2+2) = m^3$ Then let $m = nk$. $n = 0$ works so suppose $n \ne 0$. Then $3(n^2+2) = n^2k^3$. Suppose $3|n^2$. Then take $\pmod 9$ which gives a contradiction. So, $3|k^3$ or $k = 3p$. But then we rearrange to get $n^2(9p^3-1) = 2$. But this has no integer solutions. So the only answer is $n = 0$ ?? I think the fault *may* lie in \[\ 3n(n^2+2) = m^3 \] Then let $m = nk$, perphaps this should be $m^3=nk$, take the case of $8x=y^3$ this may mean that $y$ is in the form $8k$ or it may mean that $y \equiv 0 \mod 2,4$ yea, what you guys said. Edit: however because $3$ is prime you can say $m=3k$

So we agrre we have $3n(n^2+2) = m^3$. We clearly have $\gcd(n,n^2+2) \mid 2$. If it is $2$, it means $2\mid n$, so $n=2k$, thus $12k(2k^2+1) = m^3$. But then $2\mid m$, so $8\mid m^3$, forcing $2\mid 3k(2k^2+1)$, thus $2\mid k$, meaning $4\mid n$. If it is $1$, then $n$ and $n^2+2$ are coprime; we have the cases $\bullet$ $n=a^3$ and $n^2+2 = 9b^3$, so $a^6 + 2 = 9b^3$. The only cubic residues modulo $9$ are $0,\pm 1$, and so $a^6 + 2 \equiv 0 \pmod{9}$ is impossible. $\bullet$ $n=9a^3$ and $n^2+2 = b^3$, so $81a^6 = b^3 - 2$. The only cubic residues modulo $9$ are $0,\pm 1$, and so $b^3 - 2 \equiv 0 \pmod{9}$ is impossible. Thus it only remains that we must have $4\mid n$.

mavropnevma wrote: So we agrre we have $3n(n^2+2) = m^3$. We clearly have $\gcd(n,n^2+2) \mid 2$. If it is $2$, it means $2\mid n$, so $n=2k$, thus $12k(2k^2+1) = m^3$. But then $2\mid m$, so $8\mid m^3$, forcing $2\mid 3k(2k^2+1)$, thus $2\mid k$, meaning $4\mid n$. If it is $1$, then $n$ and $n^2+2$ are coprime; we have the cases $\bullet$ $n=a^3$ and $n^2+2 = 9b^3$, so $a^6 + 2 = 9b^3$. The only cubic residues modulo $9$ are $0,\pm 1$, and so $a^6 + 2 \equiv 0 \pmod{9}$ is impossible. $\bullet$ $n=9a^3$ and $n^2+2 = b^3$, so $81a^6 = b^3 - 2$. The only cubic residues modulo $9$ are $0,\pm 1$, and so $b^3 - 2 \equiv 0 \pmod{9}$ is impossible. Thus it only remains that we must have $4\mid n$. Thank you for the solution now I remember it.