# An equation concerning eisenstein integers

An interesting exercise on gaussian integers is to prove that those of the form $n-i$, with $n$ a positive integer, are multiplicatively independant. To solve this, one has to consider the equation involving the norms of elements, that is, $n^2+1=(m^2+1)^k$, which can be written as $n^2 = (m^2+1)^k-1$, and it is well known that this equation has no solutions (this is a particular case of Catalan-Mihailescu theorem, but we don't even need such a strong result as this specific form was already solved by Lebesgue).

I was wondering whether one can get the same result considering Eisenstein integers (numbers of the form $a+bj$ with $j=\frac{-1+i\sqrt{3}}{2}$. Determining whether $n-j$ and $m-j$ are multiplicatively independant leads to the equation $$n^2+n+1=(m^2+m+1)^k, \qquad(*)$$ and then we are done with the previous method of solving as it seems that there is no easy way to rewrite this equation.

It would have been nice to prove that this equation has no solution; unfortunately this is not the case since $(18^2+18+1)=(2^2+2+1)^3$, and one checks that $(2-j)^18 = (18-j^2)^6$ since $(2-j)^3$ and $18-j^2$ are conjugated.

So far, I don't know if there are other solutions (but scilab seems to answer me that this is the only one) and I can't obtain anything more than trivial results ($k$ is odd, $n$ is bounded by $m^k$ and $(m+1)^k$). So, I would like to know if this equation $(*)$ has already been studied somewhere, and/or if you have any idea to find all solutions; feel free to add any comment!

-
Do you have a link to Lebesgues proof? Maybe we can extend it. – user58512 Jan 29 '13 at 13:56
well, this proof relies on the study of the factorization of $n^2 = (m^2+1)^k-1$ in the ring of gaussian integers. I don't think that it could be extended as it is "easy" to study the factorization of n^2 but we don't have such an easy case here. The link to the article (sorry, that's in french) : archive.numdam.org/article/NAM_1850_1_9__178_0.pdf – user60066 Jan 29 '13 at 14:14

I've found that the trick to obtain a square is still working here: solving the equation (*) is related to finding solutions of $x^2=4 y^k-3$ with $k>2$. In the proof for gaussian integers we get $x^2=y^k-1=(y-1)(\underset{a=0}{\overset{k-1}{\sum}} y^a)$, but I don't see such an easy factorization for $4 y^k-3$.