Infinitely many integer solutions of a cubic equation Is it true that there are infinitely many pairs of integers $(m,n)$ such that
$m^3 + 5n^3 + m^2n = 1$? Or maybe $m^3 + 5n^3 + m^2n = -1$? The point is that I am trying to find a description of an infinite set of units of $\mathbb{Q}(\alpha)$ where $\alpha$ is a root of $x^3 - x^2 - 5 = 0$.
The only idea I have is to attack is as Pell's equation -- start from a small solution then find a suitable recursion. However, I am unlucky with the latter.
Any help appreciated!
 A: Thue proved in 1909 that if the cubic form $f(m,n)=am^3+bm^2n+cmn^2+dn^3$ has integer coefficients and nonzero discriminant, and $r$ is an arbitrary integer, then $f(m,n)=r$ has only finitely many solutions. 
The condition on the discriminant amounts to saying that the cubic polynomial $p(m)=am^3+bm^2+cm+d$ has no repeated roots. 
More information is available by searching the web for Thue's theorem on cubic forms. 
A: This isn't really an answer, but rather a long comment
Factoring yields 
\begin{align*}
m^3+5n^3+m^2n=1&\iff 5n^3+m^2n=1-m^3\\
&\iff n\cdot \underbrace{(5n^2+m^2)}_{\ge 0}=(1-m)\cdot \underbrace{(m^2+m+1)}_{\ge 0}
\end{align*}
We observe the obvious integer solution $n=0, m=1$. Wolfram Alpha suggests, that the only integer solution left is $n=1, m=-2$.
Similarly, for the second equation $$m^3+5n^3+m^2n=-1$$ the only integer solutions are $n=0, m=-1$ and $n=-1, m=2$...
A: Above equation shown below:
$m^3+5n^3+m^2n=1$
Above also has numerical solution: 
$(n,m)=[(8/13),(-11/13)]$
