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!
On
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 $n=-1, m=2$...
On
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.
I'm surprised Gerry did not correct you. Take a matrix $M$ with characteristic polynomial $x^3 - x^2 - 5.$ We might as well use the companion matrix $$ M = \left( \begin{array}{ccc} 0 & 1 & 0\\ 0&0&1 \\ 5&0&1 \end{array} \right) $$ Note that $M^3 = M^2 + 5 I.$ Next define the normform as $$ f(u,v,w) = \det \left( uI + v M + w M^2 \right) $$ or $$ f(u,v,w) = u^3 + 5 v^3 + 25 w^3 - 15 uvw + u^2 v + u^2 w -10 u w^2 + 5 v^2 w$$
The coefficient is zero for $uv^2$ and $v w^2$
This polynomial gives your infinite set. We have integer triples $(u,v,w)$ and,say, $(x,y,z).$ We get a new triple with multiplied form value using the multiplication rules $$ M^3 = M^2 + 5I \; , \; $$ $$ M^4 = M^2 + 5M + 5I \; , \; $$ and expanding $$ \left(uI + vM + wM^2 \right) \left(xI+ yM + zM^2 \right) = \left(pI + qM + rM^2 \right) $$ to calculate the new integer triple $(p,q,r)$
Let me find some units, give me a few minutes
next are units given by raising a fixed unit $\alpha - 2$ to an exponent, then finding the coefficients. The multiplication by that unit just comes out as a linear integer recursion, $$ \color{purple}{ (u,v,w) \mapsto (-2u+5w, u-2v, v-w)} $$
next are units given by raising a fixed unit $2 +\alpha + \alpha^2 $ to an exponent, then finding the coefficients. In comments just below the question, user655377 has pointed out that my $2 + \alpha + \alpha^2$ is simply the multiplicative inverse of $2 - \alpha.$
The multiplication by that unit just comes out as a linear integer recursion, $$ \color{purple}{ (u,v,w) \mapsto (2u+5v+10w, u+2v+5w,u+2v+4w)} $$ Cayley-Hamilton tells us that we have separate recurrences for the coefficients below, $$u_{n+3} = 8 u_{n+2} + 5 u_{n+1} + u_n \; , \; $$ $$v_{n+3} = 8 v_{n+2} + 5 v_{n+1} + v_n \; , \; $$ $$w_{n+3} = 8 w_{n+2} + 5 w_{n+1} + w_n \; . \; $$
These units come out with positive, and increasing, coefficients, so it is obvious they are all distinct.