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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!

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  • $\begingroup$ Since $u = (\alpha - 2)$ is a unit, then so is $\pm u^{n}$ for any integer $n$. That's a pretty explicit description! In fact, it turns out that this gives every unit of $\mathbf{Q}(\alpha)$ which follows (mostly) from Dirichlet's theorem. However, only finitely many powers of $u$ will not have an $\alpha^2$ term. That is more subtle, and follows from results of Thue. (The point is that $m/n$ will give an approximation to $\alpha$ with error of size $O(n^{-3})$, and algebraic numbers (of degree three by Thue and of all degrees by Roth) have only finitely many such approximations. $\endgroup$ – Furlo Roth Apr 15 at 13:17
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    $\begingroup$ Even better, Thue's result is effective, and in this case, the only solutions correspond to $\pm 1$ and $\pm u$. One can compute this in $\texttt{pari/gp}$ by typing $\texttt{thue(thueinit(x^3-x^2-5,1),1)}$. $\endgroup$ – Furlo Roth Apr 15 at 13:17
  • $\begingroup$ @user655377 I believe that the polynomial in two variables is just an error on the part of the OP, and that he should learn how to create the three variable polynomial coming from including $u + v \alpha + w\alpha^2.$ I did something like that in an answer. Your $(\alpha - 2)$ shows up in my computer output line 1 u: -2 v: 1 w: 0 //// Put another way, often enough I answer the question that the OP should have asked. $\endgroup$ – Will Jagy Apr 15 at 17:55
  • $\begingroup$ @user655377 I found what appears to be a distinct set of units as $\left(2 + \alpha+\alpha^2 \right)^n$ $\endgroup$ – Will Jagy Apr 15 at 18:35
  • $\begingroup$ @WillJagy, Umm... you do realize that Dirichlet's theorem says that the unit group of a non-totally real cubic field is $\mathbf{Z} \oplus \mu_2 = \pm u^{\mathbf{Z}}$, right? That might have let you to suspect that you did not find a "distinct set of units." Indeed, $u^{-1} = (\alpha - 2)^{-1} = 2 + \alpha + \alpha^2$. $\endgroup$ – Furlo Roth Apr 15 at 19:36
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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

jagy@phobeusjunior:~$ ./mse | sort -n
   -1        u: -123   v: -35   w: 44
   -1        u: -163   v: -77   w: -69
   -1        u: 19   v: 27   w: -17
   -1        u: -19   v: -9   w: -8
   -1        u: -1   v: 0   w: 0
   -1        u: 2   v: -1   w: 0
   -1        u: -2   v: -1   w: -1
   -1        u: 3   v: -12   w: 5
   -1        u: -4   v: 4   w: -1


    1        u: 123   v: 35   w: -44
    1        u: 163   v: 77   w: 69
    1        u: -19   v: -27   w: 17
    1        u: 19   v: 9   w: 8
    1        u: 1   v: 0   w: 0
    1        u: -2   v: 1   w: 0
    1        u: 2   v: 1   w: 1
    1        u: -3   v: 12   w: -5
    1        u: 4   v: -4   w: 1
jagy@phobeusjunior:~$

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, $$ \color{purple}{ (u,v,w) \mapsto (-2u+5w, u-2v, v-w)} $$

jagy@phobeusjunior:~$ ./mse 
    1        u: -2   v: 1   w: 0
    2        u: 4   v: -4   w: 1
    3        u: -3   v: 12   w: -5
    4        u: -19   v: -27   w: 17
    5        u: 123   v: 35   w: -44
    6        u: -466   v: 53   w: 79
    7        u: 1327   v: -572   w: -26
    8        u: -2784   v: 2471   w: -546
    9        u: 2838   v: -7726   w: 3017
   10        u: 9409   v: 18290   w: -10743
   11        u: -72533   v: -27171   w: 29033
   12        u: 290231   v: -18191   w: -56204
   13        u: -861482   v: 326613   w: 38013
   14        u: 1913029   v: -1514708   w: 288600
   15        u: -2383058   v: 4942445   w: -1803308
   16        u: -4250424   v: -12267948   w: 6745753
   17        u: 42229613   v: 20285472   w: -19013701
   18        u: -179527731   v: 1658669   w: 39299173
   19        u: 555551327   v: -182845069   w: -37640504
   20        u: -1299305174   v: 921241465   w: -145204565
   21        u: 1872587523   v: -3141788104   w: 1066446030
   22        u: 1587055104   v: 8156163731   w: -4208234134
   23        u: -24215280878   v: -14725272358   w: 12364397865
   24        u: 110252551081   v: 5235263838   w: -27089670223
   25        u: -355953453277   v: 99782023405   w: 32324934061
   26        u: 873531576859   v: -555517500087   w: 67457089344
   27        u: -1409777706998   v: 1984566577033   w: -622974589431
   28        u: -295317533159   v: -5378910861064   w: 2607541166464
   29        u: 13628340898638   v: 10462504188969   w: -7986452027528
   30        u: -67188941934916   v: -7296667479300   w: 18448956216497
   31        u: 226622664952317   v: -52595606976316   w: -25745623695797
   32        u: -581973448383619   v: 331813878904949   w: -26849983280519
   33        u: 1029696980364643   v: -1245601206193517   w: 358663862185468

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.

jagy@phobeusjunior:~$ ./mse 
    1        u: 2   v: 1   w: 1   normform: 1
    2        u: 19   v: 9   w: 8   normform: 1
    3        u: 163   v: 77   w: 69   normform: 1
    4        u: 1401   v: 662   w: 593   normform: 1
    5        u: 12042   v: 5690   w: 5097   normform: 1
    6        u: 103504   v: 48907   w: 43810   normform: 1
    7        u: 889643   v: 420368   w: 376558   normform: 1
    8        u: 7646706   v: 3613169   w: 3236611   normform: 1
    9        u: 65725367   v: 31056099   w: 27819488   normform: 1
   10        u: 564926109   v: 266935005   w: 239115517   normform: 1
   11        u: 4855682413   v: 2294373704   w: 2055258187   normform: 1
   12        u: 41735815216   v: 19720720756   w: 17665462569   normform: 1
   13        u: 358729859902   v: 169504569573   w: 151839107004   normform: 1
   14        u: 3083373637709   v: 1456934534068   w: 1305095427064   normform: 1
   15        u: 26502374216398   v: 12522719841165   w: 11217624414101   normform: 1
   16        u: 227794591779631   v: 107635935969233   w: 96418311555132   normform: 1
   17        u: 1957951978956747   v: 925158021493757   w: 828739709938625   normform: 1
   18        u: 16829091164768529   v: 7951966571637386   w: 7123226861698761   normform: 1
   19        u: 144650283804711598   v: 68349158616537106   w: 61225931754838345   normform: 1
   20        u: 1243305678240492176   v: 587478259811977535   w: 526252328057139190   normform: 1
   21        u: 10686525936112263927   v: 5049523838150143196   w: 4523271510093004006   normform: 1
   22        u: 91853386163905283894   v: 43401931162877570349   w: 38878659652784566343   normform: 1
   23        u: 789503024670044082963   v: 373050546753583256307   w: 334171887100798689964   normform: 1
   24        u: 6785977654115991347101   v: 3206463553681204045397   w: 2872291666580405355433   normform: 1
   25        u: 58327189742442056475517   v: 27560363094380426215060   w: 24688071427800020859627   normform: 1
   26        u: 501336909234786452622604   v: 236888273070203013203772   w: 212200201642402992344145   normform: 1
   27        u: 4309117200244617894705518   v: 2036114463587207440750873   w: 1823914261944804448406728   normform: 1
   28        u: 37037949337873317477232681   v: 17500917437143055018240904   w: 15677003175198250569834176   normform: 1
   29        u: 318350517613444415744011642   v: 150424800088150680362885369   w: 134747796912952429793051193   normform: 1
   30        u: 2736303004797166531232962059   v: 1292939102354507925435038345   w: 1158191305441555495641987152   normform: 1
   31        u: 23519214575782427646060987363   v: 11113137736713959860312974509   w: 9954946431272404364670987357   normform: 1
   32        u: 202153582147858698240396720841   v: 95520222205572369190041873166   w: 85565275774299964825370885809   normform: 1
   33        u: 1737561033066578890684711665602   v: 821020405430503260747334896218   w: 735455129656203295921964010409   normform: 1
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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.

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    $\begingroup$ I put the version with three integer variables, but in matrix slang. $\endgroup$ – Will Jagy Apr 15 at 17:51
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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 $n=-1, m=2$...

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Above equation shown below:

$m^3+5n^3+m^2n=1$

Above also has numerical solution:

$(n,m)=[(8/13),(-11/13)]$

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