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Let $a$, $b$ be (complex) numbers (satisfying some natural conditions, see below). Show that for every $(u,v,w)$ the system $$ a x + b y + z = u \\ a x^2 + b y^2 + z^2 = v \\ a x^3 + b y^3 + z^3 = w$$

has finitely many solutions $(x,y,z)$.

The conditions on $a$, $b$, $c$ : $a$, $b$, $a+1$, $b+1$, $a+b$, $a+b+1 \ne 0$ ( otherwise it is easy to produce infinitely many solutions by having some of the $x$, $y$, $z$ equal).

Note: We want to show that the above polynomial map in $(x,y,z)$ has finite fibers. It is clear how to generalize this statement. That is in fact equivalent to : the map is proper. This perhaps can be checked in this particular case and in other cases using Gröbner bases. I am interested in an approach that might work in general.

$\bf{Added:}$ With WA I found an algebraic dependence of $x$ on $u$, $v$, $w$ (as it should). The coefficient of $x^6$ is $a^2 (1 + a) (a + b) (1 + a + b)^2$ Dividing by it, we get that $x$ is integral over the algebra generated by $u$, $v$, $w$. The same works perhaps with $y$, $z$. The calculations are fairly large.

$\bf{Added:}$ Sketch of a proof: Stratify $\mathbb{C}^n$ using equality of components. Assume that some fibre is infinite. It follows that its dimension is positive. But then the fibre restricted to some stratum has positive dimension. However, using the Jacobian, we see that restricted to a stratum the map $\Phi$ is locally injective. Contradiction

orangeskid
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    so, what happens when $a=b=1?$ – Will Jagy Sep 28 '22 at 00:50
  • @Will Jagy: In that case it is easy to see there are finitely many solutions. But not true for $a=1$, $b=-1$. Yes, it is a good particular case. – orangeskid Sep 28 '22 at 01:00
  • Intuitively, it is not surprizing that in $\mathbb{R}^3$ (resp $\mathbb{R}^n$), $3$ (resp. $n$) algebraic surfaces have a finite number of intersection points... Proving it in the particular case of these equations is another thing. – Jean Marie Sep 28 '22 at 09:57
  • @Jean Marie: Yes, indeed. Solving for say $x$ from the system is possible, but quite involved. Maybe some easier way is possible, – orangeskid Sep 28 '22 at 10:40
  • How about showing that the three equations are algebraically independent in $\mathbb{C}[x,y,z]$? Then, the ring of regular functions are finite dimensional over $\mathbb{C}$ and such rings have finite number of prime ideals all of which are maximal. In addition, to show algebraic independence, we can reduce to the case $u=v=w=0$. – Acrobatic Sep 28 '22 at 14:19
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    @Acrobatic: Interesting idea! If you show that $x$, $y$ $z$ are algebraically dependent on $u$, $v$, $w$ that shows algebraic independence. I think you need in fact to show integral dependence. Compare with the case $u=x$, $v= x y$. – orangeskid Sep 28 '22 at 18:11

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There is a computer-based proof done with Mathematica 13.1. The command

Solve[{a x + b y + z == u , a x^2 + b y^2 + z^2 == v, 
a x^3 + b y^3 + z^3 == w}, {x, y, z},  Assumptions -> 
a != 0 && b != 0 && a + 1 != 0 && b + 1 != 0 && a + b != 0 && a + b + 1 != 0]

produces the solution where $x,y,z$ are expressed as roots of polynomials with coefficients from the field of rational functions $\mathbb Q (a,b,u,v,w)$. The whole output takes 79.6 MB in memory. Here is the screen of the shortened output. BTW, the execution of the command is not long.

Addition.The code

a = 1; b = -1; Reduce[{a x + b y + z == u, a x^2 + b y^2 + z^2 == v, 
  a x^3 + b y^3 + z^3 == w}, {x, y, z}]

(u^2 - v != 0 && (x == ( 4 u^3 - 4 w - Sqrt[(-4 u^3 + 4 w)^2 - 4 (6 u^2 - 6 v) (u^4 + 3 v^2 - 4 u w)])/(12 (u^2 - v)) || x == (4 u^3 - 4 w + Sqrt[(-4 u^3 + 4 w)^2 - 4 (6 u^2 - 6 v) (u^4 + 3 v^2 - 4 u w)])/(12 (u^2 - v))) && y == (-u^3 + 3 u v - 2 w)/(3 (u^2 - v)) && z == u - x + y) || (w == 0 && v == 0 && u == 0 && x == 0 && z == y) || ((v == w^(2/3) || v == -(-1)^(1/3) w^(2/3) || v == (-1)^(2/3) w^(2/3)) && v != 0 && u == w/v && x == u && z == y) || (w == 0 && v == 0 && u == 0 && -x != 0 && y == x && z == -x + y) || ((v == w^(2/3) || v == -(-1)^(1/3) w^(2/3) || v == (-1)^(2/3) w^(2/3)) && v != 0 && u == w/v && u - x != 0 && y == x && z == u - x + y)

, as we see, results in infinite number of the solutions.

user64494
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  • Thank you for your reply! I used WA to check with Groebner bases that $x$ ( and $y$, $z$ I believe) are integral over $u$, $v$, $w$ ( we use some expression in $a$, $b$ as a denominator, that we know is $\ne 0$) . Can you compare it with the case $a=1$, $b=-1$? I believe there would exist an infinite family of solutions. – orangeskid Sep 28 '22 at 18:08
  • Continue to believe... – user64494 Sep 28 '22 at 19:31