Solving a system of the form $\{ux^n + vx^n = c_n\}$ for $n=0,1,2,3$

Question

Problem:

Solve the system: $$\begin{cases} u+v=2 \\ ux+vy=1 \\ ux^2+vy^2=-1 \\ ux^3+vy^3=-5 \end{cases}$$

My approach: If we sum relations we get $u(x^3+x^2+x+1) + v(y^3+y^2+y+1) = -3$. About giving $u=v$ and $x=y$ I get nothing. Any ideas?

Answer 1

We can eliminate $u$, $v$ and $x$ so that $$ u=\frac{4y^2 - 4y + 1}{2y^2 - 2y - 1},\; v=-\frac{3}{2y^2 - 2y - 1},\; x=\frac{y + 1}{2y - 1} $$ Then the first three equations hold, and the last one is $$ 3y^2-9y+6=0. $$

Answer 2

[Disclaimer : I don't pretend that I give here a solution ; it is only an insight about the situation allowing to predict (almost without calculation and almost certainly) the fact that there exists a finite number of solutions].

In such cases with intricated non-linear equations, let us provide this insight with a geometrical viewpoint.

As the LHSs of the equations are homogeneous in $u,v$, let us set $u=2k, v=2k-2$.

Let us as well set $x_1:=x, x_2:=y$ in order that the 3 last equations can be given the following barycentric "shape" :

$$k\underbrace{\begin{pmatrix}t_1\\t_1^2\\t_3^3\end{pmatrix}}_{P_1}+(1-k)\underbrace{\begin{pmatrix}t_2\\t_2^2\\t_2^3\end{pmatrix}}_{P_2}=\underbrace{\begin{pmatrix} \ \ \ 1/2\\-1/2\\-5/2\end{pmatrix}}_{P}\tag{1}$$

$P_1,P_2$ belong to the same curve [which, besides, is a known curve, "archetype" of cubic Bézier curves in 3D]. Relationship (1) means that $P_1, P_2,P$ are aligned.

One can say that for a fixed $P_1$, and a variable $P_2$ on the curve, the set of lines $P_1P_2$ defines a 2D surface (a sort of assymetrical conical surface with apex $P_1$ as shown on the figure for a certain position of $P_1$) that in general hasn't given point $P$ on it. With other words, it is hopefully for exceptional positions of $P_1$ that it happens to be the case.

Fig. 1 : Representation of the solution where $P_1(1,1^2,1^3), P_2(2,2^2,2^3)$ are aligned with $M(1/2,-1/2,-5/2)$.

I stop there this qualitative analysis (a methodology that can be applied to other similar cases) ; it has, so I think a "didactic interest" but of course a deeper analysis is needed.

I have - since - discovered the neat results of @Yves Daoust and @Jose Ramirez.

Answer 3

From the first two equations, we draw $u,v$:

$$\begin{cases}u=\dfrac{2y-1}{y-x},\\v=\dfrac{1-2x}{y-x}.\end{cases}$$

Plugging in the last two,

$$\begin{cases}-2xy+x+y=-1,\\x^2+xy+y^2-2xy^2-2yx^2=-5.\end{cases}\\$$

and setting $s:=x+y,p:=xy$, $$\begin{cases}-2p+s=-1,\\s^2-p-2ps=-5.\end{cases}$$

After elimination of $p$, we easily get $s=3,p=2$ and $x=1,y=2$ or conversely. Finally, $u=3,v=-1$ or conversely.

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Note that the option $x=y$ must be rejected.

Answer 4

From the initial equations we have: $$ ux^2 + vy^2 = (ux + vy)(x + y) - (u + v)xy \Rightarrow -1 = x + y - 2xy, $$ and $$ ux^3 + vy^3 = (ux^2 + vy^2)(x + y) - (ux + vy)xy \Rightarrow 5 = x + y + xy; $$ if we set $a = x + y$ and $b = xy$, we have the system of equations: $$ \begin{cases} a - 2b = -1 \\ a + b = 5 \end{cases} $$ with solution $(a,b) = (3, 2)$. Hence, $x$ and $y$ are solutions to the quadratic $x^2 - 3x + 2 = 0$, yielding two cases: $(x, y) = (1, 2)$ or $(x, y) = (2, 1)$. For the first case, the second equation of the system becomes $u + 2v = 1$, which with the first one leads to $(u, v) = (3, -1)$, hence a solution to the original system is $(u,v,x,y) = (3,-1,1,2)$; for the second case, by symmetry, we get $(u,v,x,y) = (-1,3,2,1)$.