On the underdetermined system $x+y+z =a$, $x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 +b$?

Question

It is a reasonable assumption that to get the complete set of solutions to a system of $m$ equations in $n$ variables, we can just keep on eliminating variables until only one equation is left. For example, consider the system in 3 unknowns $(x,y,z)$ discussed in this post,

$x+y+z =a$

$x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 +b$

for some constant $(a,b)$.

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I. Method 1: Quartic

We can eliminate $y$ between the two and end up with a single equation, a quartic. The solution is then,

$$x =\frac{a-z+\sqrt{-r-2z+\sqrt{\beta\,}}}2$$ $$y =\frac{a-z-\sqrt{-r-2z+\sqrt{\beta\,}}}2$$

for any $z$ and where,

$$r = 3 (a - z)^2 - 3a + 2z$$ $$\beta = a^2( 9 - 16 a + 8 a^2) - 8 b - 8 (a - 1) (r + a^2 + z^2)z$$

and no need for cube roots. Using either case of $\pm\sqrt{\beta}$, there are actually two triplets $(x,y,z)$. It seems at first this is the complete solution.

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II. Method 2: Cubic

But as pointed out by Thomas Andrews in the linked post, the expressions $x^n+y^n+z^n$ and $xyz$ suggests the elementary symmetric polynomials for cubics. Re-phrasing the system as,

$x+y+z =a$

$x^3 +y^3 +z^3 +xyz = k$

$x^4 + y^4 +z^4 +b = k$

so a system of 3 equations in 3 unknowns. Solving for $(x,y,z)$, we find they are the 3 roots of a cubic in $w$,

$$16 w^3 - 16a w^2 + 4a^2 w + (a^3 - 4 k) =(3 a - 4 w)\sqrt{a^4 - 8 b - 8 k (a - 1)}$$

where we can use either case of the square root $\pm\sqrt{N}$, so again there are two triplets $(x,y,z)$.

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III. Example

Let $(a,b)=(1,-10)$

$x+y+z =1$

$x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 -10$

with quadratic solutions for infinitely many $z$,

$$x = \frac{1-z+\sqrt{-3z^2+2z+9}}2\\ y = \frac{1-z-\sqrt{-3z^2+2z+9}}2$$

and cubic solutions for infinitely many $k$,

$$4w^3 - 4w^2 - 8w - k + 7=0$$

where $(x,y,z)$ are the three roots of the cubic. The most well-known is for $k=3$,

$$(x,y,z)= \big({-2}\cos(2\pi/7),\;-2\cos(4\pi/7),\;-2\cos(6\pi/7)\big)$$

the roots of $w^3 - w^2 - 2w + 1 = 0$. Note that for appropriate $z$, then the first set of solutions are constructible numbers, whereas the second set is not.

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IV. Question

Q: Why doesn't the first method suffice to give the complete solution or, especially for more complicated systems, are we sure these two methods solve the system completely?

Answer 1

You commented

the first set of solutions are constructible numbers, needing only square roots of square roots.

Since $z$ can be any number, I think the first set of solutions includes the solutions whose $z$ is not a constructible number.

In your example where $(a,b)=(1,-10)$, I noticed that $$(x,y,z)=(-2\cos(2\pi/7),-2\cos(4\pi/7),-2\cos(6\pi/7))$$ is included in the first set of solutions.

(A proof is written later.)

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I think the first method gives the complete solution.

Let $A:=a-z,B:=-r-2z$.

$$\begin{align}&\begin{cases}x+y+z=a \\x^3+y^3+z^3+xyz=x^4+y^4+z^4+b\end{cases} \\\\&\iff\begin{cases}y=A-x \\x^3+(A-x)^3+z^3+x(A-x)z \\\qquad=x^4+(A-x)^4+z^4+b\end{cases} \\\\&\iff\begin{cases}y=A-x \\((2x-A)^2-B)^2=\beta\end{cases} \\\\&\iff\begin{cases}y=A-x \\x=\frac{A\pm\sqrt{B\pm\sqrt{\beta}}}{2}\end{cases} \\\\&\iff (x,y)=\begin{cases}(\frac{A+\sqrt{B+\sqrt{\beta}}}{2},\frac{A-\sqrt{B+\sqrt{\beta}}}{2}) \\(\frac{A+\sqrt{B-\sqrt{\beta}}}{2},\frac{A-\sqrt{B-\sqrt{\beta}}}{2}) \\(\frac{A-\sqrt{B+\sqrt{\beta}}}{2},\frac{A+\sqrt{B+\sqrt{\beta}}}{2}) \\(\frac{A-\sqrt{B-\sqrt{\beta}}}{2},\frac{A+\sqrt{B-\sqrt{\beta}}}{2})\end{cases}\end{align}$$

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Also, I think the second method gives the complete solution.

$$\begin{align}&\begin{cases}x+y+z=a \\x^3+y^3+z^3+xyz=k \\x^4+y^4+z^4+b=k\end{cases} \\\\&\iff\begin{cases}x+y+z=a \\a^3-3a(xy+yz+zx)+4xyz=k \\a^4-4a^2(xy+yz+zx) \\\qquad+2(xy+yz+zx)^2+4axyz+b=k\end{cases} \\\\&\iff(x+y+z,xy+yz+zx,xyz)=\begin{cases}(a,\frac{a^2+\sqrt{a^4-8ak-8b+8k}}{4},\frac{4k-a^3+ 3a\sqrt{a^4-8ak-8b+8k}}{16}) \\(a,\frac{a^2-\sqrt{a^4-8ak-8b+8k}}{4},\frac{4k-a^3- 3a\sqrt{a^4-8ak-8b+8k}}{16})\end{cases}\end{align}$$

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In the following, let us consider the case where $(a,b)=(1,-10)$.

Let $$p:=-2\cos(2\pi/7),q:=-2\cos(4\pi/7),r:=-2\cos(6\pi/7)$$

I think $(x,y,z)=(p,q,r)$ is included in the first set of solutions.

Let $$f_+(z):= \frac{1-z+\sqrt{-3z^2+2z+9}}2$$ $$f_-(z):= \frac{1-z-\sqrt{-3z^2+2z+9}}2$$

Then, we can prove that $$f_-(r)=p,f_+(r)=q$$

Proof :

We have $$(2p-1+r)^2=-3r^2+2r+9\tag1$$ since $(1)$ is equivalent to $$\begin{align}&4p^2+1+r^2-4p-2r+4pr=-3r^2+2r+9 \\&\iff 4(p^2+r^2)-4(p+r)+4pr-8=0 \\&\iff 4(5-q^2)-4(1-q)-\frac{4}{q}-8=0 \\&\iff q^3-q^2-2q+1=0\end{align}$$ which is true.

From $(1)$ with $2p-1+r\lt 0$ and $2q-1+r\gt 0$, we get $$\begin{align}2p-1+r&=-\sqrt{-3r^2+2r+9} \\2q-1+r&=\sqrt{-3r^2+2r+9}\end{align}$$ so we finally obtain $$f_-(r)=\frac{1-r+(2p-1+r)}{2}=p,$$ $$f_+(r)=\frac{1-r+(2q-1+r)}{2}=q.\ \square$$

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(The first set of solutions)

Let $g_{\pm}(\zeta):=\sqrt{-3\zeta^2+2\zeta\pm 9}$.

$$(x,y,z)=\bigg(\frac{1-\zeta+g_+(\zeta)}{2},\frac{1-\zeta-g_+(\zeta)}{2},\zeta\bigg),$$ $$\bigg(\frac{1-\zeta+g_-(\zeta)}{2},\frac{1-\zeta-g_-(\zeta)}{2},\zeta\bigg),$$ $$\bigg(\frac{1-\zeta-g_+(\zeta)}{2},\frac{1-\zeta+g_+(\zeta)}{2},\zeta\bigg),$$ $$\bigg(\frac{1-\zeta-g_-(\zeta)}{2},\frac{1-\zeta+g_-(\zeta)}{2},\zeta\bigg)$$

(The second set of solutions)

$x,y,z$ are the roots of the cubic equation $$w^3-w^2+\frac{1\mp 9}{4}w-\frac{4k-1\mp 27}{16}=0$$

Now, let us show that every solution in the first set is included in the second set.

Proof :

For $$(x,y,z)=\bigg(\frac{1-\zeta+g_+(\zeta)}{2},\frac{1-\zeta-g_+(\zeta)}{2},\zeta\bigg),\bigg(\frac{1-\zeta-g_+(\zeta)}{2},\frac{1-\zeta+g_+(\zeta)}{2},\zeta\bigg)$$ we have $$\begin{cases}x+y+z=1 \\xy+yz+zx=-2 \\xyz=\zeta^3 - \zeta^2 - 2\zeta\end{cases}$$ So, $x,y,z$ are the roots of the cubic equation $$w^3-w^2+\frac{1-9}{4}w-\frac{4k-1-27}{16}=0$$ where $k=4\zeta^3 - 4\zeta^2 - 8\zeta + 7$.

For $$(x,y,z)=\bigg(\frac{1-\zeta+g_-(\zeta)}{2},\frac{1-\zeta-g_-(\zeta)}{2},\zeta\bigg),\bigg(\frac{1-\zeta-g_-(\zeta)}{2},\frac{1-\zeta+g_-(\zeta)}{2},\zeta\bigg)$$ we have $$\begin{cases}x+y+z=1 \\xy+yz+zx=\frac 52 \\xyz=\zeta^3-\zeta^2+\frac 52\zeta\end{cases}$$ So, $x,y,z$ are the roots of the cubic equation $$w^3-w^2+\frac{1+ 9}{4}w-\frac{4k-1+ 27}{16}=0$$ where $k=4\zeta^3-4\zeta^2+10\zeta-\frac{13}{2}$.$\ \square$

Finally, let us show that every solution in the second set is included in the first set.

Proof :

For any $k$, we can take $\zeta$ satisfying $k=4 \zeta^3 - 4 \zeta^2 - 8 \zeta + 7$. Using such $\zeta$, the roots of the cubic equation $$w^3-w^2+\frac{1-9}{4}w-\frac{4k-1-27}{16}=0$$ i.e. $$(w - \zeta)\bigg(w^2-(1-\zeta)w + \zeta^2 - \zeta - 2\bigg)=0$$ can be written as $$w=\zeta,\frac{1-\zeta\pm g_+(\zeta)}{2}$$ Each of the $3!=6$ solutions is included in the first set.
(For example, letting $u:=\frac{1-\zeta+g_+(\zeta)}{2}$, we get $\zeta=\frac{1-u\pm g_+(u)}{2}$.)

For any $k$, we can take $\zeta$ satisfying $k=4\zeta^3-4\zeta^2+10\zeta-\frac{13}{2}$. Using such $\zeta$, the roots of the cubic equation $$w^3-w^2+\frac{1+ 9}{4}w-\frac{4k-1+ 27}{16}=0$$ i.e. $$(w-\zeta) \bigg(w^2- (1-\zeta)w+ \zeta^2 - \zeta + \frac 52\bigg) = 0$$ can be written as $$w=\zeta,\frac{1-\zeta\pm g_-(\zeta)}{2}$$ Each of the $3!=6$ solutions is included in the first set.$\ \square$