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Introduction: I found this question from Arthur Engel problem solving strategies , I thought the question was easy to solve using identities but I kept reaching dead ends

Question: Find all real solutions of the system

$ x+y+z =1 $

$x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 +1$

My approach :

Firstly,I tried subtracting $4xyz$ from both sides of the equation

So, $x^3 + y^3 +z^3 -3xyz = x^4 + y^4 +z^4 +1 -4xyz $

$ (x+y+z)(x^2+y^2+z^2-xy-yz-xz) = x^4 +y^4 +z^4 -4xyz +1$

$ (x^2+y^2+z^2-xy-yz-xz)= x^4 +y^4 +z^4 -4xyz +1 $

But I am unable to proceed further , even by using the identity $ x^2 + y^2 + z^2 = (x+y+z)^2 -2(xy + yz + zx)$ the equation just becomes too much complicated to solve then

I also tried taking specific cases but no luck in that either

Ved Patel
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    Whenever I see so many symmetric polynomials, my first inclination is to try to rewrite in terms of $s_1=x+y+z,$ $s_2=xy+xz+yz,$ and $s_3=xyz.$ – Thomas Andrews Feb 11 '25 at 16:37
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    For example: $x^4+y^4+z^4=s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2.$ – Thomas Andrews Feb 11 '25 at 16:47
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    Given that you know $s_1=1,$ you will get a quadratic equation for $s_2$ given any $s_3.$ So you can work out the cases when these three values are real. Once you know that, though, $x,y,z$ stil; might be complex, so you have to solve the cubic equation $u^3-s_1u^2+s_2u-s_3=0$ and find out if those roots are all real. – Thomas Andrews Feb 11 '25 at 16:54
  • Well as I wrote above $x^4 +y^4 + z^4 =4xyz$ and then using the simplification you gave the $4xyz$ term gets cancelled thus I get a quadratic in $s_2$ which has real roots but for $s_3$ i really dint know, I suppose that is correct? – Ved Patel Feb 11 '25 at 17:06
  • BTW, how did you jump from $x^3+y^3+z^2-3xyz$ to $(x+y+z)^3$? That isn't true in general, $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. – Macavity Feb 13 '25 at 14:02
  • The Grobner basis is $y^4 + 2y^3z + 3y^2z^2 + 2yz^3 + z^4 - 2y^3 - 4y^2z - 4yz^2 - 2z^3 + \frac{3}{2}y^2 + \frac{5}{2}yz + \frac{3}{2}z^2 - \frac{1}{2}y - \frac{1}{2}z + \frac{1}{2}, x + y + z - 1$ (computed using Sage). Choose any value for $z$, solve the quartic for $y$, and solve for $x$ using $x+y+z-1 = 0$. – David Lui May 01 '25 at 06:38

3 Answers3

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If $s_1=x+y+z,$ $s_2=xy+xz+yz,$ and $s_3=xyz$ are the primitive symmetric polynomials, then $$\begin{align} x^3+y^3+z^3&=s_1^3-3s_1s_2+3s_3\\ x^4+y^4+z^4&=s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2. \end{align}$$

Then using $s_1=1,$ your equations become:

$$1-3s_2+4s_3=2-4s_2+4s_3+2s_2^2.$$

So $2s_2^2-s_2+1=0.$

This give $s_2=\frac{1\pm\sqrt{-7}}4,$ which is not real.

Thomas Andrews
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    If the OP had a slightly different system where the RHS used $x^4+y^4+z^4-10$, then it could be solved by infinitely many reals (as I mentioned in my answer) with two solution classes: constructible (quadratics) and non-constructible (cubics). But doesn't using the elementary symmetric polynomials suggest there should be only one solution class? – Tito Piezas III May 01 '25 at 06:50
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It seems more can be said about the system,

$x+y+z =1$

$x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 +1$

which, after an adjustment, can be connected to $p=7$ and the heptagon. It is just 2 equations in 3 unknowns, hence is under-determined and should have infinitely many solutions. Interestingly, it seems there are two classes of solutions: 1st, using only square roots of square roots (quartics) and 2nd, roots of cubics.

I. Roots of quartics

Solve for $x$ in the first equation, then $y$ in the second, and we have the solution,

$$x = \frac{1-z+\sqrt{-3z^2+2z+\sqrt{-7}}}2\\ y = \frac{1-z-\sqrt{-3z^2+2z+\sqrt{-7}}}2$$

where we can choose either sign of $\pm\sqrt{-7}$, with free parameter $z$. Thus, the system has infinitely many solutions where one variable can be real ($z$), but the other two are complex.

II. Roots of cubics

As pointed out by Thomas Andrews, we can use the elementary symmetric polynomials. Rephrase the system as,

$x+y+z =1 $

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

$x^4 + y^4 +z^4 +1 = k$

and we now have 3 equations in 3 unknowns for some free parameter $k$. The $(x,y,z)$ then are the three roots of the cubic,

$$16 x^3 - 16x^2 + 4x - 4k + 1 = (4 x - 3)\sqrt{-7}$$

Because of the $\sqrt{-7},\,$ this generally yields only complex roots. For example, let $k=2$,

$$16 x^3 - 16x^2 + 4x - 7 = (4 x - 3)\sqrt{-7}$$

and the cubic has roots,

$$(x,y,z) \approx (-0.4134 - 0.7567 i,\; 0.3171+ 0.6174 i,\; 1.0962 + 0.1393 i)$$

which solve the OP's system. But they are all complex in contrast to the previous method where one of $(x,y,z)$ can be real.

III. Real roots

However, there are similar systems with all $(x,y,z)$ solvable in the reals (as the OP wished) and where the infinite number of solutions are again of two kinds: constructible numbers (which uses only square roots), and non-constructible numbers (which uses higher roots). Consider the system,

$x+y+z =1$

$x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 \color{red}{-10}$

which differs from the OP's only in the red number. Using the same approach as above, then:

A. First class (constructible)

The $(x,y,z)$ are,

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

and we can choose infinitely many rational $z$ such that $-3z^2+2z+9\geq0$.

B. Second class (non-constructible)

The $(x,y,z)$ are the three roots of the cubic,

$$x^3 - x^2 - 2x - n + 1 = 0$$

and we can choose infinitely many real $n$ such that $-27n^2 + 14n + 49\geq0$ so all roots are real. The most well-known example is $n=0$ and the cubic,

$$x^3 - x^2 - 2x + 1 = 0$$

whose roots are,

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

which appear in the heptagon and solve the given system, namely,

$x+y+z =1$

$x^3 +y^3 +z^3 +xyz = x^4 + y^4 +z^4 \color{red}{-10}$

And so on for other $n$. Thus, if this system had been the example in Arthur Engel's problem solving strategies, then both solution classes would have to be mentioned.

Tito Piezas III
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  • "It is just 2 equations in 3 unknowns, hence is under-determined and should have infinitely many solutions." This isn't necessarily true for systems of nonlinear equations, right? e.g. consider a paraboloid with a tangent plane at its vertex; that's two equations of 3 unknowns, but only finitely many solutions. – apnorton May 01 '25 at 06:17
  • @apnorton I guess I should phrase it as "... generally should have...". Mea culpa. What I find interesting about the system are the two classes of solutions. If the OP had chosen $-10$, then the real solutions are either constructible (quadratic) or non-constructible (cubic). – Tito Piezas III May 01 '25 at 06:22
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    That is certainly really interesting! :D – apnorton May 01 '25 at 06:23
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    Very Interesting solution , I could never figure this out lol – Ved Patel May 05 '25 at 19:00
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    @VedPatel What I find eye-opening about your post is that under-determined systems of equations may have more than one infinite family of solutions. Yours has a quartic and cubic family. I deal often with systems of equations and sometimes I assume I have the complete solution. This post reminded me to be careful about making that assumption. – Tito Piezas III May 06 '25 at 03:01
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You could also solve your problem by making an appropriate change of variables.
Let $\,u\,$ and $\,v\,$ be two variables such that $\;\begin{cases}x=u+v\\[3pt]y=u-v\\[3pt]z=1-2u\end{cases}\;.$

So the system $\;\begin{cases}x+y+z=1\\[3pt]x^3+y^3+z^3+xyz=x^4+y^4+z^4+1\end{cases}$

is equivalent to

$(u\!+\!v)^3\!+\!(u\!-\!v)^3\!+\!(1\!-\!2u)^3\!+\!(u\!+\!v)(u\!-\!v)(1\!-\!2u)=(u\!+\!v)^4\!+\!(u\!-\!v)^4\!+\!(1\!-\!2u)^4\!+\!1$

which is equivalent to

$18u^4\!+\!2v^4\!+\!12u^2v^2\!-\!24u^3\!-\!8uv^2\!+\!11u^2\!+\!v^2\!-\!2u\!+\!1=0$

which is equivalent to

$2\left(3u^2+v^2-2u+\dfrac14\right)^{\!2}\!+\dfrac78=0$

which cannot have real solutions.

Angelo
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