Four Equations in Four Unknowns

Question

Four Equations in Four Unknowns
Completely solve the following equation!

$$\begin{eqnarray} x&+&y&+&z&+&w&=&10 \\ x^2&+&y^2&+&z^2&+&w^2&=&30 \\ x^3&+&y^3&+&z^3&+&w^3&=&100\\ &&&&&&xyzw&=&24 \end{eqnarray}$$

This is problem 3 on page 3 of
Mathematical Quickies
270 Stimulating Problems with Solutions
by Charles W. Trigg
Dover Publications, Inc., New York
ISBN 0-486-24949-2

Here is the solution from page 78 of the book:

By inspection $(1,2,3,4)$ is a solution if the first and fourth equation and satisfies the second and third equations. Since the equations are symmetrical in $x$, $y$, $z$, $w$ the other $23$ permutations of $1,2,3,4$ are solutions also. But these are all the solutions, since the product of the degrees of the equations is 4!

I assume that the exclamation mark at the end of the last sentence is a factorial symbol because the product of the degree of the equation is 24. This seems to be a property that is similar to the fact that a univariate polynomial of degree $n$ has at most $n$ zeroes. But I can't see how to generalize this to multivariate equations.

Why does this system of equations where the product of the degree of the equations is 24 has at most 24 solutions?

Answer 1

I. Yes, you are correct that a univariate polynomial of degree $n$ has $n$ zeros (counting multiplicity). However, your system are just the roots of a univariate in disguise.

The clue is the elementary symmetric polynomials $x+y+z+w$ and $xyzw$. If these unknowns are indeed the roots of the quartic, $$F(u)=u^4+au^3+bu^2+cu+d=0$$ then,

$$\begin{aligned} x+y+z+w &=10 = -a\\ x^2+y^2+z^2+w^2 &=30 = a^2-2b\\ x^3+y^3+z^3+w^3 &=100=-a^3 + 3 a b - 3 c\\ xyzw &=24=d \end{aligned}\tag1$$

It is easy to solve for $a,b,c,d\,$ above and we get, $$F(u)=u^4 - 10u^3 + 35u^2 - 50u + 24 = 0$$ $$F(u)=(u - 1)(u - 2)(u - 3)(u - 4)=0$$ hence these, including their permutations, are all the solutions.

II. Let $n_r$ be the number of roots and $n_d$ be the product of the degrees. By Bezout's theorem, then $n_r$ is at most equal to $n_d$. The observation that your example has $n_r=n_d=24$ is just a peculiarity of the system. If we tweak it slightly,

$$\begin{aligned} x+y+z+w &=10 = -a\\ x^2+y^2+z^2+w^2 &=30 = a^2-2b\\ \color{blue}{x^5+y^5+z^5+w^5} &=100=-a^5 + 5 a^3 b - 5 a b^2 - 5 a^2 c + 5 b c + 5 a d\\ xyzw &=24=d \end{aligned}\tag2$$

Solving for $a,b,c,d,\,$ they turn out to be rational so, $$F(u)=u^4 - 10u^3 + 35u^2 - \color{blue}{\tfrac{602}{13}}u + 24 = 0$$ though $F(u)$ is no longer rationally factorable. Like the previous, by including the permutations of the $u_i$, there are again $n_r=24$ roots, but the product of the degrees is different now as $n_d=1\times2\times5\times4 = 40$ so $n_r\neq n_d$.