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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?

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  • $\begingroup$ I am interested to hear what others have to say about this question. I wouldn't have expected such a convenient result to be true myself. For clarification, are we restricting the domain of the solutions? $x^2+y^2=1$ has infinitely many solutions (the unit circle)... $\endgroup$ – JMoravitz Dec 12 '16 at 7:42
  • $\begingroup$ @JMoravitz: thank you, I corrected the typo. I cited the full problem text. I think the domain is $\mathbb{C}^4$ $\endgroup$ – miracle173 Dec 12 '16 at 7:50
  • $\begingroup$ @JMoravitz: you have two variables but one equation. If you have two equations then you may have only finitely many solutions. circle and line: product of degree is 2, number of solutions is two. Circle and ellipsis: product of degee is 4, number of solutions can be 4. So I think something like $n$ equations in $n$ variables, product of degree is $k$, then there are at most $k$ solutions. $\endgroup$ – miracle173 Dec 12 '16 at 7:54
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    $\begingroup$ This is a consequence of Bezout's Theorem. en.wikipedia.org/wiki/B%C3%A9zout's_theorem $\endgroup$ – Leon Sot Dec 12 '16 at 8:00
  • $\begingroup$ this is a nice problem.............+1 $\endgroup$ – Bhaskara-III Dec 12 '16 at 8:10
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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 $n_r\neq n_d$.

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  • $\begingroup$ Hi @Tito, Thank you for the answer. I think your first answer is what the the author of the recreational math problem has in his mind. $\endgroup$ – miracle173 Dec 13 '16 at 15:35
  • $\begingroup$ @miracle173: You're welcome. Incidentally, If you recall, it is also the case that $\sum_{n=1}^k n^3 = \big(\sum_{n=1}^k n\big)^2$ which explains his value $(1+2+3+4)^2 = 10^2 = 100$. $\endgroup$ – Tito Piezas III Dec 14 '16 at 1:33

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