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Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal

This question was asked in a past international math exam. My idea to solve it was to factorize it and then end up showing that the polynomial that was formed would directly show that at least two of the numbers $x$, $y$, $z$ are equal. However, it was to no avail.

I researched for theorems, that may help me in my quest to solving this problem, but none of them seemed applicable to this problem. Can someone please show me, a method of either finishing my thought, or if you believe that I was on the completely wrong trail of thought, can you please show me another method of solving it?

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    $\begingroup$ Which "international math exam" is this from? $\endgroup$ – Xander Henderson Dec 27 '19 at 17:38
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Solution. An identity that holds for arbitrary $x, y, z$ (in any commutative ring) is \begin{align} x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right) = \left(x-y\right) \left(x-z\right) \left(y-z\right) \label{darij1.eq1} \tag{1} \end{align} (you can check this by straightforward expanding). Thus, in your situation, $\left(x-y\right) \left(x-z\right) \left(y-z\right) = x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right) = 0$, so that one of the factors $x-y$, $x-z$ and $y-z$ must be $0$, and therefore two of the numbers $x, y, z$ must be equal.

How would you come up with this? Of course, it is much easier to expand a product of polynomials than to factor a polynomial. There is an algorithm (due to Dedekind, I think) for factoring an arbitrary polynomial over the ring of integers, but it is not pretty. Here are its main ideas: In the following, "polynomial" means "polynomial with integer coefficients".

  1. To factor a nonzero polynomial $f$ in a single variable $x$, we try to find two polynomials $g$ and $h$ (not equal to $1$ or $-1$) with the property that $f = gh$. Any two such polynomials $g$ and $h$ must satisfy $g\left(m\right) \mid f\left(m\right)$ for every integer $m$ (since $f\left(m\right) = g\left(m\right) h\left(m\right)$), and this divisibility restricts the possible values for $g\left(m\right)$ to a finite set of numbers (viz., the divisors of $f\left(m\right)$) as long as $f\left(m\right) \neq 0$. Thus, if we fix sufficiently many (= more than $\deg f$) distinct integers $m_1, m_2, \ldots, m_k$ such that $f\left(m_i\right) \neq 0$ for all $i$, then we have a finite number of options for each $g\left(m_i\right)$, and thus have only a finite number of options for $g$ (because any choice of $g\left(m_i\right)$ for all $i$ yields at most one possible polynomial $g$ via Lagrange interpolation -- here we are using the fact that $\deg g \leq \deg f$). Try all these options, and check if the resulting polynomial $g$ is a divisor of $f$. If it is (for some option), then you have found a proper divisor of $f$, and thus have made a first step towards the factorization of $f$; you can then proceed by recursion. If it is not (i.e., none of the options yields a valid polynomial $g$ with integer coefficients that is a divisor of $f$), then you have shown that $f$ is irreducible, and thus are done.

  2. So we know how to factor a polynomial in one variable. How do we factor a polynomial in $k$ variables for $k \geq 2$ ? There is a slick trick for this. Proceed by recursion on $k$; thus, we assume that we already know that we can factor any polynomial in $k-1$ variables. Now, let $f \in \mathbb{Z}\left[x_1, x_2, \ldots, x_k\right]$ be a polynomial in $k$ variables. Let $d$ be the (total) degree of $f$. Consider the $\mathbb{Z}\left[x_1, x_2, \ldots, x_{k-1}\right]$-algebra homomorphism $\mathbb{Z}\left[x_1, x_2, \ldots, x_k\right] \to \mathbb{Z}\left[x_1, x_2, \ldots, x_{k-1}\right]$ that sends $x_k$ to $x_1^{d+1}$. (It must also send each $x_i$ to $x_i$ when $i < k$, because it is a $\mathbb{Z}\left[x_1, x_2, \ldots, x_{k-1}\right]$-algebra homomorphism.) So $\Phi_d$ simply substitutes $x_1^{d+1}$ for $x_k$ in its input polynomial. This homomorphism $\Phi_d$ is injective on the set of polynomials of degree $\leq d$, because it sends distinct monomials in $x_1, x_2, \ldots, x_k$ of degree $\leq d$ to distinct monomials in $x_1, x_2, \ldots, x_{k-1}$ (indeed, it sends any monomial $x_1^{a_1} x_2^{a_2} \cdots x_k^{a_k}$ of degree $\leq d$ to the monomial $x_1^{\left(d+1\right) a_k + a_1} x_2^{a_2} x_3^{a_3} \cdots x_{k-1}^{a_{k-1}}$; but you can reconstruct the former monomial from the latter, because the exponent $\left(d+1\right) a_k + \underbrace{a_1}_{\leq d < d+1}$ can be decomposed back into its substituents $a_k$ and $a_1$ via division with remainder by $d+1$). Now, in order to factor $f$, again it suffices to find two polynomials $g$ and $h$ (not equal to $1$ or $-1$) with the property that $f = gh$, or to prove that no such $g$ and $h$ exist. In order to do so, we observe that any two such $g$ and $h$ must have degree $\leq d$, and for two polynomials $g$ and $h$ of such degree, the equality $f = gh$ is equivalent to $\Phi_d\left(f\right) = \Phi_d\left(g\right) \Phi_d\left(h\right)$ (because the map $\Phi_d$ is a ring homomorphism and is injective on polynomials of degree $\leq d$). Thus, if you can factor $\Phi_d\left(f\right)$, you can also factor $f$ (with the caveat that $\Phi_d$ is not surjective when restricted to polynomials of the appropriate degree, and thus not every divisor of $\Phi_d\left(f\right)$ can be "lifted" back to a divisor of $f$). But you can factor $\Phi_d\left(f\right)$ thanks to the induction hypothesis; so we can factor $f$.

This is a constructive (albeit rather inefficient) algorithm for factoring polynomials in any finite number of variables over $\mathbb{Z}$. If I apply it to your three-variable polynomial $x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right)$ (taking $x_1 = x$, $x_2 = y$ and $x_3 = z$), I first reduce it to the two-variable polynomial $x^2 \left(y-x^4\right) + y^2 \left(x^4-x\right) + x^8 \left(x-y\right)$ (which has degree $9$), and then reduce this result to the one-variable polynomial $x^2 \left(x^{10}-x^4\right) + x^{20} \left(x^4-x\right) + x^8 \left(x-x^{10}\right)$ (which has degree $24$). The latter one-variable polynomial can then be factored by checking combinations of divisors of values. A modern computer could well do it. Needless to say, there are much better algorithms around these days.

However, we can do much better. Fortunately, your polynomial $x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right)$ is not just a random polynomial. Rather, it is (up to sign) what you obtain if you expand the determinant \begin{align} \det\begin{pmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{pmatrix} \end{align} with respect to its last row. This determinant is the particular case (for $n = 3$ and $z_1 = x$ and $z_2 = y$ and $z_3 = z$) of the Vandermonde determinant \begin{align} \det\begin{pmatrix} 1 & 1 & \cdots & 1 \\ z_1 & z_2 & \cdots & z_n \\ z_1^2 & z_2^2 & \cdots & z_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ z_1^{n-1} & z_2^{n-1} & \cdots & z_n^{n-1} \end{pmatrix} , \end{align} which is known to equal $\prod_{1 \leq j < k \leq n} \left(z_k - z_j\right)$. The factorization of your polynomial is simply a particular case of this fact.

There is yet another way to discover the factorization \eqref{darij1.eq1}. A polynomial $f$ in three variables $x, y, z$ is divisible by the degree-$1$ polynomial $y-z$ if and only if $f$ becomes $0$ when $y$ and $z$ are set to be equal (i.e., if and only if $f\left(x,y,y\right) = 0$ as a polynomial in two variables $x$ and $y$). Thus, the polynomial $x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right)$ is divisible by $y-z$ (since setting $z=y$ in it results in $x^2 \left(y-y\right) + y^2 \left(y-x\right) + y^2 \left(x-y\right) = 0$). For similar reasons, it is divisible by $x-y$ and $x-z$ as well. But a polynomial that is divisible by each of the three degree-$1$ polynomials $x-y, x-z, y-z$ must always be divisible by their product $\left(x-y\right)\left(x-z\right)\left(y-z\right)$. (Indeed, when we are working with polynomials over the integers, this follows from the fact that $\mathbb{Z}\left[x,y,z\right]$ is a UFD. But this is still true for polynomials over arbitrary commutative rings. For a proof, see Theorem 1.2 in my Regular elements of a ring, monic polynomials and "lcm-coprimality".) So you conclude that your polynomial $x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right)$ is divisible by $\left(x-y\right) \left(x-z\right) \left(y-z\right)$. In other words, \begin{align} x^2 \left(y-z\right) + y^2 \left(z-x\right) + z^2 \left(x-y\right) = g\left(x,y,z\right) \cdot \left(x-y\right) \left(x-z\right) \left(y-z\right) \end{align} for some polynomial $g\left(x,y,z\right)$. But comparing the degrees on both sides of this equality, you see that $\deg g = 0$, and thus $g$ is just an integer constant. This constant must be $1$, as you can see by comparing the values of both sides at $\left(x,y,z\right) = \left(2,1,0\right)$. Thus, you obtain \eqref{darij1.eq1}.

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Hint: It is $(x-y)(x-z)(y-z)=0.$

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  • $\begingroup$ you were a big help thanks a lot $\endgroup$ – kenith Feb 26 '19 at 16:47
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It is the $(-1) \times$ Laplace expansion of the determinant of $$ \left[\begin{array}{ccc}1&1&1\\x&y&z\\x^2&y^2&z^2 \end{array}\right] $$ along the third row. This gives the Vandermonde determinant multiplied by $-1$ $$ -(x-y)(y-z)(z-x). $$

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  • $\begingroup$ Thank you mate you're great $\endgroup$ – kenith Feb 26 '19 at 16:47
  • $\begingroup$ @Song: the way you have the matrix, its determinant is $\mathbf + (x-y)(y-z)(z-x)$ (though that doesn't matter for the answer to OP's question) $\endgroup$ – J. W. Tanner Feb 26 '19 at 16:49
  • $\begingroup$ @kenith You're welcome! :) $\endgroup$ – Song Feb 26 '19 at 16:54
  • $\begingroup$ @J.W.Tanner Thank you for pointing out! I fixed the expression. $\endgroup$ – Song Feb 26 '19 at 16:54
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Factoring is hard and I'm not good at it.

But if:

$x^2(y-z)+y^2(z-x) + z^2(x-y) = 0$

$x^2(y-z)+y^2(z-x)= z^2(y-x)$

If $y =x$ then $y=x$. But if $y\ne x$ then $y-x \ne 0$.

$\frac {x^2(y-z) + y^2(z-x)}{y-x} = z^2$

So $\frac {xy*y - x^2z + y^2z - xy*y}{y-x} =$

$ -xy + \frac{y^2z -x^2z}{y-x} = $

$-xy + z\frac{(y-x)(y+x)}{y-x} =$

$-xy + z(y+x)=$

$-xy + zy + xz = z^2$

And so

$xz -xy = z^2 - zy$

$x(z-y) = z(z-y)$.

If $z = y$ then $z=y$ but if $z \ne y$ the $z - y\ne 0$ and

$x = z$.

So either $x=y$ or if not then either $z=y$ or if not $x = z$.

....

Which actually helps me figure out how to factor. So we could have done the following instead:

We can factor $x-y$ (or equivalently $y-x$) from $x^2(y-z) + y^2(z-x)$ via

$x^2(y-z) + y^2(z-x) =$

$xy*x - xy*y - (z*x^2 - z*y^2) = $

$(x-y)(xy - z(x+y))$

So $x^2(y-z) + y^2(z-x) + z^2(x-y) = $

$(x-y) (xy -z(x+y) + z^2)$

And we can factor $y-z$ from $xy - z(x+y) + z^2$ via

$xy - z(x+y) + z^2 = $

$xy - zx + z^2 - zy =$

$x(y-z) + z(z-y) =$

$(y-z)(x-z)$

And thus:

$x^2(y-z) + y^2(z-x) + z^2(x-y)= (x-y)(y-z)(x-z) = 0$

.... which means one of $x-y$ or $y-z$ or $x - z=0$.

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