A(x,y), B(x,y), and C(x,y) are three homogeneous real-coefficient polynomials of x and y with degree 2, 3, and 4 respectively. we know that there is a real-coefficient polinimial R(x,y) such that $B(x,y)^2-4A(x,y)C(x,y)=-R(x,y)^2$. Proof that there exist 2 polynomials F(x,y,z) and G(x,y,z) such that $F(x,y,z)^2+G(x,y,z)^2=A(x,y)z^2+B(x,y)z+C(x,y)$ if for any x, y, z real numbers $A(x,y)z^2+B(x,y)z+C(x,y)\ge 0$
Problem
Source: 2017 China TST 5 P5
Tags: algebra, polynomial, China TST
Domingues3
14.04.2017 17:03
Does this help? $A(x,y)z^2+B(x,y)z+C(x,y) = \frac{1}{4A(x,y)}[(2A(x,y)z + B(x,y))^2 + R(x,y)^2]$
TLP.39
01.03.2018 15:57
Assume that $A(x,y)z^2+B(x,y)z+C(x,y)\ge 0\,\forall x,y,z\in\mathbb{R}$.It is easy to see that $A(x,y)\ge 0\,\forall x,y\in\mathbb{R}$.
For convenience,we will write $A,B,C,F,G,R$ to represent the polynomial henceforth.
We will split up into $2$ cases.
1. $A$ is square.
Let $A=N^2$,then for any roots $(x,y)$ of $N$,we must have $B(x,y)=0$.Hence $N|B$ and $N|R$ and thus
$$Az^2+Bz+C=(Nz+\frac{B}{2N})^2+(\frac{R}{2N})^2$$2. $A$ is not square (hence irreducible on $\mathbb{R}[x,y]$.)
2.1. $B$ has a common non-constant factor with $A$.
This means that $A|B$ (since $A$ is irreducible on $\mathbb{R}[x,y]$),which imply that $A|R$ and thus $A|C$.
Let $B=KA,R=MA$ and $C=LA$
It is easy to see that $A=S^2+T^2$ for some polynomial $S,T\in\mathbb{R}[x,y]$.Hence
$$Az^2+Bz+C=A(z^2+Kz+L)=(S^2+T^2)[(z+\frac{K}{2})^2+(\frac{M}{2})^2]=(Sz+\frac{SK+TM}{2})^2+(Tz+\frac{TK-SM}{2})^2$$2.2. $B$ doesn't have a common non-constant factor with $A$.
This means that $R$ doesn't have a common non-constant factor with $A$ either.Thus $R\not\equiv 0$.
Let $G_1,F_1$ be the homogeneous real-cooficient polynomials on $x$ and $y$ with degree $1$ such that $A|B-G_1x^2$ and $A|R-F_1x^2$.
It is easy to see that $A|G_1R-F_1B$ and $A|F_1R+G_1B$.
Let $F_2=\frac{F_1B-G_1R}{2A},G_2=\frac{F_1R+G_1B}{2A}$.
Consider the polynomial $(F_1z+F_2)^2+(G_1z+G_2)^2$,by some algebraic manipulating,we get that $z=\frac{-B\pm Ri}{2A}$ are the roots of the polynomial.Hence $(F_1z+F_2)^2+(G_1z+G_2)^2=k(Az^2+Bz+C)$ for some positive real number $k$.
Finally,let $F=\frac{F_1z+F_2}{\sqrt{k}},G=\frac{G_1z+G_2}{\sqrt{k}}$ and we are done.
xdiegolazarox
01.03.2018 16:29
TLP.39 wrote: 2. $A$ is not square (hence irreducible on $\mathbb{R}[x,y]$.) why $A$ would be irreducible?
TLP.39
01.03.2018 17:09
If it is reducible but is not square,then there must exists $(x,y)\in\mathbb{R}^2$ such that $A(x,y)<0$.Taking $z\to \infty$ would lead to contradiction with the assumption $Az^2+Bz+C\ge 0$.
Iamsohappy
29.01.2020 14:37
Hi! @above, I still don't understand what you wrote. Every polynomial with certain variables that is always positive but not square is not necessarily irreducible. You can take $A(x,y)=(x^2+1)(y^2+2)$ as an example or many other polynomials of the form $A(x,y)=(r_1x^4+r_2x^8y^2+r_3y^{12}+...+r_k)(s_1x^6+...+s_u)()....()$ is a counterexample if we assume all coefficients $r_1,r_2,...,r_k,s_1,...,s_u,....$ are positive.
TLP.39
29.01.2020 15:20
@Above I also used the fact that $A$ is homogeneous with degree $2$. In this case, if $A$ is reducible, then it's of the form $(ax+by)(cx+dy)$. Either both terms are dependent which implies that $A$ is a square, or both terms are independent which implies that for some $x,y$, we have $ax+by=1$ and $cx+dy=-1$, clearly a contradiction.
Iamsohappy
31.01.2020 19:14
@above Thank you very much! Could you also explain how you said that there are homogeneous $F_1,G_1$ satisfying $A\mid B-G_1x^2$ and $A\mid R-F_1x^2$?