Symmetric Homogeneous Polynomial of Degree 3

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Symmetric Homogeneous Polynomial of Degree 3

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Source: 2025 Taiwan TST Round 3 Independent Study 1-A

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USJL

#1 h Apr 26, 2025, 12:37 PM • 1 Y

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Find all symmetric homogeneous polynomials $P(x,y,z)$ with real coefficients of degree $3$ such that $P(1,x,x^2)$ divides $P(-(x+1)^3,x,x^2)$ .

Proposed by usjl

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snorlax_snorlax

#2 h Apr 26, 2025, 12:40 PM

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When trying to bash the problem, remember to consider that the coefficient of $(x^3+y^3+z^3)$ could be 0

This post has been edited 2 times. Last edited by snorlax_snorlax, Apr 26, 2025, 12:41 PM
Reason: Grammer

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#3 h Apr 26, 2025, 1:18 PM

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USJL wrote:

Find all symmetric homogeneous polynomials $P(x,y,z)$ with real coefficients of degree $3$ such that $P(1,x,x^2)$ divides $P(-(x+1)^3,x,x^2)$ .

Proposed by usjl

Let $P(x,y,z)=ap^3+bpq+cr$ , where $p=x+y+z$ , $q=xy+yz+zx$ and $r=abc$ . Note that $G(x)=P(1,x,x^2)=a(1+x+x^2)^3+bx(1+x+x^2)^2+cx^3$ and $F(x)=P(-(x+1)^3,x,x^2)=-a(x+1)^3(x^2+x+1)^3+bx(x^2-(x+1)^4)-cx^3(x+1)^3$ . We want to find when $G(x) \mid F(x)$ .

Note that it's equivalent to $G(x) \mid F(x)+(x+1)^3G(x)=bx(x^2-(x+1)^4)+bx(1+x+x^2)^2(1+x)^3=bx(x^2+x+1)(x^4+4x^3+7x^2+6x+1)=:H(x)$ . If $b=0$ , this obviously holds. Now let $b \neq0$ , so WLOG $b=1$ .

Note that $\deg G \geqslant 5$ .
If $c=0$ , then $(1+x+x^2)^2 \mid G(x) \mid H(x)$ , but $1+x+x^2 \nmid x(x^4+4x^3+7x^2+6x+1)$ , so $c \neq 0$ . Therefore $\deg \left( gcd(G(x), 1+x+x^2) \right)=0$ and so $G(x) \mid x(x^4+4x^3+7x^2+6x+1)$ . If $a \neq 0$ , then $\deg \left(gcd(G(x),x) \right)=0$ and so $G(x) \mid x^4+4x^3+7x^2+6x+1$ , but $\deg G >4$ , so it's impossible. So $a=0$ .

Then $\deg G(x)=5$ and so $G(x)=x(x^4+4x^3+7x^2+6x+1) \cdot const$ . But it is impossible:
$G(x)=x(1+x+x^2)^2+cx^3$ , so by looking on $x^5$ we get $const=1$ , while looking on $x^4$ we get $const=\frac{1}{2}$ , impossible!

So $P(x,y,z)=s(x+y+z)^3+txyz$ for some $s,t \in \mathbb{R}$ .

This post has been edited 1 time. Last edited by NO_SQUARES, Apr 26, 2025, 2:48 PM
Reason: deg of constant polynom is zero

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USJL

#4 h Apr 26, 2025, 1:19 PM

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NO_SQUARES wrote:

USJL wrote:

Find all symmetric homogeneous polynomials $P(x,y,z)$ with real coefficients of degree $3$ such that $P(1,x,x^2)$ divides $P(-(x+1)^3,x,x^2)$ .

Proposed by usjl

Let $P(x,y,z)=ap^3+bpq+cr$ , where $p=x+y+z$ , $q=xy+yz+zx$ and $r=abc$ . Note that $G(x)=P(1,x,x^2)=a(1+x+x^2)^3+bx(1+x+x^2)^2+cx^3$ and $F(x)=P(-(x+1)^3,x,x^2)=-a(x+1)^3(x^2+x+1)^3+bx(x^2-(x+1)^4)-cx^3(x+1)^3$ . We want to find when $G(x) \mid F(x)$ .

Note that it's equivalent to $G(x) \mid F(x)+(x+1)^3G(x)=bx(x^2-(x+1)^4)+bx(1+x+x^2)^2(1+x)^3=bx(x^2+x+1)(x^4+4x^3+7x^2+6x+1)=:H(x)$ . If $b=0$ , this obviously holds. Now let $b \neq0$ , so WLOG $b=1$ .

Note that $\deg G \geqslant 5$ .
If $c=0$ , then $(1+x+x^2)^2 \mid G(x) \mid H(x)$ , but $1+x+x^2 \nmid x(x^4+4x^3+7x^2+6x+1)$ , so $c \neq 0$ . Therefore $\deg \left( gcd(G(x), 1+x+x^2) \right)=1$ and so $G(x) \mid x(x^4+4x^3+7x^2+6x+1)$ . If $a \neq 0$ , then $\deg \left(gcd(G(x),x) \right)=1$ and so $G(x) \mid x^4+4x^3+7x^2+6x+1$ , but $\deg G >4$ , so it's impossible. So $a=0$ .

Then $\deg G(x)=5$ and so $G(x)=x(x^4+4x^3+7x^2+6x+1) \cdot const$ . But it is impossible:
$G(x)=x(1+x+x^2)^2+cx^3$ , so by looking on $x^5$ we get $const=1$ , while looking on $x^4$ we get $const=\frac{1}{2}$ , impossible!

So $P(x,y,z)=s(x+y+z)^3+txyz$ for some $s,t \in \mathbb{R}$ .

Oops you are missing a couple of solutions

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