Chủ đề này có 1 trả lời

[ThEdArKlOrD]

Find all two variables polynomials $P(x,y)$ such that for any real numbers $a,b,c$, we have 

$P(ab,c^2-2)+P(ac,b^2-2)+P(bc,a^2-2)=0$

Vietnamese

Tìm tất cả các đa thức hai biến $P(x, y)$ sao cho với mọi số thực $a, b, c$ thì

$$P(ab, c^{2} - 2) + P(ca, b^{2} - 2) + P(bc, a^{2} - 2) = 0$$

Bài viết đã được chỉnh sửa nội dung bởi Ego: 16-04-2016 - 10:25

[perfectstrong]

The unique constant solution is $P \equiv 0$. Assuming that $P$ is not a constant.

\[P\left( {bc,{a^2} - 2} \right) + P\left( {ca,{b^2} - 2} \right) + P\left( {ab,{c^2} - 2} \right) = 0\,\left( 1 \right)\]

$a=b=c:=0, (1) \Rightarrow P(0,-2)=0$

$b=c:=0, (1) \Rightarrow P(0,a^2-2)=0 \forall a \Rightarrow P(0,x)=0 \forall x\, (2)$

$a:=0, (1) \Rightarrow P(bc,-2)=0 \forall b,c \Rightarrow P(x,-2)=0 \forall x\, (3)$

 

For a given $k \in \mathbb{R}$:

$b=c:= k , (1) \Rightarrow P(k^2,a^2-2)+2P(ka,k^2-2)=0$

$b=c:=-k, (1) \Rightarrow P(k^2,a^2-2)+2P(-ka,k^2-2)=0$

We observe that $P(ka,k^2-2)=P(-ka,k^2-2) \forall k\in \mathbb{R}, a \in \mathbb{R} \Rightarrow P(x,y)=P(x,-y) \forall x\in \mathbb{R}, y \ge -2\, (4)$.

 

Note that $(4)$ is also valable for all real $x,y$ because $P$ is polynominal.

 

We induce the form of $P$ from $(2),(3),(4)$: $$P(x,y)=x^{2t}(y+2)^m R(x,y)$$

where $R(0,y) \ne 0, R(x,-2) \ne 0$.

 

Replacing the form into $(1)$:

\[{b^{2t}}{c^{2t}}{a^{2m}}R\left( {bc,{a^2} - 2} \right) + {c^{2t}}{a^{2t}}{b^{2m}}R\left( {ca,{b^2} - 2} \right) + {a^{2t}}{b^{2t}}{c^{2m}}R\left( {ab,{c^2} - 2} \right) = 0\]

 

If $t=m$, we simplify by ${b^{2t}}{c^{2t}}{a^{2m}}$: \[R\left( {bc,{a^2} - 2} \right) + R\left( {ca,{b^2} - 2} \right) + R\left( {ab,{c^2} - 2} \right) = 0\]

By following the demonstration above, we'll see that $R(0,y)=0$, which makes contradiction.

 

If $t<m$, we simplify by ${b^{2t}}{c^{2t}}{a^{2t}}$: \[{a^{2m - 2t}}R\left( {bc,{a^2} - 2} \right) + {b^{2m - 2t}}R\left( {ca,{b^2} - 2} \right) + {c^{2m - 2t}}R\left( {ab,{c^2} - 2} \right) = 0\]

Take $a = 0,b = c \Rightarrow 2{b^{2m - 2t}}R\left( {0,{b^2} - 2} \right) = 0 \Rightarrow R\left( {0,y} \right) = 0:contradict$

 

If $t>m$, we simplify by ${b^{2m}}{c^{2m}}{a^{2m}}$:\[{b^d}{c^d}R\left( {bc,{a^2} - 2} \right) + {c^d}{a^d}R\left( {ca,{b^2} - 2} \right) + {a^d}{b^d}R\left( {ab,{c^2} - 2} \right) = 0\]

where $d=2t-2m$.

For $a = 0: \,\,{b^d}{c^d}R\left( {bc, - 2} \right) = 0 \Rightarrow R\left( {x, - 2} \right) = 0:contradict$

 

Hence, $P\equiv 0$ is the solution.