Математикадан 43-ші халықаралық олимпиада, 2002 жыл, Глазго

Барлық $x,y,z,t$ нақты сандары үшін $$\left( f(x)+f(z) \right)\left( f(y)+f(t) \right)=f(xy-zt)+f(xt+yz)$$ орындалатындай барлық $ f: \mathbb{R}\to \mathbb{R}$ функцияларын табыңыздар.

Комментарий/решение

$$ \textbf{Решение}$$

$$ (f(z)+f(x))(f(t)+f(y))=f(xy-zt)+f(xt+yz) \qquad (1)$$

$$ x=y=z=t=0 \Rightarrow 4f^2(0)=2f(0) \Rightarrow$$

$$\Rightarrow 2f(0)(2(0)-1)=0\Rightarrow \begin{cases} f(0)=0 \\ f(0)=\frac{1}{2} \end {cases}$$

$$ x \leftrightarrow z , \quad y \leftrightarrow t \Rightarrow$$

$$ \Rightarrow (f(z)+f(x))(f(t)+f(y))=f(zt-xy)+f(zy+xt) \quad (2)$$

$$ (2)-(1) \Rightarrow f(zt-xy)=f(xy-zt) \quad (*)$$

$$ t=1, x=y=0 \Rightarrow f(z)=f(-z) , \forall z \in \mathbb{R} \quad (3)$$

$$ z=t=0 \Rightarrow \begin{cases} f(x)f(y)=f(xy) \\ (f(x)+\frac{1}{2})(f(y)+\frac{1}{2})=f(xy) \end{cases}\Rightarrow $$

$$ \Rightarrow \begin{cases} f(x)f(y)=f(xy) \\ \frac{f(x)+f(y)}{2} + f(x)f(y)+ \frac{1}{4}=f(xy) \end{cases}$$

$$ \textbf{1)} \frac{f(x)+f(y)}{2} + f(x)f(y)+ \frac{1}{4}=f(xy) \quad y=0 \Rightarrow f(x) \equiv 0 \Rightarrow \varnothing$$

$$ \textbf{2)}P(x,y): \quad f(x)f(y)=f(xy) $$

$$ P(\pm \sqrt{x}, \pm \sqrt{x}):\quad f((\pm \sqrt{x})(\pm \sqrt{x}))=f(x)=f(\pm \sqrt{x})f(\pm \sqrt{x})=\left( f(\pm \sqrt{x})\right)\geq 0$$

$$ P(0,0): \quad \left( f(0) \right)^2 = f(0) \Rightarrow f(0)=0, \quad f(0)=1$$

$$ P(x,1): \quad f(x\cdot 1)=f(x)=f(x)f(1) \Rightarrow f(x) \equiv 0 \quad f(1)=1$$

$$ P\left( x, \frac{1}{x} \right): \quad f(x)f\left(\frac{1}{x} \right)=f\left(x\cdot \frac{1}{x} \right) = f(1) =1 \Rightarrow f\left( \frac{1}{x} \right) =\frac{1}{f(x)} \quad x\ne 0$$

$$ \Rightarrow f(x)=\pm x^n $$

$$ (3) \Rightarrow n=2k \Rightarrow f(x)=\pm x^{2k}$$

$$ (1) \Rightarrow (x^{2k}+z^{2k})(t^{2k}+y^{2k})=(xy-zt)^{2k}+(xt+yz)^{2k}$$

$$ (xt)^{2k}+(xy)^{2k}+(zt)^{2k}+(zy)^{2k} = (xy-zt)^{2k}+(xt+yz)^{2k}\Leftrightarrow$$

$$ \Leftrightarrow \left((xy)^k-(zt)^k \right)^2 +\left((xt)^k+(zy)^k \right)^2 =(xy-zt)^{2k}+(xt+yz)^{2k}\Rightarrow k=1$$

$$ \Rightarrow f(x)=\pm x^2 \qquad f_1(x)=-x^2 \quad f_2(x)=x^2 $$

$$ x=y=z=0 \Rightarrow f(t)=\frac{1}{2}, \forall t \in R$$

$$ \color{blue} { f(x)= x^2 ,\quad f(x) =0 , \quad f(x)=0,5 \quad \forall x \in \mathbb{R}}$$