ประมาณนี้หรือป่าว
\[\begin{array}{l}
P\left( {x,y} \right)\quad \Rightarrow \quad f\left( {xf\left( x \right) + f\left( y \right)} \right) = {\left( {f\left( x \right)} \right)^2} + y\\
\exists a,f\left( a \right) = 0\\
P\left( {a,x} \right)\quad \Rightarrow \quad f\left( {af\left( a \right) + f\left( x \right)} \right) = {\left( {f\left( a \right)} \right)^2} + x\\
\quad \quad \quad \quad \Rightarrow \quad f\left( {f\left( x \right)} \right) = x\\
P\left( {f\left( x \right),y} \right)\quad \Rightarrow \quad f\left( {f\left( x \right)f\left( {f\left( x \right)} \right) + f\left( y \right)} \right) = {\left( {f\left( {f\left( x \right)} \right)} \right)^2} + y\\
{\kern 1pt} \quad \quad \quad \quad \Rightarrow \quad f\left( {xf\left( x \right) + f\left( y \right)} \right) = {x^2} + y\\
P\left( {f\left( x \right),y} \right) = P\left( {x,y} \right)\\
\quad \quad \quad \quad \quad \Rightarrow f\left( x \right) = x
\end{array}\]
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