Question Number 83035 by bshahid010@gmail.com last updated on 27/Feb/20
Commented by Tony Lin last updated on 27/Feb/20
$${f}\left(\mathrm{1}\right)=\mathrm{4}{f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\left(\sqrt{{f}\left(\mathrm{0}\right)}+\sqrt{{f}\left(\mathrm{1}\right)}\right)^{\mathrm{2}} \\ $$$$={f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)+\mathrm{2}\sqrt{{f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{9}}{\mathrm{4}}{f}\left(\mathrm{1}\right)=\mathrm{9}{f}\left(\mathrm{0}\right)=\mathrm{9} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{4}{f}\left(\mathrm{1}\right)=\mathrm{16}{f}\left(\mathrm{0}\right)=\mathrm{16} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by mr W last updated on 27/Feb/20
$${with}\:{x}={y}=\mathrm{0}, \\ $$$${f}\left(\mathrm{1}\right)=\left(\mathrm{2}\sqrt{{f}\left(\mathrm{0}\right)}\right)^{\mathrm{2}} =\mathrm{4}{f}\left(\mathrm{0}\right)=\mathrm{4} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${with}\:{y}=\mathrm{0}, \\ $$$${f}\left({x}+\mathrm{1}\right)=\left(\sqrt{{f}\left({x}\right)}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${let}\:{g}\left({x}\right)=\sqrt{{f}\left({x}\right)} \\ $$$$\left({g}\left({x}+\mathrm{1}\right)\right)^{\mathrm{2}} =\left({g}\left({x}\right)+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${g}\left({x}+\mathrm{1}\right)={g}\left({x}\right)+\mathrm{1} \\ $$$${let}\:{g}\left({x}\right)={Ax}+{B} \\ $$$${Ax}+{A}+{B}={Ax}+{B}+\mathrm{1} \\ $$$$\Rightarrow{A}=\mathrm{1} \\ $$$${g}\left({x}\right)={x}+{B} \\ $$$${g}\left(\mathrm{0}\right)=\mathrm{0}+{B}=\mathrm{1}\:\Rightarrow{B}=\mathrm{1} \\ $$$$\Rightarrow{g}\left({x}\right)={x}+\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({g}\left({x}\right)\right)^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$