Question Number 150311 by naka3546 last updated on 11/Aug/21
$${x},{y}\:\in\:\mathbb{R} \\ $$$${Find}\:\:{all}\:{functions}\:\:{that}\:\:{satisfy}\:\:{this}\:\:{condition}\:: \\ $$$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right)\:\centerdot\:{f}\left({y}\right)\:−\:{f}\left({x}\:\centerdot\:{y}\right)\:+\:\mathrm{1} \\ $$$$ \\ $$$${Find}\:\:{all}\:{functions}\:\:{that}\:\:{satisfy}\:\:{this}\:\:{condition}\:: \\ $$$${f}\left({f}\left({x}\right)\right)\:=\:{f}\left({x}\right)\:+\:{x} \\ $$
Answered by aleks041103 last updated on 11/Aug/21
$$\mathrm{1}\Rightarrow{f}\left({x}+\mathrm{0}\right)={f}\left({x}\right)={f}\left(\mathrm{0}\right).{f}\left({x}\right)−{f}\left(\mathrm{0}\right)+\mathrm{1} \\ $$$$\left(\mathrm{1}−{f}\left(\mathrm{0}\right)\right){f}\left({x}\right)=−\left(\mathrm{1}−{f}\left(\mathrm{0}\right)\right) \\ $$$${If}\:{f}\left(\mathrm{0}\right)\neq\mathrm{1}\:{then}\:{f}\left({x}\right)=−\mathrm{1}={const}.\:{but} \\ $$$${this}\:{doesn}'{t}\:{satisfy}\:\mathrm{2}. \\ $$$${This}\:{leaves}\:{f}\left(\mathrm{0}\right)=\mathrm{1},\:{which}\:{trivially}\:{satisfies} \\ $$$$\mathrm{1}.\:{when}\:{y}=\mathrm{0}. \\ $$$${Then}\:{by}\:\mathrm{2}. \\ $$$${f}\left({f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{1}\right)={f}\left(\mathrm{0}\right)+\mathrm{0}=\mathrm{1} \\ $$$${f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)+\mathrm{1}\:\rightarrow\:{contradiction} \\ $$$$ \\ $$$$\Rightarrow{No}\:{such}\:{functions}\:{exist}! \\ $$