Question Number 169391 by mathlove last updated on 29/Apr/22
Commented by infinityaction last updated on 29/Apr/22
$$\mathrm{0} \\ $$
Commented by mathlove last updated on 29/Apr/22
$${any}\:{solution}? \\ $$
Answered by mr W last updated on 30/Apr/22
$$\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{x}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−{y} \\ $$$${x}+{y}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${xy}=\mathrm{1}−\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{1}−{xy}=\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{xy}+\left({xy}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} +\left({xy}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{xy}=\mathrm{0}\:\Rightarrow{x}={y}=\mathrm{0} \\ $$$$\Rightarrow{x}+{y}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by mathlove last updated on 04/May/22
$${thank}\:{sir} \\ $$