Question Number 171226 by SLVR last updated on 10/Jun/22
$${Possible}\:{number}\:{of}\:{order}\:{pairs} \\ $$$${satisfy}\:\mathrm{16}^{{x}^{\mathrm{2}} +{y}} +\mathrm{16}^{{y}^{\mathrm{2}} +{x}} =\mathrm{1}\:{is} \\ $$
Commented by SLVR last updated on 11/Jun/22
$${can}\:{any}\:{one}\:{me}\:\:\:{please} \\ $$
Commented by SLVR last updated on 10/Jun/22
$${i}\:{mean}\:{pairs}\:\left({x},{y}\right)\:{is}\:{how}\:{many}? \\ $$$${kindly}\:{help}\:{me}\:{out} \\ $$
Commented by mr W last updated on 11/Jun/22
$$\mathrm{2}^{\mathrm{4}\left({x}^{\mathrm{2}} +{y}\right)} +\mathrm{2}^{\mathrm{4}\left({y}^{\mathrm{2}} +{x}\right)} =\mathrm{1} \\ $$$$\mathrm{4}\left({x}^{\mathrm{2}} +{y}\right)=\mathrm{4}\left({y}^{\mathrm{2}} +{x}\right)=−\mathrm{1} \\ $$$$\Rightarrow{x}={y}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by SLVR last updated on 11/Jun/22
$${So}\:\:\:{kind}\:{of}\:{you}\:{professor}. \\ $$$${Greatly}\:{thank}\:{you} \\ $$