Question Number 69111 by Fawole last updated on 20/Sep/19
Commented by Rasheed.Sindhi last updated on 20/Sep/19
$$\left(\mathrm{2},\mathrm{11}\right),\left(\mathrm{11},\mathrm{2}\right),\left(\mathrm{5},\mathrm{10}\right)\:\&\:\left(\mathrm{10},\mathrm{5}\right). \\ $$
Commented by Fawole last updated on 20/Sep/19
$${Show}\:{workings}\:{pleasesir} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Sep/19
$${m}^{\mathrm{6}} +{n}^{\mathrm{6}} +\mathrm{375}{m}^{\mathrm{2}} {n}^{\mathrm{2}} =\mathrm{1953125} \\ $$$$\Rightarrow{m}^{\mathrm{6}} +{n}^{\mathrm{6}} +\mathrm{3}{m}^{\mathrm{2}} {n}^{\mathrm{2}} \left(\mathrm{125}\right)=\left(\mathrm{125}\right)^{\mathrm{3}} \\ $$$${Compare}\:{it}\:{with}: \\ $$$$\:\:\:\left({m}^{\mathrm{2}} \right)^{\mathrm{3}} +\left({n}^{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{3}{m}^{\mathrm{2}} {n}^{\mathrm{2}} \left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\Rightarrow{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{125} \\ $$$$\:\:\:\:\:{m}=\sqrt{\mathrm{125}−{n}^{\mathrm{2}} } \\ $$$${Only}\:{for}\:{n}=\mathrm{2},\mathrm{11},\mathrm{5},\mathrm{10}\:\:\:{m}\in\mathbb{Z}^{+} \\ $$$${Hence} \\ $$$${The}\:{solutions}\:{are} \\ $$$$\left({m},{n}\right)=\left(\mathrm{2},\mathrm{11}\right),\left(\mathrm{11},\mathrm{2}\right),\left(\mathrm{5},\mathrm{10}\right),\left(\mathrm{10},\mathrm{5}\right) \\ $$
Commented by mr W last updated on 21/Sep/19
$${nice}\:{solution}! \\ $$
Commented by Rasheed.Sindhi last updated on 21/Sep/19
$$\mathcal{T}{h}\alpha{nks}\:{Sir}! \\ $$