Question Number 69571 by Ajao yinka last updated on 25/Sep/19
Answered by MJS last updated on 25/Sep/19
$${m}^{\mathrm{6}} +\mathrm{375}{n}^{\mathrm{2}} {m}^{\mathrm{2}} +{n}^{\mathrm{6}} −\mathrm{1953125}=\mathrm{0} \\ $$$${m}=\sqrt{{t}} \\ $$$${t}^{\mathrm{3}} +\mathrm{375}{n}^{\mathrm{2}} {t}+{n}^{\mathrm{6}} −\mathrm{1953125}=\mathrm{0} \\ $$$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{method}\:\mathrm{leads}\:\mathrm{to} \\ $$$${t}=\mathrm{125}−{n}^{\mathrm{2}} \\ $$$$\Rightarrow\:{m}=\sqrt{\mathrm{125}−{n}^{\mathrm{2}} } \\ $$$$\mathrm{pairs}\:\mathrm{of}\:{m},\:{n}\:\mathrm{are}\:\mathrm{2},\:\mathrm{11}\:\mathrm{or}\:\mathrm{5},\:\mathrm{10}\:\mathrm{or}\:\mathrm{10},\:\mathrm{5}\:\mathrm{or}\:\mathrm{11},\:\mathrm{2} \\ $$
Commented by Ajao yinka last updated on 26/Sep/19
$${nice}\:{one} \\ $$