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Question Number 69571 by Ajao yinka last updated on 25/Sep/19

Answered by MJS last updated on 25/Sep/19

m^6 +375n^2 m^2 +n^6 −1953125=0  m=(√t)  t^3 +375n^2 t+n^6 −1953125=0  Cardano′s method leads to  t=125−n^2   ⇒ m=(√(125−n^2 ))  pairs of m, n are 2, 11 or 5, 10 or 10, 5 or 11, 2

$${m}^{\mathrm{6}} +\mathrm{375}{n}^{\mathrm{2}} {m}^{\mathrm{2}} +{n}^{\mathrm{6}} −\mathrm{1953125}=\mathrm{0} \\ $$$${m}=\sqrt{{t}} \\ $$$${t}^{\mathrm{3}} +\mathrm{375}{n}^{\mathrm{2}} {t}+{n}^{\mathrm{6}} −\mathrm{1953125}=\mathrm{0} \\ $$$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{method}\:\mathrm{leads}\:\mathrm{to} \\ $$$${t}=\mathrm{125}−{n}^{\mathrm{2}} \\ $$$$\Rightarrow\:{m}=\sqrt{\mathrm{125}−{n}^{\mathrm{2}} } \\ $$$$\mathrm{pairs}\:\mathrm{of}\:{m},\:{n}\:\mathrm{are}\:\mathrm{2},\:\mathrm{11}\:\mathrm{or}\:\mathrm{5},\:\mathrm{10}\:\mathrm{or}\:\mathrm{10},\:\mathrm{5}\:\mathrm{or}\:\mathrm{11},\:\mathrm{2} \\ $$

Commented by Ajao yinka last updated on 26/Sep/19

nice one

$${nice}\:{one} \\ $$

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