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Question Number 75987 by Ajao yinka last updated on 21/Dec/19

Answered by MJS last updated on 22/Dec/19

m=(√x)∧n=(√y)  x^3 +(375y)x+(y^3 −1953125)=0  Cardano with p=375y∧q=y^3 −1953125  ⇒  x=125−y  ⇒  m=(√(125−y))∧n=(√y);  ((m),(n) )∈N^2 ⇒  ⇒ y=j^2 ∧125−y=k^2 ; j, k∈N ⇒  ⇒ y∈{4, 25, 100, 11} ⇒  ⇒  ((m),(n) )∈{ ((2),((11)) ),  ((5),((10)) ),  (((10)),(5) ),  (((11)),(2) )}

$${m}=\sqrt{{x}}\wedge{n}=\sqrt{{y}} \\ $$$${x}^{\mathrm{3}} +\left(\mathrm{375}{y}\right){x}+\left({y}^{\mathrm{3}} −\mathrm{1953125}\right)=\mathrm{0} \\ $$$$\mathrm{Cardano}\:\mathrm{with}\:{p}=\mathrm{375}{y}\wedge{q}={y}^{\mathrm{3}} −\mathrm{1953125} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{125}−{y} \\ $$$$\Rightarrow \\ $$$${m}=\sqrt{\mathrm{125}−{y}}\wedge{n}=\sqrt{{y}};\:\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}\in\mathbb{N}^{\mathrm{2}} \Rightarrow \\ $$$$\Rightarrow\:{y}={j}^{\mathrm{2}} \wedge\mathrm{125}−{y}={k}^{\mathrm{2}} ;\:{j},\:{k}\in\mathbb{N}\:\Rightarrow \\ $$$$\Rightarrow\:{y}\in\left\{\mathrm{4},\:\mathrm{25},\:\mathrm{100},\:\mathrm{11}\right\}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}\in\left\{\begin{pmatrix}{\mathrm{2}}\\{\mathrm{11}}\end{pmatrix},\:\begin{pmatrix}{\mathrm{5}}\\{\mathrm{10}}\end{pmatrix},\:\begin{pmatrix}{\mathrm{10}}\\{\mathrm{5}}\end{pmatrix},\:\begin{pmatrix}{\mathrm{11}}\\{\mathrm{2}}\end{pmatrix}\right\} \\ $$

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