Question Number 216769 by ArshadS last updated on 19/Feb/25
$${Solve}\:{for}\:{integer}\:{k},{m}\:{and}\:{n}: \\ $$$${k}^{\mathrm{2}} {m}−{n}^{\mathrm{2}} =\mathrm{8} \\ $$
Answered by mehdee7396 last updated on 20/Feb/25
$$``\:{m}\:''\:{must}\:{be}\:\:{positive} \\ $$$$\left(\sqrt{{m}}{k}+{n}\right)\left(\sqrt{{m}}{k}−{n}\right)=\mathrm{8}=\mathrm{4}×\mathrm{2} \\ $$$${if}\:\:{k},{n}>\mathrm{0}\Rightarrow\sqrt{{m}}{k}+{n}=\mathrm{4}\:\:\&\:\:\sqrt{{k}}{m}−{n}=\mathrm{2} \\ $$$$\Rightarrow{n}=\mathrm{1}\Rightarrow\sqrt{{m}}{k}=\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\sqrt{{m}}=\mathrm{1}\:\&\:{k}=\mathrm{3}\Rightarrow{m}=\mathrm{1}\:\&\:{k}=\mathrm{3\&}{n}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\sqrt{{m}}=\mathrm{3}\:\&\:{k}=\mathrm{1}\Rightarrow{m}=\mathrm{9}\:\&\:{k}=\mathrm{1\&}{n}=\mathrm{1} \\ $$$${if}\:\:{k},{n}<\mathrm{0}\Rightarrow<{k}=−\mathrm{3}\:\:\&\:{n}=−\mathrm{1} \\ $$$$\left.{Ans}\right) \\ $$$$\left({m},{n},{k}\right):\left(\mathrm{1},\mathrm{1},\mathrm{3}\right),\left(\mathrm{9},\mathrm{1},\mathrm{1}\right),\left(\mathrm{1},−\mathrm{1},−\mathrm{3}\right),\left(\mathrm{9},−\mathrm{1},−\mathrm{3}\right) \\ $$$$ \\ $$
Commented by ArshadS last updated on 20/Feb/25
$$\:\mathcal{T}{hanks}!\:\:{Some}\:{solutions}\:{out}\:{of}\:{many}! \\ $$$${Can}\:{you}\:{or}\:{anyone}\:{else}\:\:{get}\:{general}\:{solution}/{s}\:? \\ $$