Question Number 223414 by BaliramKumar last updated on 24/Jul/25
$${is}\:{it}\:{possible}\:{to}\:{prove}\:{that}\:{mn}\left({m}+{n}\right)\left({m}−{n}\right)\: \\ $$$${divisible}\:{by}\:\mathrm{6}\:{always}\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by mehdee7396 last updated on 24/Jul/25
$${let}\:\:\:{d}={mn}\left({m}+{n}\right)\left({m}−{n}\right) \\ $$$${case}\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{m}\in{E}\:\:{or}\:\:\:{n}\in{E}\Rightarrow\:\mathrm{2}\mid{mn}\Rightarrow\mathrm{2}\mid{d} \\ $$$$\left.\mathrm{2}\right)\:{m},{n}\in{O}\Rightarrow\mathrm{2}\mid{m}+{n}\:\&\:\mathrm{2}\mid{m}−{n}\Rightarrow\:\mathrm{2}\mid{d} \\ $$$$\Rightarrow\forall{m},{n}\:\Rightarrow\mathrm{2}\mid{d}\:\:\left({i}\right) \\ $$$${case}\:\mathrm{2} \\ $$$$\left.\mathrm{1}\right)\:{m}=\mathrm{3}{k}\:\:{or}\:\:{n}=\mathrm{3}{k}'\Rightarrow\mathrm{3}\mid{mn}\Rightarrow\mathrm{3}\mid{d} \\ $$$$\left.\mathrm{2}\right)\:{m}=\mathrm{3}{k}\pm\mathrm{1}\:\:\&\:\:\:{n}=\mathrm{3}{k}'\pm\mathrm{1}\Rightarrow\mathrm{3}\mid{m}−{n}\Rightarrow\mathrm{3}\mid{d} \\ $$$$\left.\mathrm{3}\right){m}=\mathrm{3}{k}+\mathrm{1}\:\:\&\:\:\:{n}=\mathrm{3}{k}'−\mathrm{1}\Rightarrow\mathrm{3}\mid{m}+{n}\Rightarrow\mathrm{3}\mid{d} \\ $$$$\Rightarrow\forall{m},{n}\:\Rightarrow\mathrm{3}\mid{d}\:\:\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\overset{\left(\mathrm{2},\mathrm{3}\right)=\mathrm{1}} {\Rightarrow}\mathrm{6}\mid{d} \\ $$$$ \\ $$