Question Number 121870 by mnjuly1970 last updated on 12/Nov/20 | ||
$$\:\:\:\:\:\:{number}\:\:{theory}: \\ $$ $$\:\:\:\:\:\:{m},{n}\:\in\:\mathbb{N}\:\:,\:\:\left({m},{n}\right)=\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:{prove}\::\:\:\:{m}^{\varphi\left({n}\right)} +{n}^{\varphi\left({m}\right)} \overset{{mn}} {\equiv}\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\varphi\left({n}\right)=\mid\left\{{x}\in\mathbb{N}\mid\:{x}<{n}\:,\:\left({x},{n}\right)=\mathrm{1}\right\}\mid \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$ | ||
Answered by mindispower last updated on 12/Nov/20 | ||
$${m}^{\varphi\left({n}\right)} =\mathrm{1}\left[{n}\right]....{euler}\:{theorm}\:{t} \\ $$ $${n}^{\varphi\left({m}\right)} =\mathrm{1}\left[{m}\right] \\ $$ $$\Rightarrow{n}\mid{m}^{\varphi\left({n}\right)} −\mathrm{1},{m}\mid{n}^{\varphi\left({m}\right)} −\mathrm{1} \\ $$ $$\Rightarrow{nm}\mid\left({m}^{\varphi\left({n}\right)} −\mathrm{1}\right)\left({n}^{\varphi\left({m}\right)} −\mathrm{1}\right) \\ $$ $$\Leftrightarrow{nm}\mid\left({m}^{\varphi\left({n}\right)} .{n}^{\varphi\left({m}\right)} −\left({n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} −\mathrm{1}\right)\right) \\ $$ $$\Leftrightarrow{mn}\mid{mn}\left({m}^{\varphi\left({n}\right)−\mathrm{1}} .{n}^{\varphi\left({m}\right)−\mathrm{1}} −\left({n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} −\mathrm{1}\right)\:\right. \\ $$ $$\Leftrightarrow{mn}\mid−\left({n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} −\mathrm{1}\right) \\ $$ $$\Rightarrow{n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} \equiv\mathrm{1}\left({mn}\right) \\ $$ | ||
Commented bymnjuly1970 last updated on 13/Nov/20 | ||
$${good}\:\:{very}\:{good}\:{mr}\:{power}.. \\ $$ $${thank}\:{you}.. \\ $$ | ||