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Question Number 121870 by mnjuly1970 last updated on 12/Nov/20

      number  theory:        m,n ∈ N  ,  (m,n)=1           prove :   m^(ϕ(n)) +n^(ϕ(m)) ≡^(mn) 1                           ϕ(n)=∣{x∈N∣ x<n , (x,n)=1}∣               .m.n.

$$\:\:\:\:\:\:{number}\:\:{theory}: \\ $$ $$\:\:\:\:\:\:{m},{n}\:\in\:\mathbb{N}\:\:,\:\:\left({m},{n}\right)=\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:{prove}\::\:\:\:{m}^{\varphi\left({n}\right)} +{n}^{\varphi\left({m}\right)} \overset{{mn}} {\equiv}\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\varphi\left({n}\right)=\mid\left\{{x}\in\mathbb{N}\mid\:{x}<{n}\:,\:\left({x},{n}\right)=\mathrm{1}\right\}\mid \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$

Answered by mindispower last updated on 12/Nov/20

m^(ϕ(n)) =1[n]....euler theorm t  n^(ϕ(m)) =1[m]  ⇒n∣m^(ϕ(n)) −1,m∣n^(ϕ(m)) −1  ⇒nm∣(m^(ϕ(n)) −1)(n^(ϕ(m)) −1)  ⇔nm∣(m^(ϕ(n)) .n^(ϕ(m)) −(n^(ϕ(m)) +m^(ϕ(n)) −1))  ⇔mn∣mn(m^(ϕ(n)−1) .n^(ϕ(m)−1) −(n^(ϕ(m)) +m^(ϕ(n)) −1)   ⇔mn∣−(n^(ϕ(m)) +m^(ϕ(n)) −1)  ⇒n^(ϕ(m)) +m^(ϕ(n)) ≡1(mn)

$${m}^{\varphi\left({n}\right)} =\mathrm{1}\left[{n}\right]....{euler}\:{theorm}\:{t} \\ $$ $${n}^{\varphi\left({m}\right)} =\mathrm{1}\left[{m}\right] \\ $$ $$\Rightarrow{n}\mid{m}^{\varphi\left({n}\right)} −\mathrm{1},{m}\mid{n}^{\varphi\left({m}\right)} −\mathrm{1} \\ $$ $$\Rightarrow{nm}\mid\left({m}^{\varphi\left({n}\right)} −\mathrm{1}\right)\left({n}^{\varphi\left({m}\right)} −\mathrm{1}\right) \\ $$ $$\Leftrightarrow{nm}\mid\left({m}^{\varphi\left({n}\right)} .{n}^{\varphi\left({m}\right)} −\left({n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} −\mathrm{1}\right)\right) \\ $$ $$\Leftrightarrow{mn}\mid{mn}\left({m}^{\varphi\left({n}\right)−\mathrm{1}} .{n}^{\varphi\left({m}\right)−\mathrm{1}} −\left({n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} −\mathrm{1}\right)\:\right. \\ $$ $$\Leftrightarrow{mn}\mid−\left({n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} −\mathrm{1}\right) \\ $$ $$\Rightarrow{n}^{\varphi\left({m}\right)} +{m}^{\varphi\left({n}\right)} \equiv\mathrm{1}\left({mn}\right) \\ $$

Commented bymnjuly1970 last updated on 13/Nov/20

good  very good mr power..  thank you..

$${good}\:\:{very}\:{good}\:{mr}\:{power}.. \\ $$ $${thank}\:{you}.. \\ $$

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