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

$$numbertheory:$$ $$m,n\in \mathbb{N},(m,n)=1$$ $$prove:m^{\phi \left(n\right)}+n^{\phi \left(m\right)}\stackrel{mn}{\equiv }1$$ $$\phi \left(n\right)=\mid \{x\in \mathbb{N}\mid x<n,(x,n)=1\}\mid$$ $$.m.n.$$

Answered by mindispower last updated on 12/Nov/20

$$m^{\phi \left(n\right)}=1\left[n\right]....eulertheormt$$ $$n^{\phi \left(m\right)}=1\left[m\right]$$ $$\Rightarrow n\mid m^{\phi \left(n\right)}-1,m\mid n^{\phi \left(m\right)}-1$$ $$\Rightarrow nm\mid (m^{\phi \left(n\right)}-1)(n^{\phi \left(m\right)}-1)$$ $$\iff nm\mid (m^{\phi \left(n\right)}.n^{\phi \left(m\right)}-(n^{\phi \left(m\right)}+m^{\phi \left(n\right)}-1))$$ $$\iff mn\mid mn(m^{\phi \left(n\right)-1}.n^{\phi \left(m\right)-1}-(n^{\phi \left(m\right)}+m^{\phi \left(n\right)}-1)$$ $$\iff mn\mid -(n^{\phi \left(m\right)}+m^{\phi \left(n\right)}-1)$$ $$\Rightarrow n^{\phi \left(m\right)}+m^{\phi \left(n\right)}\equiv 1\left(mn\right)$$

Commented bymnjuly1970 last updated on 13/Nov/20

$$goodverygoodmrpower..$$ $$thankyou..$$

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