Question Number 163845 by mnjuly1970 last updated on 12/Jan/22
$$ \\ $$$$\:\:\:\:\:\:\:{a},\:{b}\:\in\:\mathbb{N}\:\:,\:\:{gcd}\left(\:{a}\:,\:\:{b}\:\right)=\mathrm{1} \\ $$$$\:\:\:\:\:{prove}\:\:{that}\: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\:\:\varphi\:\left({b}\:\right)} +\:{b}^{\:\varphi\:\left({a}\:\right)} \:\overset{\:{ab}} {\equiv}\:\mathrm{1} \\ $$$$\:\:\:\:\:\varphi\::\:\:\:\mathscr{E}{uler}\:{phi}\:\:{function}\:… \\ $$$$ \\ $$
Answered by mindispower last updated on 14/Jan/22
![a^(ϕ(b)) ≡1[b]⇔a^(ϕ(b)) −1≡0[b] b^(ϕ(a)) ≡1[a]⇔b^(ϕ(a)) −1≡0[a].. fermat Theorem ⇒(a^(ϕ(b)) −1)(b^(ϕ(a)) −1)≡0[ab] ⇔a^(ϕ(b)) b^(ϕ(a)) −(a^(ϕ(b)) +b^(ϕ(a)) −1)≡0[ab].....E ϕ:N^∗ →N^∗ ⇒ϕ(a),ϕ(b)≥1⇒ab∣a^(ϕ(b)) b^(ϕ(a)) a^(ϕ(b)) b^(ϕ(a)) ≡0[ab] E⇔b^(ϕ(a)) +a^(ϕ(b)) −1≡[ab]⇔b^(ϕ(a)) +a^(ϕ(b)) ≡1[b]](https://www.tinkutara.com/question/Q164114.png)
$${a}^{\varphi\left({b}\right)} \equiv\mathrm{1}\left[{b}\right]\Leftrightarrow{a}^{\varphi\left({b}\right)} −\mathrm{1}\equiv\mathrm{0}\left[{b}\right] \\ $$$${b}^{\varphi\left({a}\right)} \equiv\mathrm{1}\left[{a}\right]\Leftrightarrow{b}^{\varphi\left({a}\right)} −\mathrm{1}\equiv\mathrm{0}\left[{a}\right]..\:\:\:{fermat}\:{Theorem} \\ $$$$\Rightarrow\left({a}^{\varphi\left({b}\right)} −\mathrm{1}\right)\left({b}^{\varphi\left({a}\right)} −\mathrm{1}\right)\equiv\mathrm{0}\left[{ab}\right] \\ $$$$\Leftrightarrow{a}^{\varphi\left({b}\right)} {b}^{\varphi\left({a}\right)} −\left({a}^{\varphi\left({b}\right)} +{b}^{\varphi\left({a}\right)} −\mathrm{1}\right)\equiv\mathrm{0}\left[{ab}\right]…..{E} \\ $$$$\varphi:\mathbb{N}^{\ast} \rightarrow\mathbb{N}^{\ast} \Rightarrow\varphi\left({a}\right),\varphi\left({b}\right)\geqslant\mathrm{1}\Rightarrow{ab}\mid{a}^{\varphi\left({b}\right)} {b}^{\varphi\left({a}\right)} \\ $$$${a}^{\varphi\left({b}\right)} {b}^{\varphi\left({a}\right)} \equiv\mathrm{0}\left[{ab}\right] \\ $$$${E}\Leftrightarrow{b}^{\varphi\left({a}\right)} +{a}^{\varphi\left({b}\right)} −\mathrm{1}\equiv\left[{ab}\right]\Leftrightarrow{b}^{\varphi\left({a}\right)} +{a}^{\varphi\left({b}\right)} \equiv\mathrm{1}\left[{b}\right] \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 14/Jan/22
$${thanks}\:{alot}\:{sir}\:{power}\:.. \\ $$
Commented by mindispower last updated on 18/Jan/22
$${withe}\:{Pleasur}\:{sir} \\ $$