a-b-N-gcd-a-b-1-prove-that-a-b-b-a-ab-1-Euler-phi-function-

Question Number 163845 by mnjuly1970 last updated on 12/Jan/22

a, b \in N, g c d (a, b) = 1

p r o v e t h a t

a^{φ (b)} + b^{φ (a)} \overset{a b}{\equiv} 1

φ : E u l e r p h i f u n c t i o n \dots

Answered by mindispower last updated on 14/Jan/22

a^{φ (b)} \equiv 1 [b] \Leftrightarrow a^{φ (b)} - 1 \equiv 0 [b]

b^{φ (a)} \equiv 1 [a] \Leftrightarrow b^{φ (a)} - 1 \equiv 0 [a] . . f e r m a t T h e o r e m

\Rightarrow (a^{φ (b)} - 1) (b^{φ (a)} - 1) \equiv 0 [a b]

\Leftrightarrow a^{φ (b)} b^{φ (a)} - (a^{φ (b)} + b^{φ (a)} - 1) \equiv 0 [a b] \dots . . E

φ : N^{*} \to N^{*} \Rightarrow φ (a), φ (b) ⩾ 1 \Rightarrow a b ∣ a^{φ (b)} b^{φ (a)}

a^{φ (b)} b^{φ (a)} \equiv 0 [a b]

E \Leftrightarrow b^{φ (a)} + a^{φ (b)} - 1 \equiv [a b] \Leftrightarrow b^{φ (a)} + a^{φ (b)} \equiv 1 [b]

Commented by mnjuly1970 last updated on 14/Jan/22

t h a n k s a l o t s i r p o w e r . .

Commented by mindispower last updated on 18/Jan/22

w i t h e P l e a s u r s i r