a-b-N-b-1-a-a-1-b-Z-Prove-that-a-b-a-b-

Question Number 207436 by CrispyXYZ last updated on 15/May/24

a, b \in N_{+}, \frac{b + 1}{a} + \frac{a + 1}{b} \in Z . Prove that

(a, b) ⩽ \sqrt{a + b .}

Commented by A5T last updated on 15/May/24

W h a t i s m e a n t b y (a, b) ?

Commented by CrispyXYZ last updated on 15/May/24

the greatest common divisor of a, b

Answered by Berbere last updated on 16/May/24

\frac{b^{2} + a^{2} + a + b}{a b} \in Z

a b ∣ {(a + b)}^{2} - a b + (a + b); a ∣ b \Leftrightarrow a = k b

\Rightarrow a b ∣ (a + b) {(1 + (a + b))}^{'} . . (E)

l e t d = (a, b) \Rightarrow a = {d a}^{'}; b = {d b}^{'}

E \Leftrightarrow d^{2} a^{'} b^{'} ∣ ({d a}^{'} + {d b}^{'}) (1 + d (a^{'} + b^{'}))

d^{2} ∣ d^{2} a^{'} b^{'} \Rightarrow d^{2} ∣ ({d a}^{'} + {d b}^{'}) + d^{2} {(a^{'} + b^{'})}^{2}

\Rightarrow d^{2} ∣ d (a^{'} + b^{'}) \Leftrightarrow d^{2} ∣ {d a}^{'} + {d b}^{'} \Leftrightarrow d^{2} ∣ a + b

\Rightarrow d^{2} ⩽ a + b \Rightarrow d ⩽ \sqrt{a + b}

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