Cho các số tự nhiên a, b thỏa mãn [tex]\frac{a+1}{b}+\frac{b+1}{a}\in \mathbb{Z}[/tex].
Chứng minh: [tex](a,b) \leq \sqrt{a+b}[/tex]
Ta có $\frac{a+1}{b}+\frac{b+1}{a}\in \mathbb{Z}$
$\Leftrightarrow \frac{a^2+b^2+a+b}{ab} \in \mathbb{Z}$
$\Leftrightarrow ab | a^2+b^2+a+b$
Gọi $(a,b)=d (d \in \mathbb{N}^*)$
Thì $a=dx, b=dy (x;y \in \mathbb{N}^*; (x,y)=1)$
Khi đó, $d^2xy|d^2x^2+d^2y^2+dx+dy$
$\Leftrightarrow xy | x^2+y^2+\frac{x+y}{d}$
$\Rightarrow x^2+y^2+\frac{x+y}{d} \in \mathbb{N}^*$
$\Rightarrow \frac{x+y}{d} \in \mathbb{N}^*$
$\Rightarrow d|x+y$
$\Rightarrow d^2|a+b$
$\Rightarrow a+b \geq d^2=(a,b)^2$
$\Rightarrow (a,b) \leq \sqrt{a+b}$