Ta thấy: [tex](2x+1)(2x+3)\vdots 4xy-1[/tex]
Mà [tex](2x+1,2x+3)=1[/tex] nên có 2 trường hợp:
+ [tex]2x+1\vdots 4xy-1\Rightarrow 4xy+2y\vdots 4xy-1\Rightarrow 2y-1\vdots 4xy-1[/tex]
Đặt [tex]2y+1=t(4xy-1) (t /geq 1)\Rightarrow (4tx-2)y=1+t\Rightarrow y=\frac{t+1}{4tx-2}[/tex]
Vì [tex]y\geq 1\Rightarrow \frac{t+1}{4tx-1}\geq 1\Rightarrow t+1\geq 4tx-1\Rightarrow x\leq \frac{t+2}{4t}[/tex]
Dễ thấy: [tex]x=1\Rightarrow 15\vdots 4y-1\Rightarrow 4y-1\in \left \{ 1;3;5;15 \right \}\Rightarrow y\in \left \{ 1;4 \right \}[/tex]
[tex]x> 1[/tex] . Ta suy ra [tex]\frac{t+2}{4t}\geq 1[/tex] (vì nếu [tex]\frac{t+2}{4t}< 1[/tex] thì [tex]x\leq \frac{t+2}{4t}< 1[/tex] )
[tex]\Rightarrow t+2\geq 4t\Rightarrow 3t\leq 2\Rightarrow t< 1[/tex] (vô lí)
+ [tex]2x+3\vdots 4xy-1\Rightarrow 2x+3\geq 4xy-1\Rightarrow y\leq \frac{x+2}{2x}[/tex]
Dễ thấy [tex]\frac{x+2}{2x}\geq 1\Rightarrow x+2\geq 2x\Rightarrow x\leq 2[/tex]
Xét x = 1 [tex]\Rightarrow y\in \left \{ 1;4 \right \}[/tex]
Xét x = 2 [tex]\Rightarrow 7\vdots 8y-1\Rightarrow 8y-1\in \left \{ 1;7\Rightarrow 8y\in \left \{ 2;8 \right \} \right \}\Rightarrow y=1[/tex]
Vậy [tex](x,y)=(1,1),(1,4),(2,1)[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
