Tìm x, y nguyên thỏa mãn: (y+2)x+1= y^2

Đáp án:

(y+2)x+1= y^2

=> xy+2x+1=y ²

=> x ²/4+2x+16-15=y ²-xy+x ²/4

=> (x/2+4) ²-15=(y-x/2) ²

=> (x-y+4)(y+4)=15

x,y nguyên => x-y+4, y+4 nguyên

=> y+4 ∈{-15,-5,-3,-1,1,3,5,15}

=> y ∈{-19, – 9,-7,-5,-3,-1,1,11}

y=-19=> x= -24

y=-9=> x=-16

y=-7=> x=-16

y=-5=> x=-10

y=-3=> x=8

y=-1=> x=0

y=1=>x=0

y=11=>x= 8