Question

已知x、y、z是互不相等的正整数,是xyz|(xy-1)(yz-1)(zx-1),求x、y、z

Answer 1

(xy-1)(yz-1)(zx-1)=(xyz)²-xyz(x+y+z)+xy+xz+yz-1
=>xyz|((xyz)²-xyz(x+y+z)+xy+xz+yz-1)
=>xyz|(xy+xz+yz-1)
不妨设x>y>z
则xy+xz+yz-1z,∴y≥2,得xy≥2x>x+y-1,矛盾
∴z=2,得2xy|(xy+2x+2y-1)
=>2x+2y-1≥xy =>(x-2)(y-2)≤3
而x-2>y-2=y-z>0
∴(x-2)(y-2)≥2,即(x-2)(y-2)=2或3
若(x-2)(y-2)=2,则x-2=2,y-2=1
得x=4,y=3,但此时不成立2xy|(xy+2x+2y-1)
∴(x-2)(y-2)=3,则x-2=3,y-2=1
∴此时有x=5,y=3,z=2,满足xyz|(xy-1)(yz-1)(zx-1)
∴(x,y,z)=(2,3,5)或(2,5,3)或(3,2,5)或(3,5,2)或(5,2,3)或(5,3,2)