1x+1y+1z=1x+y+z是否有整數解?
已知xyz≠0,則
1x+1y+1z=1x+y+z是否有整數解?
因為yz+xz+xyxyz=1x+y+z
(x+y+z)(xy+yz+zx)−xyz=0
x(xy+yz+zx)+y(xy+yz+zx)+z(xy+yz+zx)−xyz=0
(x2y+xyz+x2z)+(xy2+y2z+xyz)+(xyz+yz2+xz2)−xyz=0
x2y+xyz+x2z+xz2+y2z+xy2+xyz+yz2+xyz−xyz =0
x2y+xyz+x2z+xz2+y2z+xy2+xyz+yz2=0
(x2y+xyz)+(x2z+xz2)+(y2z+xy2)+(xyz+yz2)=0
xy(x+z)+xz(x+z)+y2(z+x)+yz(x+z)=0
(x+z)(xy+xz+y2+yz)=0
(x+z)[x(y+z)+y(y+z)]=0
(x+z)(y+z)(x+y)=0
所以x=-z或y=-z或 x=-y,即
(x,y,z)=(-s,t,s)或(t,-s,s)或(-s,s,t),其中s和t都是非0整數。
因此 1x+1y+1z=1x+y+z有非0的整數解,
例如(-2,3,2),(4,-6,6),(7,-7,-5)...都是其整數解。
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