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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+xyzxyz =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=-zy=-zx=-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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