若X,Y,Z为大于2的整数,且XY≡2(modZ),XZ≡2(modY),YZ≡2(modX),求X+Y+Z的值

Question

若X,Y,Z为大于2的整数,且XY≡2(modZ),XZ≡2(modY),YZ≡2(modX),求X+Y+Z的值

Answer 1

不妨设x ≥ y ≥ z ≥ 3,由xy = 2 ≠ 0 (mod z)有 y > z,同理x > y.
由条件,xyz | (xy-2)(yz-2)(zx-2) = x²y²z²-2xyz(x+y+z)+4(xy+yz+zx)-8,∴xyz | 4(xy+yz+zx)-8.
∵x,y,z ≥ 3,∴4(xy+yz+zx)-8 > 0,∴0 < (4(xy+yz+zx)-8)/(xyz) = 4(1/x+1/y+1/z)-8/(xyz) < 4.
∴4(1/x+1/y+1/z)-8/(xyz) = 1,2,3.
若4(1/x+1/y+1/z)-8/(xyz) = 3,12/z > 4(1/x+1/y+1/z) > 3,∴ z < 4,只有z = 3.
若4(1/x+1/y+1/z)-8/(xyz) = 2,12/z > 4(1/x+1/y+1/z) > 2,∴ z < 6,3 ≤ z ≤ 5.
若4(1/x+1/y+1/z)-8/(xyz) = 1,12/z > 4(1/x+1/y+1/z) > 1,∴ z < 12,3 ≤ z ≤ 11.
将z的值依次代入,解关于x,y的不定方程,并检验.得到以下解:
x = 22,y = 8,z = 3;
x = 14,y = 10,z = 3;
x = 18,y = 5,z = 4;
x = 11,y = 6,z = 4;
x = 82,y = 14,z = 6;
x = 26,y = 22,z = 6;
x+y+z的可能取值为33,27,21,102,54.
做得比较麻烦,不排除计算错误的可能.