若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除

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若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除若x,y,z

若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除
若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除

若x,y,z均是正整数,试说明(z^2-x^2-y^2)^2-4x^2y^2能被x+y+Z整除
(z^2 - x^2 - y^2)^2 - 4x^2 * y^2
= (z^2 - x^2 - y^2 - 2xy)(z^2 - x^2 - y^2 + 2xy)
= (z^2 -(x+y)^2)(z^2 - (x-y)^2)
= (z-x-y)(z+x+y)(z-x+y)(z-x+y)
所以(z^-x^-y^)^-4x^y^能被x+y+z整除.

=(z2-x2-y2+2xy)(z2-x2-y2-2xy)
=(z2-(x-y)2)(z2-(x+y)2)
=(z-(x-y))(z+(x-y))(z-(x+y))(z+(x+y))
=(z-x+y)(z+x-y)(z-x-y)(z+x+y)
除以x+y+z
=(z-x+y)(z+x-y)(z-x-y)
因为x,y,z是整数,所以上式(z-x+y)(...

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=(z2-x2-y2+2xy)(z2-x2-y2-2xy)
=(z2-(x-y)2)(z2-(x+y)2)
=(z-(x-y))(z+(x-y))(z-(x+y))(z+(x+y))
=(z-x+y)(z+x-y)(z-x-y)(z+x+y)
除以x+y+z
=(z-x+y)(z+x-y)(z-x-y)
因为x,y,z是整数,所以上式(z-x+y)(z+x-y)(z-x-y)是整数
所以可以被x+y+z整除

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