已知x,y,z,均为正整数,且满足x²+z²=10,z²+y²=13,x²<10z²<10 y²<13x²+z²=10,z²=10-x²z²+y²=1310-x²+y²=13y²-x²=3y=2 x=1z=3(x-y)²的值=(1-2)²=1
∵z²+y²-(x²+z²)=3,∴(y-x)(y+x)=3,∵x,y,z是正整数,x+y≥2∴y+x=3,所以y-x=1,解得y=2,x=1,所以(x-y)²=1.
x=1,y=2,z=3,(x-y)(x-y)=1