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Algebra Difficulty 3.1 AMC 10/12 Find the answer

Which of the following conditions is sufficient to guarantee that integers xx, yy, and zz satisfy the equation
x(x−y)+y(y−z)+z(z−x)=1?x(x-y)+y(y-z)+z(z-x) = 1?

Pick one

Solution

It is obvious xx, yy, and zz are symmetrical. We are going to solve the problem by Completing the Square.
x2+y2+z2−xy−yz−zx=1x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx = 1
2x2+2y2+2z2−2xy−2yz−2zx=22x ^ 2 + 2y ^ 2 + 2z ^ 2 - 2xy - 2yz - 2zx = 2
(x−y)2+(y−z)2+(z−x)2=2(x-y)^2 + (y-z)^2 + (z-x)^2 = 2
Because x,y,zx, y, z are integers, (x−y)2(x-y)^2, (y−z)2(y-z)^2, and (z−x)2(z-x)^2 can only equal 0,1,10, 1, 1. So one variable must equal another, and the third variable is 11 different from those 22 equal variables. So the answer is D\boxed{D}.
~isabelchen

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