Problem:
Which of the following conditions is sufficient to guarantee that integers x,y, and z satisfy the equation
x(x−y)+y(y−z)+z(z−x)=1?
Answer Choices:
A. x>y and y=z
B. x=y−1 and y=z−1
C. x=z+1 and y=x+1
D. x=z and y−1=x
E. x+y+z=1
Solution:
It is obvious x, y, and z are symmetrical. We are going to solve the problem by Completing the Square.
x2+y2+z2−xy−yz−zx=12x2+2y2+2z2−2xy−2yz−2zx=2(x−y)2+(y−z)2+(z−x)2=2
Because x,y,z are integers, , (x−y)2, (y−z)2, and (z−x)2 can only equal 0,1,1. So one variable must equal another, and the third variable is 1 different from those 2 equal variables. So the answer is (D) x=z and y−1=x.
The problems on this page are the property of the MAA's American Mathematics Competitions