I have chosen 3 different whole numbers less than 10, and have found several simple combinations that lead to perfect squares. Calling the numbers x,y, and z, the following combinations all yield a perfect square as the answer. (A perfect square is a number that has a whole number square root).

(x^2)y + (y^2)z + (z^2)x

x+y+z

z-y-x

xyz

(x^2)(z-1)

There are also several more complicated arrangements that lead to perfect squares, such as

x((z^2)-1)+z((y^2)-3)-x(yz-xy)

2xz+x+z

x((z^2)+x)+z(y^2)-(x^2)(z-y)

Given that these perfect squares are all different, and range between 0 and 100 (inclusive), can you determine x,y, and z?

Since x, y, z are less than 10, then x+y+z is ≤ 27. Now, since it's a perfect square, we must have x+y+z = 25, 16, 9, 4, 1 or 0. We have that z-y-x is also a perfect square, and it must be 4, 1 or 0, since z<10.

Let's suppose z-y-x=4. Then since z-y-x plus z+y+x gives 2z, x+y+z must be even. x+y+z must be 16, because it's smaller than 27 and different from 4. Then 2z=20. So z=10. But that doesn't work, since z<10. So z-x-y isn't 4.

Let's suppose that z-x-y=0. Then, x+y+z must be even, since z-x-y plus x+y+z is 2z, which is even. So x+y+z is 4 or 16. Let's suppose it's 16. The 2z=16, so z=8. But since (x²)(z-1) is a square, then 7x² is a square. But that isn't true, since 7 is not a perfect square. So x+y+z=4. Then 2z=4, so z=2. Since xyz is a square, 2xy is a square, different from 0. We know x+y=2, if x or y is 2, the other one is 0. But that wouldn't work. Then they both must be 1. But then xyz isn't a square.

Posted by Fernando on 2003-03-29 05:49:56