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?

So we must have z-y-x=1. Then z+y+x must be odd, since adding z-y-x plus z+y+x gives 2z, which is even. Then z+y+x = 9 or 25 (it cannot be 1, since all squares are different). Let's say that x+y+z=25. Then 2z=26 and z=13. But that doesn't work, since z<10. So z-y-z cannot be 1.
So then x+y+z=9. Then 2z=10 and z=5. We also know that xyz is a square so x or y must be divisible by 5. Since they are less than 10, one of them must be either 5 or 0. If one of them is 5, then x+y+z>10. So one of them must be 0. Since x+y+z=9, then {x,y}={0,4}. If x=0, then 2xz+x+z=5, which isn't a perfect square. Then x=4, and y=0. So (x,y,z)=(4,0,5). Now:


(x^2)y + (y^2)z + (z^2)x = 100
x+y+z = 9
z-y-x = 1
xyz = 0
(x^2)(z-1) = 64
x((z^2)-1)+z((y^2)-3)-x(yz-xy) = 81
2xz+x+z = 49
x((z^2)+x)+z(y^2)-(x^2)(z-y) = 36


So (4,0,5) is indeed a solution and there are no others...

Posted by Fernando on 2003-03-29 05:52:27