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?

x = 4
y = 0
z = 5

(x^2)(z-1) is a square. So, (x)√(z-1) = whole number. since x is whole number, √(z-1) must be whole. Therefore, (z-1) is a square less than 10.

The values of z can be 5, 2, 1. z is greatest of all three numbers chosen since (z - y - x) is a square. So, z cannot be 1 (x and y cant be assigned values). Similary if z is 2, then x+y+z = 3 which is not a square.

Once we have z = 5, then x can be either 4 or, 1 to satisfy "2xz+x+z is a square". If x = 1, then xyz cannot be satisfied.

Therefore x is 4.

Now, "xyz is a square" can only be satisfied if y = 0.

Posted by Rajat on 2003-03-29 08:01:09