Let 'x' and 'y' are the numbers we are looking for.

x + y = 100

Thus x^2 + y^2 = x^2 + (100 - x)^2 = 2*x^2 - 200*x +10^4

We need to find such 'x' so that the function f(x) = 2*x^2 - 200*x +10^4 has minimum.

Lets find the vertex form of the function using process called completing the square

f(x) = 2*x^2 - 200*x + 10000

= 2*(x2 - 100x + 2500 ) + 5000

= 2*(x - 50)^2 + 5000

f(x) = 2*(x - 50)^2 + 5000 is vertex form

The vertex is x = 50, f(x) = 5000 - the minimum of the function f

Thus for x = 50 and y = 100 - x = 50 the sum of the squares is minimum.

Another way to solve this problem is by finding the roots of the first derivative of f(x):

f(x) = 2*x^2 - 200*x + 10000

First derivative of f(x) is 4*x - 200

4*x - 200 = 0

x = 50

y = 100 - 50 = 50