The Sum of the squares of two positive integers is 100. The difference of the squares is 28. What is the smaller of the two numbers?
Positive squares:
4, 9, 16, 25, 36, 49, 64, 81
Which two of those fit the other clues?
36, 64
To find the solution, let's use algebraic reasoning.
Let's represent the two positive integers as "x" and "y". We can set up equations based on the given information.
The sum of the squares of two positive integers is 100:
x^2 + y^2 = 100
The difference of the squares is 28:
x^2 - y^2 = 28
We can use these two equations to solve for "x" and "y".
First, let's simplify the second equation by factoring the difference of squares:
(x + y)(x - y) = 28
Now, let's look for possible values of "x" and "y" that satisfy both equations.
Since "x" and "y" are positive integers and the sum of their squares is 100, we know that (x, y) = (10, 0) or (0, 10) is not a valid solution.
So, let's solve the simplified equation:
(x + y)(x - y) = 28
Possible factor pairs of 28 are:
1 x 28 = 28
2 x 14 = 28
4 x 7 = 28
Let's substitute these pairs into the equation and see if they satisfy the sum of squares equation (x^2 + y^2 = 100):
For (x + y) = 1 and (x - y) = 28, we get x = 14.5 and y = -13.5, which are not positive integers.
For (x + y) = 2 and (x - y) = 14, we get x = 8 and y = 6, which satisfies the sum of squares equation.
For (x + y) = 4 and (x - y) = 7, we get x = 5.5 and y = -1.5, which are not positive integers.
Therefore, the solution is x = 8 and y = 6. And the smaller of the two numbers is 6.