) The sum of squares of two consecutive even integers is 244. Find the numbers.

Let's denote the two consecutive even integers as x and x+2.

The problem states that the sum of the squares of these two numbers is 244. We can write this as an equation:

x^2 + (x+2)^2 = 244

Expanding the equation gives:

x^2 + x^2 + 4x + 4 = 244

Simplifying the equation gives:

2x^2 + 4x - 240 = 0

Dividing the entire equation by 2 to simplify further gives:

x^2 + 2x - 120 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a):

x = [-2 ± sqrt((2)^2 - 41(-120))] / (2*1) x = [-2 ± sqrt(4 + 480)] / 2 x = [-2 ± sqrt(484)] / 2 x = [-2 ± 22] / 2

This gives us two possible solutions: x = 10 or x = -12.

However, since we're looking for positive even integers, we discard x = -12.

So, the two consecutive even integers are 10 and 12.