Question

Find three even consecutive integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 20.

Answer 1

Let the three consecutive even integers be \( x \), \( x + 2 \), and \( x + 4 \).

According to the problem, the sum of the squares of the first two integers is equal to the square of the third integer plus 20. We can write this as an equation:

\[
x^2 + (x + 2)^2 = (x + 4)^2 + 20
\]

Now, let's expand both sides of the equation:

1. Left side:
\[
x^2 + (x + 2)^2 = x^2 + (x^2 + 4x + 4) = 2x^2 + 4x + 4
\]

2. Right side:
\[
(x + 4)^2 + 20 = (x^2 + 8x + 16) + 20 = x^2 + 8x + 36
\]

Now setting both sides equal gives us:

\[
2x^2 + 4x + 4 = x^2 + 8x + 36
\]

Next, let's simplify this equation by moving all terms to one side:

\[
2x^2 + 4x + 4 - x^2 - 8x - 36 = 0
\]
\[
x^2 - 4x - 32 = 0
\]

Now we can apply the quadratic formula where \( a = 1 \), \( b = -4 \), and \( c = -32 \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1}
\]
\[
x = \frac{4 \pm \sqrt{16 + 128}}{2}
\]
\[
x = \frac{4 \pm \sqrt{144}}{2}
\]
\[
x = \frac{4 \pm 12}{2}
\]

Calculating the two possible values for \( x \):

1. \( x = \frac{16}{2} = 8 \)
2. \( x = \frac{-8}{2} = -4 \)

So the two cases for the integers based on these values of \( x \) are:

**Case 1**: For \( x = 8 \):
- The integers are \( 8, 10, 12 \).

**Case 2**: For \( x = -4 \):
- The integers are \( -4, -2, 0 \).

The solution consists of both sets of integers that satisfy the conditions of being even and consecutive. Thus:

1. Even consecutive integers: \( 8, 10, 12 \)
2. Even consecutive integers: \( -4, -2, 0 \)

Therefore, the three consecutive even integers satisfying the condition are \( \{ 8, 10, 12 \} \) and \( \{ -4, -2, 0 \} \).