The difference of the squares of two consecutive even integers is 68. What are these numbers?

Let the two consecutive even integers be \( x \) and \( x + 2 \).

The difference of the squares of these two integers is given as:

\[
(x + 2)^2 - x^2 = 68
\]

First, expand the left-hand side:

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

Substitute this into the equation:

\[
(x^2 + 4x + 4) - x^2 = 68
\]

Simplify:

\[
4x + 4 = 68
\]

Now, subtract 4 from both sides:

\[
4x = 64
\]

Divide by 4:

\[
x = 16
\]

Thus, the two consecutive even integers are \( 16 \) and \( 18 \).

To verify, calculate the difference of their squares:

\[
18^2 - 16^2 = 324 - 256 = 68
\]

The integers are indeed \( 16 \) and \( 18 \).

Would you like more details, or do you have any questions?

Here are 5 related questions to expand on this topic:
1. What would be the result if the difference of the squares were 100 instead of 68?
2. How does the solution change if the integers are consecutive odd integers?
3. Can you solve a similar problem with three consecutive integers?
4. What happens if we change the condition to the sum of the squares instead of the difference?
5. How can we generalize this problem for \( n \) consecutive even integers?

**Tip:** For any consecutive even integers, the difference of their squares is always divisible by 4.

This page provides a step-by-step solution to the problem of finding two consecutive even integers whose difference of squares is 68. It includes key algebraic concepts and a practical application of the difference of squares theorem. Suitable for students in Grades 7-9.

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