Show that the difference between the squares of two consecutive odd numbers is a multiple of 8.

Let's denote two consecutive odd numbers as \( n \) and \( n+2 \).

The squares of these two consecutive odd numbers are:

\[
n^2 \quad \text{and} \quad (n+2)^2
\]

The difference between these squares is:

\[
(n+2)^2 - n^2
\]

Now, let's expand \( (n+2)^2 \):

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

So, the difference becomes:

\[
(n^2 + 4n + 4) - n^2 = 4n + 4
\]

We can factor out the common factor 4:

\[
4(n + 1)
\]

Since \( n \) is an odd number, \( n + 1 \) is an even number. Let \( n + 1 = 2k \), where \( k \) is an integer.

Substituting this into the expression:

\[
4(n + 1) = 4(2k) = 8k
\]

This shows that the difference between the squares of two consecutive odd numbers is \( 8k \), which is clearly a multiple of 8.

Thus, the difference between the squares of two consecutive odd numbers is always a multiple of 8.

Would you like further details or have any questions? Here are some related questions for you to explore:

1. What happens if you take the difference between squares of consecutive even numbers?
2. How would you prove that the difference between squares of any two consecutive integers is an odd number?
3. Can this method be used to show that the difference between squares of any two consecutive numbers is related to their sum?
4. What can you say about the sum of the squares of two consecutive odd numbers?
5. How does this property of squares relate to the concept of arithmetic progressions?

**Tip:** When proving mathematical properties, look for patterns and try to factor expressions to identify common multiples.

Explore the proof that the difference between the squares of two consecutive odd numbers is always a multiple of 8. This problem involves key algebraic concepts and properties of odd and even numbers. Perfect for students in Grades 9-10 looking to deepen their understanding of number theory.

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