prove that the difference of squares of any two odd numbers is divisible by 4. (first list at least 3 examples to support this statement.)

3^2 - 1^2 = 8 = 4*2

5^2 - 3^2 = 25 - 9 = 4*4
7^2 - 3^2 = 40 = 4*10

For any integer n, 2n+1 is odd.
Consider a pair of integers m and n.
The difference of the squares of odd integers can be written
(2n+1)^2 - (2m+1)^2 = 4n^2 - 4m^2
- 4(n+m)

= 4(n^2 - m^2 + n - m)
which is clearly evenly divisible by 4.

To prove that the difference of squares of any two odd numbers is divisible by 4, we need to show that the result is always an integer multiple of 4.

Let's take three examples:

Example 1:
3^2 - 1^2 = 9 - 1 = 8
The difference between the squares of 3 and 1 is 8, which is divisible by 4 because 8 ÷ 4 = 2.

Example 2:
5^2 - 3^2 = 25 - 9 = 16
The difference between the squares of 5 and 3 is 16, which is divisible by 4 because 16 ÷ 4 = 4.

Example 3:
7^2 - 5^2 = 49 - 25 = 24
The difference between the squares of 7 and 5 is 24, which is divisible by 4 because 24 ÷ 4 = 6.

Now, let's explain why this pattern holds for any two odd numbers:

1. Any odd number can be represented as 2n + 1, where n is an integer.
2. When we square an odd number, we get (2n + 1)^2 = 4n^2 + 4n + 1.
3. Similarly, the square of another odd number can be represented as (2m + 1)^2 = 4m^2 + 4m + 1, where m is an integer.
4. The difference between the squares of these two odd numbers is [(2n + 1)^2 - (2m + 1)^2].
5. Simplifying this expression, we get [4n^2 + 4n + 1 - 4m^2 - 4m - 1].
6. Combining like terms, we get 4(n^2 - m^2) + 4(n - m).
7. The expression clearly has a common factor of 4.
8. As a result, the difference between the squares is always divisible by 4.

Therefore, we can conclude that the difference of squares of any two odd numbers is divisible by 4.