To prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8, let's proceed with the detailed step-by-step solution:
Step 1: Define the consecutive odd numbers
Let \(\( n \)\) be an arbitrary integer.
The smaller odd number can be expressed as \(\( 2n - 1 \)\).
Since we are looking at consecutive odd numbers, the next odd number will be two steps away from the smaller odd number. Hence, the larger odd number can be expressed as \(\( 2n + 1 \)\).
Step 2: Express the squares of these numbers
Now, let's find the squares of these two numbers.
The square of the smaller odd number \(\( 2n - 1 \)\) is:
\(\[
(2n - 1)^2
\]\)
The square of the larger odd number \(\( 2n + 1 \)\) is:
\(\[
(2n + 1)^2
\]\)
Step 3: Calculate the difference between the squares
We need to find the difference between the squares of these two consecutive odd numbers. Therefore, we subtract the square of the smaller number from the square of the larger number.
The difference will be:
\(\[
(2n + 1)^2 - (2n - 1)^2
\]\)
Step 4: Simplify the expression
Now, we'll simplify this difference:
First, let's expand the squares:
\(\[
(2n + 1)^2 = (2n + 1)(2n + 1) = 4n^2 + 4n + 1
\]\)
\(\[
(2n - 1)^2 = (2n - 1)(2n - 1) = 4n^2 - 4n + 1
\]\)
Subtracting these two expressions:
\(\[
(2n + 1)^2 - (2n - 1)^2 = (4n^2 + 4n + 1) - (4n^2 - 4n + 1)
\]\)
Combine like terms:
\(\[
= 4n^2 + 4n + 1 - 4n^2 + 4n - 1
\]\)
\(\[
= 4n + 4n
\]\)
\(\[
= 8n
\]\)
Step 5: Conclusion
We have shown that the difference between the squares of any two consecutive odd numbers simplifies to \(\( 8n \)\), which is clearly a multiple of 8.
Therefore, the difference between the squares of any two consecutive odd numbers is indeed always a multiple of 8.
