Q13. Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8

## Step1:<br />By definition, any odd number on a number line can be expressed as \(2n - 1\) or \(2n + 1\), where 'n' is any integer and each term accounts for one possible sequence of odd numbers (that is, {1,3,5,7...} or {2, 4, 6, 8...}). <br /><br />For this particular task, we need to focus on consecutive odd numbers. Let's assign typical consecutive odd numbers such as 'n' & 'n+2'.<br /><br />In other words, we can algebraically represent two consecutive odd numbers like: \(2n - 1\) (old version) and \(2n + 1\) (next version). Notice the interval is due to twice the value in between any two consecutive odd numbers.<br /><br />## Step2:<br />To prove our initial claim, we need to find the difference of the squares of these two consecutive odd integers...<br /><br />\((2n + 1)^2 - (2n - 1)^2 \). Instead of just multiplying this out as is, notice that this formula falls under the Difference of Squares pattern \((a–b)(a + b)\). Here, \(a = 2n + 1\) and \(b = 2n - 1\).<br /><br />## Step3:<br />Next, we continue with factoring the difference of squares expression. Notice that we are not interested in \(2n + 1 - (2n - 1) = 2\); that's another possibility. But rather factoring by adding both 'a' and 'b' in this context...<br />\(a + b = 2n + 1 + 2n - 1 = 4n\). Meaning (\((2n + 1) + (2n - 1)\) equals \(4n\)<br /><br />### \((2n + 1)^2 - (2n - 1)^2 = 4n *(2n + 1 + 2n - 1) = 4n *(4n) = 16n^2\)<br /><br />The final expression shows that the difference of squares of any two consecutive odd integers is equal to \(16n^2\), which is a multiple of 8. Thus our initial claim is proved algebraically.