The difference of the squares of two consecutive odd integers is divisible by

### Concept & Formula The difference between the squares of any two consecutive odd integers is always a multiple of 8. We can prove this algebraically. Any odd integer can be written in the form $(2n + 1)$, where $n$ is an integer. Therefore, two consecutive odd integers can be represented as $(2n - 1)$ and $(2n + 1)$. ### Step-by-Step Solution * **Step 1:** Formulate the equation for the difference of their squares. $$ (2n + 1)^2 - (2n - 1)^2 $$ * **Step 2:** Expand the squares using the algebraic identities $(a+b)^2$ and $(a-b)^2$. $$ (4n^2 + 4n + 1) - (4n^2 - 4n + 1) $$ * **Step 3:** Simplify the expression. $$ 4n^2 + 4n + 1 - 4n^2 + 4n - 1 = 8n $$ * **Step 4:** Analyze the result. The resulting expression is $8n$. Since $n$ is an integer, $8n$ is always perfectly divisible by 8. ### Exam Strategy & Shortcut For property-based questions like this, the fastest method under time pressure is **Value Substitution**. Pick the smallest positive consecutive odd integers: $1$ and $3$. Their squares are $1^2 = 1$ and $3^2 = 9$. Difference: $9 - 1 = 8$. 8 is divisible by 8. To be absolutely certain, test the next pair: $3$ and $5$. Squares: $9$ and $25$. Difference: $25 - 9 = 16$. 16 is also divisible by 8. ### Common Pitfall Students often confuse "odd integers" with general integers and use $n$ and $(n+1)$. This would yield $(n+1)^2 - n^2 = 2n + 1$, which is always odd, leading to confusion when checking the even options provided. Always map "odd" explicitly to $(2n+1)$ or test using actual odd numbers. ### Final Answer **Therefore, the correct answer is 8.**