The difference between the square of any two consecutive integers is equal to

### Concept & Formula The difference of squares of two consecutive integers is governed by the algebraic identity for the difference of squares: $$a^2 - b^2 = (a - b)(a + b)$$ When $a$ and $b$ are consecutive integers, the difference $(a - b)$ is always $1$. ### Step-by-Step Solution * **Given:** Let the two consecutive integers be $n$ and $(n + 1)$. * **Calculation:** $$(n + 1)^2 - n^2 = [(n + 1) - n] \times [(n + 1) + n]$$ $$= (1) \times (2n + 1)$$ $$= 2n + 1$$ * The result $2n + 1$ is exactly the sum of the two integers $n$ and $(n + 1)$. ### Exam Strategy & Shortcut **Pick and Verify:** Immediately test with small numbers. If the integers are $3$ and $4$: $4^2 - 3^2 = 16 - 9 = 7$. The sum of the two numbers is $3 + 4 = 7$. Since $7 = 7$, option (a) is confirmed. This takes less than 10 seconds. ### Common Pitfall Students often select "an even number" because they forget that the sum of two consecutive integers ($Even + Odd$) is always odd. Verify with an example before concluding. ### Final Answer Therefore, the correct answer is **sum of two numbers**.