The Difference Between Squares of Consecutive Integers

Have you ever pondered the intriguing relationship between the squares of consecutive integers? Let's explore a specific example that highlights the mathematical elegance in these relationships. This article will delve into the problem: 'What are two consecutive integers whose squares differ by 11?' We'll break down the algebraic solution step-by-step and verify it with a practical example.

Solving the Problem Algebraically

Let's assume we have two consecutive integers: x and x 1. We are given that the difference between the squares of these integers is 11. We can express this as:

((x 1)^2 - x^2 11)

Expanding the left side, we get:

((x 1)^2 - x^2 x^2 2x 1 - x^2 11)

Upon simplifying:

(2x 1 11)

Subtract 1 from both sides:

(2x 10)

Divide by 2:

(x 5)

Therefore, the integers are 5 and 6.

Verification of the Solution

To verify our solution, we can check the difference between the squares of 5 and 6:

(6^2 - 5^2 36 - 25 11)

This confirms that our solution is correct.

Generalizing the Problem

The difference between the squares of two consecutive integers is always an odd number. This is because the difference can be expressed as:

((n 1)^2 - n^2 2n 1)

For any integer n, (2n 1) is an odd number.

Let's take the example given at the beginning:

When (n 5), the integers are 5 and 6, and indeed, (6^2 - 5^2 11).

Conclusion

Through this exploration, we have seen the elegance of algebraic expressions and the simple yet powerful method of solving for the unknowns in the context of consecutive integers. The solution to the specific case of the difference between the squares of two consecutive integers leading to 11 is 5 and 6, and this method can be applied to a wide variety of similar problems.

Whether you are a math student, a teacher, or simply curious about mathematics, understanding such relationships deepens your appreciation for the beauty and logic inherent in numbers.