What are two consecutive integers such that the sum of their squares is 265?

The two consecutive integers whose squares sum to 265 are 11 and 12.

Finding Consecutive Integers with a Specific Sum of Squares

Let's explore how to find these consecutive integers.

The problem states that we need to find two consecutive integers, let's call them n and n+1, such that the sum of their squares equals 265. This can be expressed mathematically as:

n² + (n+1)² = 265

Expanding the equation:

n² + (n² + 2n + 1) = 265

Combining like terms:

2n² + 2n + 1 = 265

Subtracting 265 from both sides to set the equation to zero:

2n² + 2n - 264 = 0

Dividing the entire equation by 2 to simplify:

n² + n - 132 = 0

Now, we need to factor this quadratic equation. We are looking for two numbers that multiply to -132 and add up to 1. Those numbers are 12 and -11. Therefore, the factored equation is:

(n + 12)(n - 11) = 0

Setting each factor to zero gives us two possible solutions for n:

n + 12 = 0 => n = -12
n - 11 = 0 => n = 11

If n = 11, then n + 1 = 12. If n = -12, then n + 1 = -11.

Let's verify these solutions:

• Case 1: n = 11, n+1 = 12
  11² + 12² = 121 + 144 = 265 (Correct)
• Case 2: n = -12, n+1 = -11
  (-12)² + (-11)² = 144 + 121 = 265 (Correct)

Therefore, there are two possible pairs of consecutive integers that satisfy the condition. The integers 11 and 12 and the integers -12 and -11.

Summary

• The question asks for two consecutive integers whose squares sum to 265.
• The answer includes both the positive solution (11 and 12) and the negative solution (-12 and -11).
• The reference provided only gave the positive solution (11 and 12).