Find the odd pair: Three options are consecutive perfect-square pairs (n^2, (n+1)^2). Identify the pair that is NOT of this form.

Introduction / Context:Squares feature heavily in number-classification problems. Here, most options form pairs of consecutive perfect squares, i.e., (n^2, (n+1)^2). The task is to detect the one pair that does not follow this structure.

Given Data / Assumptions:

• Pairs: 8–15, 25–36, 49–64, 81–100.
• Perfect squares to recall: 5^2 = 25, 6^2 = 36; 7^2 = 49, 8^2 = 64; 9^2 = 81, 10^2 = 100.

Concept / Approach:Check whether each pair corresponds to (n^2, (n+1)^2) for some integer n. Non-squares or mismatched pairs will betray the odd one out.

Step-by-Step Solution:

Verification / Alternative check:Observe that legitimate pairs differ by (n+1)^2 − n^2 = 2n + 1 (an odd number). The listed square pairs differ by 11, 15, and 19 respectively. The pair 8–15 differs by 7 and neither term is a perfect square, confirming it as the odd one out.

Why Other Options Are Wrong:They are correct examples of consecutive square pairs, so they are not the odd item.

Common Pitfalls:Confusing “consecutive integers” with “consecutive squares,” or checking only the difference without confirming each term is a perfect square.

Final Answer:8 - 15 is the odd pair.