2023 AMC 10B Problem 9

P1-10~2 minPrint

Number Properties Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Bound the quantity above/below or estimate to pin it down

The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$ . How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$ ?

Show solution without answering

Key insight

Neighboring squares n squared and (n+1) squared always differ by 2n + 1. So count the values of n that make 2n + 1 at most 2023.

Solution

Start with a few small pairs to see how the differences grow.

Small cases.

pair	$1, 4$	$4, 9$	$9, 16$	$16, 25$	$25, 36$
difference	$3$	$5$	$7$	$9$	$11$

The differences are the odd numbers $3, 5, 7, 9, 11, \ldots$ , going up by $2$ each time.
2. The general rule. Call the smaller square $n^2$ , so the next one is $(n+1)^2$ . Then
$(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1.$
Check: $n = 4$ gives $16$ and $25$ , and $2 \cdot 4 + 1 = 9$ . It matches.
3. Set up the condition. We want $2n + 1 \le 2023$ , so $2n \le 2022$ and $n \le 1011$ .
4. Count. The squares must be positive, so $n$ starts at $1$ . The allowed values are $n = 1, 2, 3, \ldots, 1011$ , and each one gives exactly one pair. The last one works: $2 \cdot 1011 + 1 = 2023$ , which is allowed because the problem says "less than or equal".

That is $1011$ pairs.

The answer is $\boxed{\textbf{(B)}\ 1011}$ .

Why this works

The gap between neighboring squares is always the odd number $2n + 1$ , so "the gap is at most $2023$ " turns into a simple limit on $n$ . Naming each pair by one number, its smaller square root $n$ , turns "count the pairs" into "count the whole numbers from $1$ up to some limit".

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

A pair of consecutive positive squares is $n^2$ and $(n+1)^2$ with $n \ge 1$ . Their difference is
$(n+1)^2 - n^2 = 2n + 1.$
We need $2n + 1 \le 2023$ , i.e. $n \le 1011$ . Each $n = 1, 2, \dots, 1011$ gives exactly one pair (and $n = 1011$ works since $2 \cdot 1011 + 1 = 2023$ ), so there are $1011$ pairs.

The answer is $\boxed{\textbf{(B)}\ 1011}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 11: Special Factorizations
Introduction to Algebra, Chapter 20: Special Functions (§20.3)
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.4)