2023 AMC 10B Problem 9

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

1. Small cases.

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 $\boxed{\textbf{(B)}\ 1011}$.

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".

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 $\boxed{\textbf{(B)}\ 1011}$.