# 2023 AMC 10B Problems/Problem 9

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## Problem

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 $2023$?

$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

## Solution 1

Let x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$. Thus, $x \le 1011$. So there are $\boxed{\text{(B)}1011}$ numbers that satisfy the equation.

~andliu766

A very similar solution offered by ~darrenn.cp and ~DarkPheonix has been combined with Solution 1.

Minor corrections by ~milquetoast

Note from ~milquetoast: Alternatively, you can let $x$ be the square root of the larger number, but if you do that, keep in mind that $x=1$ must be rejected, since $(x-1)$ cannot be $0$.

## Solution 2

The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$, which is $2^2-1^2$. The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to $2023$ is $2023$, which is $1012^2-1011^2$. These numbers are in the form $(x+1)^2-x^2$, which is just $2x+1$. These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\boxed{\text{(B)}1011}$.

~Aopsthedude

~Math-X

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)