\[ \text{Find } 2011^2 - 2010^2 \text{?} \]

To solve \( 2011^2 - 2010^2 \), we can use the difference of squares formula, which states that \( a^2 - b^2 = (a - b)(a + b) \). Here, let \( a = 2011 \) and \( b = 2010 \). So we have: \[ 2011^2 - 2010^2 = (2011 - 2010)(2011 + 2010) = (1)(4021) = 4021. \] Thus, \( 2011^2 - 2010^2 = 4021 \). Fun fact: This neat trick works for any two consecutive integers! The difference in their squares will always equal the sum of the two. So whether it's \( 10^2 - 9^2 \) or \( 1000^2 - 999^2 \), you can apply this same formula. And, talking about squares, if you enjoy puzzle games, you might find square numbers fascinating! They pop up in all sorts of math-related games and logic puzzles. You could even start a little fun study on patterns among square numbers—there's always something new to discover!