Prove that the difference between the squares of two consecutive integers is equal to the sum of the two integers.

Sure, let's prove this step by step.

Let's denote the two consecutive integers as n and n+1.

The square of n is n^2 and the square of n+1 is (n+1)^2.

The difference between the squares of these two consecutive integers is (n+1)^2 - n^2.

We can expand (n+1)^2 to get n^2 + 2n + 1.

So, the difference between the squares of the two consecutive integers is (n^2 + 2n + 1) - n^2.

Simplifying this, we get 2n + 1.

The sum of the two consecutive integers is n + (n + 1), which simplifies to 2n + 1.

So, we have shown that the difference between the squares of two consecutive integers (2n + 1) is equal to the sum of the two integers (2n + 1). Therefore, the statement is proven.