Prove what kind of number you get when you add 1 to the product of two consecutive odd numbers.

1. Problem Description

Prove what kind of number you get when you add 1 to the product of two consecutive odd numbers.

2. Solution Steps

Let $2n+1$ be an odd number, where $n$ is an integer. Then the next consecutive odd number is $2n+3$.

The product of these two consecutive odd numbers is $(2n+1)(2n+3)$.

$(2n+1)(2n+3) = 4n^2 + 6n + 2n + 3 = 4n^2 + 8n + 3$

Adding 1 to the product gives $4n^2 + 8n + 3 + 1 = 4n^2 + 8n + 4$.

We can factor this expression as $4(n^2 + 2n + 1)$.

We can further factor the expression inside the parentheses: $4(n+1)^2 = [2(n+1)]^2$.

Since $n$ is an integer, $n+1$ is also an integer, and $2(n+1)$ is an even integer.

Therefore, $[2(n+1)]^2$ is the square of an even integer. Since it is the square of $2(n+1)$, it is a perfect square.

3. Final Answer

Adding 1 to the product of two consecutive odd numbers results in the square of an even number (a perfect square).