Show that the product of $latex (m + 1)(m+2)$ is an even number.

This proof requires elementary knowledge in algebra and some manipulation of symbols. It is assumed that you already know or have proved that the sum of two even numbers is even, the sum of two odd integers is odd, and the sum of an odd number and an even number is odd. This proof is designed to be read by middle school and high school students, so it is written with details.

**Proof 1**

The numbers $latex m + 1$ and $latex m + 2$ are consecutive numbers. There are two cases possible, the smaller number is odd or even.

If the smaller number $latex m + 1$ is odd, then $latex m + 2$ is even. Now, odd multiplied by even is even.

For the second case, if the smaller number is even, then the second number is odd. Now, even multiplied by odd is always even.

Therefore, in any case the product of $latex (m + 1)(m + 2)$ is even.

**Proof 2**

The product of $latex (m + 1)(m+2) = m^2 + 3m + 2$. We can have two cases form, $latex m$ is odd, and $latex m$ is even.

If $latex m$ is even, then $latex m^2$ is even, since the square of an even number is even. Also, $latex 3m$ is even because they are a product of an even number $latex m$ and another integer. Since 2 is also even, $latex m^2 + 3m + 2$ even. Remember even number + even number + even number = even number.

For the second case $latex m$ is odd.

If m is odd, then we can represent $latex m$ as $latex 2k + 1$. If we substitute $latex k + 1$ to $latex m$, we have

$latex (2k + 1)2 + 3(2k + 1) + 2$

Since $latex 2k + 1$ is odd, its square$latex (2k + 1)2$ is also odd. Now, $latex 3(2k + 1)$ is the product of an odd number and another odd number which means that it is also odd. This means that we have only three terms. Odd Number + Odd Number + Even Number = Even Number.