Even and odd numbers are used in arithmetic, especially in the proof of properties. In this post, we show how to prove that a natural number is odd or even. Before doing this, let us give some definitions.

## Even and odd numbers

We say that a natural number $a$ is * even* if it can be written as $a=2k$ where $k$ is a natural number. On the other hand, $a$ is called an

*number if we can write $a$ as $a=2k+1,$ where $k$ is a natural number. The following are important properties of this kind of number:*

**odd**- The sum, the difference, and the product of two even numbers is still even number.
- The sum or the difference of an even number with another odd number is an odd number
- The product of an even number with another odd number is an even number

## How to prove that a natural number is even or odd

Let $n$ be a natural number. We introduce three examples in which we prove the parity of numbers.

Let $a=6n+3$. Remark that $a=6n+2+1=2(3n+1)+1=2k+1$, with $k=3n+1\in \mathbb{N}$. Thus $a$ is odd.

Let $b=4n+6$. Immediately we observe that $b=2(2n+3)=2k,$ where $k=2n+3\in\mathbb{N}$. Thus $b$ is even.

Let $c=(2n+1)^2+2n-1$. We can write

\begin{align*}c&=(2n+1)^2+2n-1\cr &= 4n^2+4n+1+2n-1\cr &= 4n^2+6n\cr &= 2(2n^2+3n).\end{align*} This shows that $c$ is an even number.

## Exercises on even and odd numbers

We give more examples of even and odd numbers.

**Exercise:** From the following list of numbers, determine which are even: $$ 27+15,\quad 5^2,\quad \frac{378}{3},\quad 15^2-8.$$

**Solution:** On a

- $27+15=42=2 \times 21$, it is an even number.
- $5^2=5\times 5=25$, it is an odd number.
- $\frac{378}{3}=126=2\times 63,$ it is an even number
- $15^2-8=225-8=217=2\times 108+1$, it is an odd number.

**Exercise:** Show that the square of an even number is even.

**Proof:** Let $N$ be an even number. Then this number takes the form $N=2n$ with $n$ natural number. Taking the square on both sides of the last equality, we obtain $N^2=(2n)^2=4n^2=2 (2n^2)$. Thus $N^2$ is an even number.

**Exercise:** Show that the product of two consecutive integers is even.

Proof: Let $k$ be an integer and select $N=k(k+1),$ product of two consecutive integers. Two possibilities for $k$: $k$ even or odd. First, assume that $k$ is even. Then $k=2n$ with $n$ an integer number. We replace in $N,$ we obtain $N=2n(2n+1)=2 p$ with $p=2n^2+n$ is an integer. Thus $N$ is even. Second, assume that $k$ is odd. Then $k=2n+1$ with $n$ is integer. We then have $N=(2n+1)(2n+1+1)=(2n+1)(2n+2)$. This can be rewritten as $N=2q$ with $q=(2n+1)(n+1)$. Thus $N$ is an even number.

**Exercise:** We consider two odd natural integers $p$ and $q$. Prove Show that $N=p^2+q^2+8$ is divisible by $8$.

**Proof:** To show that $N$ is divisible by $8,$ it suffices to find an integer number $d$ such that $N=8d$. In fact, by assumption there exist two integers such that $p=C$ and $q=2m+1$. Then \begin{align*} N&=(2n+1)^2+(2m+1)^2+6\cr &=4n^2+4n+1+4m^2+4m+1+8\cr &= 4n(n+1)+4m(m+1)+8.\end{align*} According to the previous exercise we know that the product of two consecutive integers is even. Thus we can write $n(n+1)=2k_1$ and $m(m+1)=2k_2$ with $k_1$ and $k_2$ are two integers. This gives \begin{align*} N=8k_1+8k_2+8=8 (k_1+k_2+1)=8d,\end{align*} with $d=k_1+k_2+1$ is an integer.