# The Sum of Two Prime Numbers Greater than 2 is Even

In the previous posts, we have gone quite deep in delving and proving some complicated theorems. In this post, we go again the the basics. In this short proof, we show that the sum of two prime numbers, both greater than 2 is even.

Theorem

The sum of two primes, both greater than 2, is always even.

Proof

Let $p$ and $q$ be prime numbers both greater than 2. Then, both of them are odd numbers. This means that we can let $p = 2r + 1$ and $q = 2s + 1$ where $r$ and $s$ are positive integers.  Adding, we have $p + q = (2r + 1) + (2s + 1) = 2r + 2s + 2$

Factoring out 2, we have $2s + 2r + 2 = 2(s + r + 1)$

Since $2(s + r + 1)$ is divisible by $2$, it is even. Therefore, $p + q = 2(s + r + 1)$

is even.

Therefore, the sum of two primes both greater than 2 is even.

## 3 thoughts on “The Sum of Two Prime Numbers Greater than 2 is Even”

1. m asif on said:

One mistake in the proof. you write p=2q+1 and next use p=2r+1 . So replace p=2q+1 by p=2r+1.

• Guillermo Bautista on said:

Thank m asif for pointing out the mistake. Changed it already.