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

$latex p + q = (2r + 1) + (2s + 1) = 2r + 2s + 2$

Factoring out 2, we have

$latex 2s + 2r + 2 = 2(s + r + 1)$

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

$latex p + q = 2(s + r + 1)$

is even.

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