When we multiply an even number by itself, let see what happens. Let us have some examples.

$latex 2 \times 2 = 4$

$latex 4 \times 4 = 16$

$latex 6 \times 6 = 36$

$latex 8 \times 8 = 64$

$latex 10 \times 10 = 100$

Notice that if we multiply even numbers by itself, the product is even. Now how do we prove that this is always true?

**Theorem:** The square of an even number is even.

**Proof:**

Let $latex n$ be an even number.

Then $latex n = 2k$ for some integer $latex k$. This only means that any even number $latex n$ can be represented as the product of $latex 2$ and some integer $latex k$.

Now, $latex n^2 = (2k)^2 = 4k^2$.

Notice that $latex 4k^2 = 2(2k^2)$ and we can represent $latex 2k^2$ as $latex t$ which is an integer making $latex 2(2k^2) = 2t$

Therefore, $latex 4k^2 = 2t$ and clearly $latex 2t$ is an even integer.

This means that $latex 4k^2$, which is the square of an even number is even.

Therefore, if $latex n$ is even, its square is also even and we are done.