Let’s prove the statement “The square of any odd integer is also odd.”

First we restate the theorem in a way that will lead us towards the proof, by introducing a variable $n$ into the statement of the theorem to represent “any (given) odd integer”:

Theorem: If $n$ is odd, then $n^2$ is also odd.

In order to prove this, we will need the definition of what it means for a integer to be odd:

Definition: An integer $n$ is odd if $n = 2k+1$ for some integer $k$.

Proof of theorem: Assume $n$ is an odd integer.  Therefore, by definition, $n = 2k+1$ for some $k$.  Then $$n^2 = n*n = (2k+1)(2k+1) = 4k^2 + 2k + 2k + 1 = 4k^2 + 4k + 1 = 2( 2k^2 + 2k ) + 1$$ 

This shows that $n^2$ is odd, since $n^2 = 2j + 1$, where $j = 2k^2 + 2k$, so $n^2$ satisfies the definition of being odd.