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 
into the statement of the theorem to represent “any (given) odd integer”:
Theorem: If is odd, then 
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 is odd if 
for some integer 
.
Proof of theorem: Assume is an odd integer. Therefore, by definition, for some . Then
![\[n^2 = n*n = (2k+1)(2k+1) = 4k^2 + 2k + 2k + 1 = 4k^2 + 4k + 1 = 2( 2k^2 + 2k ) + 1\]](https://openlab.citytech.cuny.edu/mat2440-ganguli-fall2018/wp-content/uploads/ql-cache/quicklatex.com-27a7a225cad1af15df928b9078665b1e_l3.png)
This shows that is odd, since 
, where 
, so satisfies the definition of being odd.
