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@Tombert
Created January 21, 2015 20:20
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Title4
A generalization for all even numbers is
$$
2k
$$
where k is an integer. Therefore, a generalization for all odd numbers would be.
$$
2k+1
$$
To prove that squaring an odd integer is odd, let's square our expression.
$$
\left( 2k + 1 \right)^{2} = 4k^2 + 4k + 1
$$
Let's factor out $4k$
$$
4k\left(k+1\right) + 1
$$
$$
2\left(k\left(k+1\right)\right) + 1
$$
Since $\left(k\left(k+1\right)\right)$ is still an integer, and we're working with generalizations, we can just merge that back into a generic constant $k$, and we're back to our original odd-form.
$$
2k + 1
$$
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