Parity of Integer equals Parity of its Square

Theorem
Let $$p \in \Z$$ be an integer.

Then $$p$$ is even iff $$p^2$$ is even.

Proof
We take it as read that all integers are either even or odd.


 * Let $$p$$ be even.

Then by definition it can be expressed as $$p = 2 k$$ for some $$k \in \Z$$.

Thus:
 * $$p^2 = \left({2 k}\right)^2 = 4 k^2 = 2 \left({2 k^2}\right)$$

and so $$p^2$$ is even.


 * Now, suppose $$p$$ is not even. That is, $$p$$.

Then by definition it can be expressed as $$p = 2 k + 1$$ is odd for some $$k \in \Z$$.

Thus:
 * $$p^2 = \left({2 k + 1}\right)^2 = 4 k^2 + 4 k + 1 = 2 \left({2 k^2 + 2 k}\right) + 1$$

and so $$p^2$$ is odd.

Therefore, if it is not the case that a number is even, then it is not the case that its square is even.

Conversely, if it is the case that a number is even (and also a perfect square), then it is the case that its square root is even.