Proving the integer that p is even and its square is even as well

[cbarker1]

Dear Everyone,I would like some help to get start with a proof. A problem states, "if p is an integer, show that p is even iff p^2 is even."I know that p is the an integer.

Let p be an integer.

$p=2m$, where m is an integer.Thank you for your help

CBarker

 

[mathmari]

To show that the integer $p$ is even iff $p^2$ is even, we have to show the following two implications:

1. Let $p \in \mathbb{Z}$. If $p$ is even, then $p^2$ is even.
2. Let $p \in \mathbb{Z}$. If $p^2$ is even, then $p$ is even.

To show the first one we do the following:

$p$ is an even integer, so $p=2m$, for some $m \in \mathbb{Z}$.
Squaring both sides we get $p^2=(2m)^2=4m^2=2(2m^2)$.
Since $m$ is an integer, we have that $2m^2$ is also an integer.
So, $p^2=2n$, with $n=2m^2 \in \mathbb{Z}$.
So, $p^2$ is even. Can you continue and show the second implication?

 

[cbarker1]

P is not even. If p is not even, then it is odd; therefore there exists an integer n such that
$p=2n+1$
${p}^{2}={\left(2n+1\right)}^{2}\implies p^2=4n^2+4n+1$
$p^2=2\left(2n^2+2n\right)+1$
$p^2=2k+1$, where $k=2n^2+2n$ is some integer.
${p}^{2}$ is odd, contrary to the hypothesis. Since p is not even, it leads to a contradiction. p is even. QED

Is it correct?

 

[mathmari]

It is correct! (Yes)