The given statement is
"If $x$ is an integer and $x^2$ is even, then $x$ is also even."
Let $p:$ If $x$ is an integer and $x^2$ is even. and
$q:$ $x$ is even.
$\Rightarrow\:$ ~$p:$ $x$ is an integer and $x^2$ is not even.
~$q$: $x$ is not even.
To prove that the above statement is true, we have to prove
If $x$ is not even then $x^2$ is not even for $x \in Z$
Let $x$ be an odd integer. (not even)
$\Rightarrow\: x=2n+1$ for any $n\in Z$
$\Rightarrow\:x^2=(2n+1)^2$
$\Rightarrow\:x^2=4n^2+4n+1$
We know that $4n^2$ is an even number.
$4n$ is also an even number.
$\Rightarrow\:4n^2+4n$ is even. (Because sum of two even numbers is even.)
$\therefore$ $4n^2+4n+1$ is odd. (Because sum of an even number and an odd number is odd.)
$\Rightarrow\:x^2$ is odd. (not even.)
$i.e.,$ If $x$ is not even then $x^2$ is not even.
$\therefore$ The given statement is true.