If $x^2$ is divisible by $4$ then $x$ is even? I am studying discrete mathematics as course and I have to prove this 

"If $x^2$ is divisible by $4$ then $x$ is even". 

I am wondering  how to prove it using the contrapositive of this statement. 
Please guide me on this proof approach.
Thanks in advance.. :)
 A: To prove a statement $p \rightarrow q$, we can instead prove its contrapositive $\lnot q \rightarrow \lnot p$, because the two forms of implication are equivalent: an implication is true if and only if its contrapositive is true: $$p\rightarrow q \equiv \lnot q \rightarrow \lnot p.$$
Let $p$ denote "$4$ divides $x^2$."  


*

*$\lnot p: \;$ "$4$ does not divide $x^2$", or equivalently, "$x^2$ is not divisible by $4$."


Let $q$ denote "$x$ is even."


*

*$\lnot q:\;$ "$x$ is not even", or equivalently, "$x$ is odd."


We want to prove $p\rightarrow q$ by proving $\lnot q \rightarrow \lnot p$.

Proof:
Suppose $x$ is not even (that is, suppose $x$ is odd). Then $x = 2k+1$ for some integer $k$.
And if $x= 2k+1$, it follows that $$x^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1$$ 
Clearly, $4$ does not divide $x^2 = 4(k^2+k) +1$, because  $4$ does not divide $1$.
Having proved "$x$ not even" $\implies$ "$x^2$ is not divisible by $4$," we have proved its equivalent:

"If $x^2$ is divisible by 4, then $x$ must be even,"

as desired.
A: If $x$ were odd, then $x=2k+1$ for some integer $k$, in which case $\,x^2=\;\ldots\;\;?$ 
A: Hints:
If not. Let $x=2k+1$. Then $x^2=(2k+1)^2=4k^2+4k+1$. 
