To prove by contradiction you assume the opposite of what is stated, ie that $\displaystyle n^2$ is odd.
As you are starting with an even number, then instead of $\displaystyle n$, lets use $\displaystyle 2n$.
$\displaystyle (2n)^2 = 4N$, therefore if $\displaystyle n \in Z$(intergers), $\displaystyle n^2$ is always even.
This is how I would do it anyway, I am sure there are different methods out there.
Craig
ps does anyone know how to do the 'member of' sign in your maths formatting thing?
Edit: solved