Suppose not. [We take the negation of the given statement and
suppose it to be true.] Assume, to the contrary, that ∃ an integer n such
that n^{2} is odd and n is
even. [We must deduce the contradiction.] By definition of even, we have

n = 2k for some integer k.

So, by substitution we have

n . n = (2k) . (2k)

= 2 (2.k.k)

Now (2.k.k) is an integer because products of integers are
integer; and 2 and k are integers. Hence,

n . n = 2 . (some integer)

or
n^{2} = 2. (some integer)

and so by definition of n^{2} even, is even.

So the conclusion is since n is even, n^{2},
which is the product of n with itself, is also even. This contradicts the
supposition that n^{2} is odd. [Hence, the supposition is false and the
proposition is true.]