Lemma. If a^{2} is an even nonnegative integer, then so is a.

Proof. The proof is by a contrapositive argument. We will show that if a is not an even nonnegative integer, then a^{2} is not an even nonnegative integer.

Suppose a is a nonnegative integer that is not even. Then by definition, a can be written in the form

a = 2k + 1

for some nonnegative integer k, and therefore

a^{2} = (2k + 1)^{2} = 4k^{2} + 4k + 1 for some nonnegative integer k

Let j = 2k^{2} + 2k. Then j is also a nonnegative integer and by substitution

a^{2} = 4k^{2} + 4k + 1 = 2(2k^{2} + 2) + 1 = 2j + 1

and so by definition a^{2} is an odd nonnegative integer, and therefore not and even nonnegative integer.

This completes the proof. □