Use proof by contradiction to prove that if for an integer a, a² is even, then so is a.

By seeking contradiction,

Let a² be even and a be odd.

So, a will be of form,

a = (2n + 1) for any integer n.

⇒ a² = (2n + 1)²

⇒ a² = 4n² + 4n + 1

⇒ a² = 4(n² + n) + 1

Let (n² + n) = k

⇒ a² = 4k + 1

For any integer k, on multiplying by 4 makes it even and adding 1 to an even no. makes it odd.

So, a² will be an odd number, which is not possible.

So, our assumption is wrong i.e. a is not odd.

Hence, proved that for an integer a if a² is even, then so is a.