Prove that $n^2$ is even if and only if $n$ is even

Question

I am practising exam questions and have come across the following.

Prove that $n^2$ is even if and only if $n$ is even

A contrapositive proof comes to mind, since it must be the case that if $n$ isn't odd then $n$ is even. So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$ which is odd, hence $n^2$ must be even.

Is this correct?

Answer 1

For an if-and-only-if proof, you need to show both ways - you have only done $n$ even $\implies$ $n^2$ even, which is correct. You must show that

• if $n=2k$ for some integer $k$, $n^2=4k^2=2(2k^2)$ is even, which is true

• if $n^2=2t$ for some integer $t$ but $n=2s+1$ is odd for some integer $s$, $$(2s+1)^2=2t\implies 4s^2+4s-2t=2(2s^2+2s-t)=-1$$ is a contradiction as the parities on LHS and RHS are different

Answer 2

You've proved that if $n^2$ is even then so is $n$. So you've proved the "only if" direction, but not the "if".

So you've just got to show that the square $n^2$ of an even number $n=2k$ is even, and you've got the "if" as well. Put them together and you're there.

Answer 3

$n>1;$

$D_n:= n^2-n=n(n-1); $

$D_n$ is even, since either $n$ or $(n-1)$ is even.

1) $n$ is even $\Rightarrow$

$n^2 = D_n +n$ is even;

(Sum of $2$ even numbers)

2)$n^2$ is even $\Rightarrow$

$n= n^2-D_n$ is even;

(Difference of $2$ even numbers)

Used: Sum and difference of $2$ even numbers is even(Why?)