MHB Prove Even Integer is Multiple of 4: Contradiction Approach

[cbarker1]

Every even integer that is the square of an integer is a multiple of four.

Prove by Contritidiction.

Assume that n is even and n is square.

I am lost to do next.

 

[I like Serena]

Every even integer that is the square of an integer is a multiple of four.

Prove by Contritidiction.

Assume that n is even and n is square.

I am lost to do next.

... and assume n is not a multiple of 4 (setting up the contradiction).

Then n must be divisible by 2 but not by 4...

What can we say about the number that was squared?
Can we tell if it's even or odd?

 

[HOI]

Lemma: if $n^2$ is even then n is even.
Proof by contradiction- Suppose n is not even. Then it is of the form n= 2k+ 1 for some integer k. Then $n^2= (2k+1)^2= 4k^2+ 4k+ 1= 2(2k^2+ 2k)+ 1$ so is odd, not even.

Theorem: if $n^2$ is even then $n^2$ is a multiple of 4.
Proof- by the lemma, since $n^2$ is even, n is even. That is, n= 2k for some integer, k. Then $n^2= (2k)^2= 4k^2$.