Help proving $odd(m^2+n^2) \implies odd((m+n)^2)$

Question

I need some help proving the following poposition:

$\forall m,n \in \mathbb{N}\;\text{odd}(m^2+n^2) \implies \text{odd}((m+n)^2)$

Since $\text{odd}$ can be defined as:

$\text{odd}(x): \exists k \in \mathbb{Z}\; x = 2k+1$

I can write the above statement as:

$\exists k_1 \in \mathbb{Z} \; m^2-n^2 = 2k_1+1 \implies \exists k_2 \in \mathbb{Z}\; (m+n)^2=2k_2+1$

After I get to this, I'm totally stuck. To prove this, I think I need to somehow write that $m^2-n^2$ can be written in the form $2k+1$, but using $(m+n)^2$. Is this the write thinking? If so, how would I go about doing that?

Answer 1

Hints:

• Even + Odd = Odd
• $2n$ is always Even
• $(n+m)^2=n^2+m^2+2(nm)$

Answer 2

$(m+n)^2=m^2+n^2+2mn\equiv m^2+n^2 \pmod 2$

So $m^2+n^2$ odd implies $(m+n)^2$ odd.