Help proving $\text{odd}(m^2-n^2)\implies \text{odd}((m+n)^2)$

Question

I need some help proving the following proposition:

$\text{odd}(m^2-n^2)\implies \text{odd}((m+n)^2)$

Since odd can be defined as:

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

I can write the above statement as:

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

I can write the right side $(m+n)^2$ as $m^2+n^2+2mn$ which looks kind of similar to the left side, but with opposite signs and a $2mn$ on the end.

Can I say that since $m^2-n^2$ is odd, then $m^2+n^2$ is also odd? And then since $2mn$ is always going to be even, that $m^2+n^2 + 2mn$ is Odd $+$ Even which is Odd?

Answer 1

$(m+n)^2=m^2+n^2+2mn=m^2-n^2+2(n^2+mn)$

Answer 2

$m^2-n^2$ is odd iff $m,n$ have different parity, which happens iff $m+n$ is odd, in turn iff $(m+n)^2$ is odd.

Answer 3

$m^{2} - n^{2}= (m-n) (m+n) \equiv 1$ (mod 2) by definition as being odd.

Thus $(m+n) \not \equiv 0$ (mod 2)

Thus $(m+n) \equiv 1$ (mod 2)

Thus $(m+n)(m+n) \equiv 1$ (mod 2)

Thus $(m+n)^{2} \equiv 1$ (mod 2), and is therefore odd.