case: a and b are both even -> find m and n in terms of k and l by algebra manipulation.

case: a and b are odd -> same as above

(\leftarrow)

AFSOC (\exists m, n \in \mathbb{Z})(a=m+n \land b=m-n) but a, b have opposite parities. WLOG Assuming one is odd, the other is even, manipulate algebraic equations into 2(k+l)+1-2m to show k+l-m=-\frac{1}{2} and that is a contradiction since -\frac{1}{2} is not an integer.

Claim: for all integers m and n m^2(n^2-2n) is odd iff m and n are both odd.
Proof: using definition only. easy, same as above