[Math] For $a, b \in \mathbb Z,\;$ if $\;a^2(b^2-2b)$ is odd, then a and b are odd. Proof check.

Question

Suppose $a,b$ are integers, if $a^2(b^2-2b)$ is odd, then a and b are odd, is my solution the best way?
PS: I know this is easy but do i need to expand the final answer? because im practicing for exams and time is of the essence and i want it to be direct as possible.
P -> Q right? now contraposition Q' -> 'P. negation of (a is odd and b is odd) is a is even or b is even so its correct to assume both is even since it still gives a boolean value of True I dont get why my first solution was wrong 🙁
Method of proof: Contraposition
$$ a = 2k $$
$$ b = 2j $$
$$ a^2(b^2-2b) = 4k^2(4j^2 – 4j) $$
$$ 2[k^2(4j^2-4j)]$$ by definition it is even

EDIT: why is my answer wrong? ive made a and b even which is 'Q, then ive proved Q' -> P'

EDIT #2: is this correct now?
Method of proof: Contraposition
$$ a = 2k $$
$$ b = 2j + 1 $$
$$ a^2(b^2-2b) = 4k^2((2j+1)^2 – 4j+2) $$
$$ 2[k^2((2j+1)^2 – 4j+2)]$$ by definition it is even (this is the most efficient solution? for time limits in tests?)

EDIT #3: Hope this is final
Method of proof: Contraposition
$$ a = 2k $$
$$ b = 2j + 1 $$
$$ a^2(b^2-2b) = 4k^2((2j+1)^2 – 4j+2) $$
$$ 2[k^2((2j+1)^2 – 4j+2)]$$ by definition it is even when a is even

and

$$ a = 2k+1 $$
$$ b = 2j $$
$$ a^2(b^2-2b) = (2k+1)^2(4j^2 – 4j) $$
$$ 2[((2k+1)^2(2j^2 – 2j)]$$ by definition it is even when b is even

is this an efficient way now? also i still dont get why i have to prove 'A V 'B biconditionally.

Answer 1

You show: If both numbers $a$ and $b$ are even, then the expression is even. This is not what is asked for.

I suggest to rather use that a product is odd iff all factors are odd.