For $a, b \in \mathbb Z,\;$ if $\;a^2(b^2-2b)$ is odd, then a and b are odd. Proof check. 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.
 A: 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.
A: You have assumed that $a$ and $b$ are both even, and showed that if so, $a^2(b^2-b)$ is also even.  By contraposition, this shows that if $a^2(b^2-b)$ is not even, then it is not true that $a$ and $b$ are both even.  That is, you have shown only that at least one of a and b is not, not that both of them are odd.
Contraposition will still work, but you need to show that if $a$ is even then $a^2(b^2-b)$ is even, and also that if $b$ is even $a^2(b^2-b)$ is even.

I'm going to try to explain your mistake another way.  
You want to show:
$$a^2(b^2-b)\text{ is odd}\implies (a\text{ is odd}\wedge b\text{ is odd})$$
You are trying to do this by controposition, which means you want to prove the contrapositive:
$$\lnot\left\{a\text{ is odd}\wedge b\text{ is odd}\right\}\implies \lnot \left\{a^2(b^2-b)\text{ is odd}\right\}.$$
So you need to start by assuming $$\lnot\left\{a\text{ is odd}\wedge b\text{ is odd}\right\}$$
but instead, you started by assuming $$\lnot a\text{ is odd}\land \lnot b\text{ is odd}$$
and then when you tried to correct your proof you started by assuming $$\lnot a\text{ is odd}\land b\text{ is odd}.$$
Neither of these is going to so what you want.
A: As you've now managed to get to a complete proof, I'll write up a slightly slicker version for you (I won't do all the algebraic manipulation, but you should be able to fill that in yourself).
As pointed out by several people, if we're going to use the contrapositive, we start by assuming that $a$ and $b$ are not both even, or equivalently, that at least one is odd.
So first we assume that $a=2k$ is even. Then:
$$a^2(b^2-2b)=4k^2(b^2-2b)$$
is even as required. (Note that I have made no assumption about $b$ here.)
Then we assume that $b=2j$ is even. Then:
$$a^2(b^2-2b)=4a^2(j^2-j)$$
is even as required. (This time I make no assumptions on $a$).
A: Hint: $\ a^2b^2\! - 2a^2b = 1\!+\!2n\,\Rightarrow\, \color{#c00}{a^2b^2}= 1\!+\! 2(n\!+\!a^2b),\,$ so $\,\color{#c00}a\,$ or $\,\color{#c00}b\,$ even $\rm\,\Rightarrow\, \color{#c00}{even} = odd.$
The same idea shows $\ a_1\cdots a_n\! + 2k = 1\! +\! 2n\ $ then all $\,a_i\,$ are odd. So a product of integers is odd iff each factor is odd. This is a special case of the fact that a product is invertible iff each factor is invertible (this fundamental fact is true in any ring). 
A: We already have some nifty proofs, however since this is a parity problem, it seems reasonable to prove it using ${\bf parity-operators}$. so let's do just that.
The problem can be rewritten, by contrapositive, as: the expression is even provided $a$ or $b$ is even. So let's prove just that.
[We goto contrapositive since I prefer working with $\mathsf{even}$ instead of $\mathsf{odd}$.]
$
\hspace{0.5cm}\mathsf{even}(a^2(b^2-2b)) 
\\\equiv \mathsf{even}(a^2b^2-2ab) \hspace{4.3cm}\text{Using arithmetic}
\\\equiv \mathsf{even}(ab)^2 \equiv \mathsf{even}(-2a^2b)
    \hspace{3cm}\text{Using property: $\mathsf{even}(n+m) \equiv \mathsf{even}(n) \equiv \mathsf{even}(m)$}
\\\equiv \mathsf{even}(ab)^2 \equiv \mathsf{true}
   \hspace{4.2cm}\text{ Multiples of 2 are even}
\\\equiv \mathsf{even}(ab)^2 \hspace{5.2cm}\text{`True' is identity of `$\equiv$'}
\\\Leftarrow \mathsf{even}(ab) \hspace{5.2cm}\text{Using property: $\mathsf{even}(n^2) \Leftarrow \mathsf{even}(n)$}
\\\Leftarrow \mathsf{even}(a) \text{ or } \mathsf{even}(b)
\hspace{4cm}\text{Using property: $\mathsf{even}(nm) \Leftarrow \mathsf{even}(n) \text{ or } \mathsf{even}(m)$}
$
Best regards,
Moses
