I can't find what't wrong with this proof, it's from a discrete mathematics slide 
This is a proof in Discrete Mathematics.
 A: The problem lies in the spurious use of biimplications. If we read the left-to-right implications, we get a correct proof of the statement 

If $a$ is odd, then $a^2$ is odd

However, the reverse implication is not correct. We cannot conclude that if $a^2$ is odd, then $a^2 = 2(2k^2+2k)+1$ for some $k \in \mathbb{N}$. All we know is that then $a^2 = 2K+1$ for some $K \in \mathbb{N}$.
A: The problem is when you go back from
$a^2$ is odd
To
$a^2 = 2(2k^2+2k)+1$
That does not (immediately) follow!
A: To be correct, you have to write 
Let $n\in \Bbb N $.

$n $ is odd $\implies \exists k\in \Bbb N \;\; n=2k+1$

$$\implies \exists k\in \Bbb N \;: n^2=2 (2k^2+2k)+1$$
$$\implies \exists K=2 (2k^2+2k)\in \Bbb N \; :$$
$$ n^2=2K+1$$
$\implies n^2$ is odd.
Conversly,

$n$ is even $\implies \exists k\in \Bbb N \; : n=2k $

$$\implies \exists k\in\Bbb N \; : n^2=4k^2$$
$$\implies \exists K=2k^2\in \Bbb N \;: n^2=2K $$
$\implies n^2$ is even.
we conclude that
$n $ is odd $\iff n^2$ is odd.
A: It should be: $a$ is odd $\iff a=2k-1, k\in N.$
