Let k be an integer. Disprove: “The equation $x^2 − x − k = 0$ has no integer solution if and only if $k$ is odd.”

You seem to be struggling with the logic involved here. The statement is $$ k\text{ is odd}\iff x^2-x-k=0\text{ has no integer solutions} $$ Yes, it seems that making $k$ odd makes the equation have no integer solutions (in other words, the $\implies$ direction is true). Your proof of this looks fine.

However, the statement also claims that if there are no integer solutions, then $k$ is odd (the $\Longleftarrow$ direction). The contrapositive of this claim is that making $k$ even will always result in an integer solution. This is disproven by providing a single counterexample.

Hint:

$x^2-x-k=0 \Rightarrow x=\frac{1 \pm \sqrt{4k+1}}{2}$

So $x$ can only be an integer if $4k+1$ is the square of an odd number.

This is true when $k=2$, in which case $\sqrt{4k+1}=3$, and when $k=6$, in which case $\sqrt{4k+1}=5$.

But what about when $k=4$ ?

Lets assume that equation has integer solutions, so $ x \in Z $

$x^2-x = k$

$x (x-1) = k$

$x-1$ and $x$ are two consecutive integer numbers, their product is even $\Rightarrow k$ is even