This problem mobilizes addition, substraction and multiplication rules of odd and even integers so knowing your rules will be the key to helping you solve this correctly.
First of all, \(k^2 - t^2\) can be rewritten as \((k-t)*(k+t)\)
If you're unsure about the notation, just develop \((k-t)*(k+t)\).
Now, assuming that \(k^2 - t^2\) = \((k-t)*(k+t)\) is odd, then according to the following rule :
odd * odd = odd (1)
We'll get \(k-t\) is odd and \(k+t\) is odd, which is extremely helpful since, if you notice the answer choices, all of them revolve around \(k+t\). So let's go through them one by one :
I. \(k+t+2\).
Using parenthesis to isolate \(k+t\), we get \((k+t)+2\) which is a sum involving an odd number and an even number. So, according to the following rule :
odd + even = odd (2)
Which means that \(k+t+2\) is odd. So answer I is not possible. (Since we're looking for an even result)
II. \(k^2 + 2kt + t^2\)
Now this answer choice may seem intimidating, but it actually isn't. Since \(k^2 + 2kt + t^2\) is equal to \((k+t)^2\). And since \(k+t\) is odd, then its square will be odd as well (rule 1). So answer II is also not possible.
III.\(k^2 + t^2\)
Once again, this answer choice may seem intimidating since you have no data on k nor t. But, looking at answer choice II., \(k^2+t^2\) is actually equal to \((k+t)^2 - 2kt\).
This is a difference between an odd number \((k+t)^2\) and an even number \(2kt\), so according to rule 2, the result will be odd. So answer III. is also not possible.
As such, the only correct answer to this question is answer A.
Hope that helped.