If k and t are integers and k^2 - t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2
II. k^2 + 2kt + t^2
III. k^2 + t^2

A) None
B) I Only
C) II Only
D) III Only
E) I, II, and III

Answer is A.

I get stuck beyond the point where I know that (even - odd) or (odd - even) produces an odd number. Is plugging in numbers the way to go with this? I want to understand if I am missing any theory here...

K^2-T^2 = Odd

So one of them is odd, and the other even.

If K^2 is odd then K is odd, Likewise if T^2 is odd then T is odd (You can interhange the two if you like).

Therefore (Proving the cases where each of these can be odd):

For 1: K+T = Odd and so K+T+2 will be odd

For 2:
K^2+2KT+T^2 = (K+T)^2
K+T = Odd=> (K+T)^2 = Odd (OxO=O)

For 3: O+E is O

i could swear to every deity with which i'm familiar that i've already answered this question here, but i can't find the thread.

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your observation is dead on.
there are 2 cases:
(a) k^2 is odd and t^2 is even --> k is odd and t is even,
(b) vice versa.

you can then take each of these cases and apply elementary rules of arithmetic with odds and evens:
(i) = odd + even + even (or even + odd + even) = odd
(ii) = odd + even + even (or even + even + odd) = odd
(iii) = odd + even (or even + odd) = odd
so, answer = (a).

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note that number plugging is supremely quick and easy here, because ANY acceptable pair of numbers will IMMEDIATELY yield the correct answer to the problem. for instance, if you realize that k = 2 and t = 1 (most people's first choice) works, then you'll immediately find that all three quantities are odd, and you'll be finished.

they're being really nice to you here: the problem is "which MUST be even" - and they're going so far as to give you counterexamples that not only don't have to be even, but actually MUST be odd. that's amazingly friendly; there are a lot of hearts and flowers in this problem.