If k and t are integers, and k^2-t^2 is and 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

Answer Choices:
A. None
B. I Only
C. II Only
D. III Only
E. I,II and III

I selected E- which was wrong => but can you explain my error in thinking

Posted: Sun Aug 10, 2008 7:38 pm


If k and t are integers, and k^2-t^2 is and 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

Answer Choices:
A. None
B. I Only
C. II Only
D. III Only
E. I,II and III

I would go with A. k^2 - t^2 is an odd integer, that is only possible when either k^2 is odd and t^2 is even or if k^2 is even and t^2 is odd
because even - even = even, and odd - odd = even
so lets pick numbers,
k = 3, t = 2 , it can be vica versa as well, i.e k = 4, t =3
thus 9 -4 = 5

1. k+t+ 2
1st case: 3+2+2 = 7 --> odd
2nd case: 4+3+2 = 9 --> odd

2 k^2+2kt+t^2 which is same as stating (k+t) ^ 2
1st case: (3+2)^2 = 25 -- > odd
2nd casE: ( 4+ 3) ^ 2 = 49 -- > odd

3. k^t + t^ 2
1st case: 9 + 4 = 13 -- > odd
2nd case: 16 + 9 = 25 -- > odd

Thus, my answer is A (None). Can you please confirm thats the right answer. Thanks !!

yeah, you can actually prove that ALL of these expressions MUST, ironically, be odd. this is a rather strange happening on a problem where even one odd result would be enough to settle the question - remember, the question is whether the quantities must be even, not whether they can - but that's life.

the original expression, k^2 - t^2, is a difference of squares. you should know the factorization of this expression without even thinking: it's (k - t)(k + t).
this means that both k - t and k + t are odd integers.

(i)
if k + t is odd, then k + t + 2, which is greater by exactly 2, is also odd.

(ii)
this is the square of k + t, which is itself odd, so it's also odd.

(iii)
this expression can't be factored, but it can be gotten by taking k^2 - t^2 (which is odd) and adding 2(t^2) (which must be even). therefore, it's odd + even = odd.