PP777 wrote:
Is (a^2 - b^2)/2 an odd integer?
(1) a-b is odd
(2) a^2 + b^2 is odd
(1) a-b is odd. We are not given that a/b are integers. If we assume that a/b are integers, then a-b will be odd if one of a/b is odd and the other is even. So in this case a^2 - b^2 will also be odd, and thus not divisible by 2. This would make (a^2 - b^2)/2 a decimal, and thus not an integer. So the answer to the question asked would be NO.
But it is also possible that a/b are not integers, rather decimals. Eg, a=4.5 and b=1.5; this satisfies the first statement in that a-b = 3 (odd), and here a^2 - b^2 is 4.5^2 - 1.5^2 = 18. 18 divided by 2 is an odd integer (9), and so in this case the answer to the question asked would be YES.
This means statement 1 is not sufficient.
Its also important to note that a-b will be odd either when both a/b are integers (one odd, one even) OR when they have some decimal values, but a/b cannot have irrational values.(2) a^2 + b^2 is odd. If a/b are integers, then a^2 + b^2 will be odd if one of a/b is odd and other is even. And in that case a^2 - b^2 will also be odd, and thus not divisible by 2. This would make (a^2 - b^2)/2 a decimal, and thus not an integer. So the answer to the question asked would be NO.
But it is possible that a/b are not integers. Lets take a=(3)^1/2 and b=1; in this case a^2 - b^2 = 3-1 = 2; and here (a^2 - b^2)/2 = 2/2 = 1, odd integer. So in this case answer to the question asked is YES.
This means statement 2 is also not sufficient.
Its also important to note that a^2 + b^2 will be odd either when both a/b are integers (one odd, one even) OR when at least one of them has irrational values, but a/b cannot have decimal values. Combining the two statements, this formula comes to mind: the common info from both statements is that both a/b have to be integers, and one of them will be odd while the other will be even. In that case a^2 - b^2 is also odd, and thus not divisible by 2. So (a^2 - b^2)/2 cannot be an integer at all, thus the answer is a definite NO.
Hence
C answer.