If m and n are integers and x*y is defined as (1)^x + (1)^y + (1)^x

We are told in the stem that \(m*n = (-1)^m + (-1)^n + (-1)^n * (-1)^m\)

(1)

\(m*n = 2 + [(-1)^m * (-1)^n]\) which when compared to the above, we know that \((-1)^m + (-1)^n = 2\). This tells us that \(m\) and \(n\) are even numbers. And while we might not know the values of \(m\)&\(n\), we do not need to know them to solve \((-1)^n * (-1)^m\), as it can be written as \((-1)^{m*n}\). As we know that the exponent will be even we know that \((-1)^{even} = 1\)

Therefore \(m*n = 2 + 1\)
\(m*n = 3\)

SUFFICIENT


(2)

\((-1)^n * (-1)^m = 1\) tells us that \(m+n\) is even, but not more. While their sum is even, \(m\) could equal 5 and \(n\) could equal -3, which would make \(m*n = (-1) + (-1) + 1 = -1\), whereas if they are both even it would yield \(m*n = 1 + 1 + 1 = 3\)

INSUFFICIENT

Answer A

2. says m+n = even and not m*n . which means both are even and sufficient. I think the answer is D

You're right, I made a slight mistake. It's meant to be m+n. However, let m = 5 and let n = -3.

\((-1)^5 * (-1)^{-3} = (-1)^2 = 1\) that works. Now completing the rest of the equation:
\(m*n = (-1)^5 + (-1)^{-3} + 1\)

\(m*n = -1 + \frac{1}{(-1)^{3}} + 1\)

\(m*n = -1 \) which very different to when working with just even numbers.

Let n=2 and m=2 will make it:

\(m*n = (-1)^2 + (-1)^2 + (-1)^2 * (-1)^2\)

\(m*n = 1 + 1 + 1 = 3\)

Could you please explain how \((-1)^n * (-1)^m\) can be written as \((-1)^{m*n}\)