the answer should be E

if a = 1 then no matter what x,y are, the statement 1 or 2(b=1) always holds.

if a is not equal to 1, then statement 1 is sufficient. because even if a is (1.1),

a^x * a^y = 1
{above will hold only for y is always = -x} or {x and y both are 0} {in either case we know x+y = even}


however I wonder why the statement A and B are distinct is given ??

Hi,
Consider a^n = 1, where n is an integer.
if a=1, n can be any number
if a = -1, n should be even
if a is not 0,1,-1, then n = 0, which is even.
Btw, any information that (x+y) is integer?
I am assuming that (x+y) is an integer
From(1):
a^(x+y) = 1. I will denote (x+y) by n.
So,
if a=1, n can be any number
if a = -1, n should be even
if a is not 0,1,-1, then n = 0, which is even.
Not sufficient
From(2):
same as above
Not sufficient

From(1) and (2):
Both a and b can't be 1. So, (x+y) has to be even.

Hence, C

I came pretty close Frank.