The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?

(1) a# + a = 0
(2) (a#)*a + a^2 = 0

Why didn't you checked with a=-1, because that option will give you different results

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depending on the value of m, a# can be either -a or a. if we can derive the nature of m (odd or even or -ve or +ve), that will suffice.

(1) a# + a =0 , a# = -a, that means m was odd in this case. Lets say m= 1, a=4, so a# = (-1)^1*4 = -4, (a#)#= (-1)(-4)=4, (a#)# -a =0. Again if we take a = -4, a# will be 4 and (a#)# will be -4. so (a#)# - a = -4 -(-4) =0. Sufficient.

(2) a {(a#) + a} = 0. since a can not be 0, {(a#) + a} =0, so, a# = -a, which is just like the previous data. Sufficient.