Case 1: a is not a member of {-1, 0, 1}

a = xa^2 where x is an int # definition of divisibility
1 = xa   #algebra
1/x = a  #algebra

a is not an integer, thus the P (a is an integer and a^2|a) is false,
thus the proof isn't violated.

Case 2:  a is a member of {-1,0,1}
a=xa^2
(a,x) = (-1,-1), (0,0), and (1,1)


So neither of the two cases disproves the proof.