Let b and x be positive integers. If b is the greatest divisor of x that is less than x, is the sum of the divisors of x, which are less than x itself and greater than one, greater than 2b?

(1) b^2 = x
(2) 2b = x

A)Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B)Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D)EACH statement ALONE is sufficient.
E)Statements (1) and (2) TOGETHER are NOT sufficient.

Does the question anywhere seem to tell that x is the perfect square of a prime number ??

for 1)

If we take x = 25 and b = 5 then sum of divisors of x other than 1 and x is 5, which is not > 2b. So NO.
If we take x = 36 and b = 6 then the sum will be 2+3+4+6+9+12+18 > 2b . SO YES

INSUFFICIENT.

for 2)

If x = 4 , b = 2 then sum of divisors = 2, which is not > 4. SO NO.
If x = 18 , b = 9 then sum of divisors = 2+3+6+9= 20 > 18. SO YES

INSUFFICIENT.

Taking both together,

b^2 = 2b => b = 0 or b = 2. given b is positive so b = 2 and x = 4 and so sum of divisors of 4 = 2, which is not > 4.
so NO.

SUFFICIENT.

The solution says x is perfect squesre of prime number so in (1) we can take x as 4,9,25 all of which give same answer and hence sufficient.
I dont understand how it is only prime.
Please explain.