Answer:
C. 5
Step-by-step explanation:
Remember that for any integer 
, the integers 
are both divisors (or factors) of . First, we will prove that n is a square, and then we will compute the factors of the n².
In this case, the integer n has exactly two different divisors greater than 1. It's impossible that 
, since 1 doesn't have positive factors greater than 1. Then 
, therefore itself is one of the required divisors. Denote by 
the other divisor greater than 1, and note that to satisfy the condition on the divisors, 
.
Because a divides n, there exists some integer 
such that 
. We must have that 
, if not, then 
, which implies that 
, which contradicts the part above.
Now, and, by definition of divisibility, k divides n. Then k must be equal either to n or a, since we can't have three different divisors of this kind. If 
then 
and by cancellation, 
which is a contradiction. Therefore 
and 
.
We have that 
. We can write n as 
. From the first equation, a divides n and a³ divides n. From the second equation, a² divides, and from the last one, 1 divides n and a⁴=n divides n. Thus n has exactly 5 positive factors.
