Nis a positive integer with exactly two different divisors greater than 1, how many positive factors does n^2 have? a. 4 b. 5 c. 6 d. 8 e. 9

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.

x + 4/7

step-by-step explanation:

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