Question

If positive integer number n has exactly 8 divisors, what is the least number of divisors can n^2 have?

Answer 1

Answer:

The simpler case is the number:

n = a^7, where a is prime.

We use this so we avoid having cross numbers by multipliying different prime numbers. because for example lets compare:

2^2 = 4 has 3 divisors, 4, 2 and 1.

(4*4) = 16 has the divisors: 1, 2, 4, 8 and 16. (5 divisors).

Now let's mix primes:

2*3 = 6 has 3 divisors, 3, 2 and 1.

6*6 = 36

(3*2*1)*(3*2*1) = 36

then the divisors are:

3*3, 3*2, 3*1, 2*2, 2*1, 1*1, 3*2*2, ..., etc.

You can clearly see that this has a lot more divisors than the previous case.

Then returning to the answer:

a^7 can be divided by a, a^2, a^3, ..., and by itself, a^7 (and 1, so we have exactly 8 divisors).

Now, we can write:

n^2 = (a^7)^2 = (a^(2*7)) = a^14.

By the same logic that was used above, a^14 has 14 + 1 divisors.

Then the smallest number of divisors possible is 15.