start from the prime power factorisation:

The above product expression only includes primes above 1 to avoid double counting on the next step.

By definition any part of the above product is a divisor of n. The full set of divisors are formed by varying the powers of each prime factor (other than 1) between 0 and

, ie the divisors are:

The total number of divisors that can be constructed by varying those powers is

. The construction includes all possible divisors since the prime factorisation of n is unique. No divisor is constructed twice since the prime factorisation of each divisor is unique. So D is the correct total of unique divisors. NB: the total (D) counts both

*n* and

*1* as divisors of n. If you dont consider those "real" divisors then you can subtract 1 or 2 from the total.

Getting the number of square divisors is a simple extension. Define

A number is square iff its prime factorsation has even powers, so the possible square divisors are

Anything constructed in this way is a divisor of n since

. Using the same arguments as before the total number of square divisors is

Now it's just a probability problem. We want P(square then non-square) + P(non-square then square).