Total number of divisors is a prime Which numbers have prime number of divisors?
For example, $16$ has $1$, $2$, $4$, $8$, $16$, a total of $5$ divisors, $5$ being a prime.
I found that primes and the power of primes such that $p^{q-1}$, where $p$ and $q$ are prime numbers, all have prime number of divisors. Is this property
limited to only these numbers?
 A: If $n$ has at least two different prime divisors, $n=ab$ with $\gcd(a,b)=1$, $a,b>1$, then $\tau(n)=\tau(a)\tau(b)$ is not prime (because $\tau(a),\tau(b)>1$).
If $n$ is a power of a prime,  $n=p^k$, then Indeed, $\tau(p^k)=k+1$ and this is prime iff $k+1$ is prime. Thus exactly the numbers of the form $n=p^{q-1}$ where $p,q$ re prime have the desired property.
A: Can you prove that if $n=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ then the number of divisors of $n$ is given by $(a_1+1)(a_2+1)\cdots (a_k+1).$ For this to be prime, you are forced to have $n=p^a$ i.e. $n$ is a power of a prime. Then $n$ has a prime number of divisors as long as $a$ is one less than any prime number (not just $p$).
A: If the number is a prime or is of the form $p^{q-1}$ where both p and q are prime, then and then only the Number will have prime number of Divisors.
NOTE :- I know this is a current codechef problem.  Even after knowing this you may still  experience TLE.
A: If $n=p_1^{k_1}p_2^{k_2}\cdots p_h^{k_h}$ for ($p_i$ prime), then the number of divisors will be $(k_1+1)(k_2+1)\cdots(k_h+1)$.
So you are almost right.  You need only one prime $p_1$ in the factorization and its exponent $k_1$ must be a prime minus one.
