Product Perfect Powers

$\text {Let } N = p^{2×2017 - 1} = p^{4033} \text {, where p is prime.}$

The number of the positive divisors N has:

$\phi {N} = \phi (p^{4033}) = 4033 + 1 = 4034 = 2 × 2017$

Now, these divisors can be organised into 2017 pairs in such a way, that:

$\forall (a_i , b_i), i \in (1, 2, \ldots , 2017) : b_i = \frac {N}{a_i} \Rightarrow N = a_i × b_i$

Therefore, the product of all positive divisors of N is the product of all 2017 pairs, each equal to N: $(N)^{2017} = N^{2017}$ .

$\text {Hence, the numbers in the form of } p^{4033} \text { are possible values of N. }$

Since p can be any prime number, and there are infinitely many primes, therefore our answer should be:

$\boxed { \text {Infinitely many} }$

For any non-perfect square ,we have the product of its divisors= $N^{\tau(N/2)}$ .

Hence,we just need $N$ to have $2017*2$ divisors which can easily be found by taking $N=(prime)^{4033}$ .