Hilbert Primes Golf

Haskell, 46 bytes

(foldr(\a b->a:[x|x<-b,mod x a>0])[][5,9..]!!)

An anonymous function.

The core is foldr(\a b->a:[x|x<-b,mod x a>0])[][5,9..], which iterates through the arithmetic progression 5,9,13,..., removing multiples of each one from the list to its right. This produces the infinite list of Hilbert primes. Then, !! takes the nth element.

I has tried making (\a b->a:[x|x<-b,mod x a>0]) pointfree but didn't find a shorter way.

CJam, 36 33 32 23 bytes

5ri{_L+:L;{4+_Lf%0&}g}*

Try it online

The latest version is actually much more @MartinBüttner's than mine. The key idea in his suggested solution is to use two nested loops to find the n-th value that meets the condition. I thought I was being clever by using only a single loop in my original solution, but it turns out that the added logic cost more than I saved by not using a second loop.

Explanation

5       Push first Hilbert prime.
ri      Get input n and convert to integer.
{       Loop n times.
  _       Push a copy of current Hilbert prime.
  L       Push list of Hilbert primes found so far (L defaults to empty list).
  +       Prepend current Hilbert prime to list.
  :L      Store new list of Hilbert primes in variable L.
  ;       Pop list off stack.
  {       Start while loop for finding next Hilbert prime.
    4+      Add 4 to get next Hilbert number.
    _       Copy candidate Hilbert number.
    L       Push list of Hilbert primes found so far.
    f%      Element wise modulo of Hilbert number with smaller Hilbert primes.
    0&      Check for 0 in list of modulo values.
  }g      End while loop.
}*      End loop n times.

Pyth, 21 bytes

Lh*4bye.fqZf!%yZyT1hQ

Try it online: Demonstration or Test Suite

Explanation:

Lh*4bye.fqZf!%yZyT1Q    implicit: Q = input number
L                       define a function y(b), which returns
 h*4b                      4*b + 1
                        this converts a index to its Hilbert number
       .f          hQ   find the first (Q+1) numbers Z >= 1, which satisfy:
           f      1        find the first number T >= 1, which satisfies:
            !%yZyT            y(Z) mod y(T) == 0
         qZ                test if the result is equal to Z 

                        this gives a list of indices of the first Q Hilbert Primes
      e                 take the last index
     y                  apply y and print