On m = sqrt(sqrt(n) + sqrt(kn+1))

Someone asked how to find the integer solutions of 
                  _____________________
           m =   /sqrt(n) + sqrt(k*n+1) 

This could be viewed as a Pell equation with an extra condition on
the solution.  We have  m^2 = sqrt(n) + sqrt(kn+1) in integers, so
we know there exist integers x,y such that 

         n = y^2         kn + 1 = x^2         x + y = m^2

Eliminating n from the first two equations gives the Pell equation

                     x^2 - ky^2 = 1

For any given k we're looking for solutions x,y such that x+y is a 
square.  Of course, for any positive integer k there are infinitely
many solutions to the Pell equation, but solutions with x+y=square 
are rare.  For example, with k=8 the values of x+y for x,y satisfying
the Pell equation are

    /16+11sqrt(2)\           q       /16-11sqrt(2)\           q
   ( ------------ )(3+sqr(8))    +  ( ------------ )(3-sqr(8))
    \     8      /                   \      8     /

This gives the sequence 4, 23, 134, 781, 4552..., which satisfies
the second order recurrence s[j] = 6s[j-1] - s[j-2].  So the question
is whether this sequence contains any squares after the initial value
4.  Recall the proof that the only square Fibonacci numbers are 0, 1,
and 144 (see J. Cohn, "On square Fibonacci numbers", J. London Math
Soc., 39, 1964, p537-540.)

In general, the problem reduces to finding the square terms of a
general second-order recurring sequence, like the Fibonacci sequence.
The best approach might be to apply Cohn's method of proof to the
general second-order recurrence.