Only by adding the condition *0<x*x’<p^2* then the
following answer can be given.

If *p* is prime but not a Mersenne or Fermat prime
then there are exactly *(p-1)/2* cases (and one less for Mersenne
and Fermat primes) where *t*p+1*=*2^r*q= 1 (mod p), t=1,2...p-1*;
and we could say that* q*_{ }is the root-inverse of* 2^r,
r>0 *and* q *odd.

As *q* can be factored into *f*_{1}*f_{2}*..f_{n}
then* f*_{2}*...f_{n }is a derivative-inverse
of *f*_{1}*2^n, and to complete the answer the mirrors
need to be considered

Example, for
prime 17 there exist (17-1)/2-1=7 cases where the inverses of opposite
parity are listed below, with the derivatives and mirrors.

1- (2,9) and
derivative (6,3) and mirrors (3,6) (9,2)

2- (4,13) and
mirror (13,4)

3- (2,43);

4- (8,15) and
derivatives (5,24)(3,40), and mirror (15,8);

5- (2,77) and
derivatives (14,11),(4,22) and mirror(11,14);

6- (4,47) ;

7- {2,111) and
derivatives (6,37), (3,34);

For the requirement that both *0<x<p and 0<x’<p *as
the example of the problem implies, then the problem can only be
analyzed by computational filtering the above result.

The surprising result of such computational filtering to
the indicated parity and *p* prime is that for**:-
**

whatever parity** ***Np(all) = p-2* (exact
relationship)

opposite parity,* Np(op) ≈ p/2 *(see
graph)

even-even parity*,***
Np(ee) ≈ p/4**

odd-odd parity*, Np(oo) = Np(ee)-1 *(exact
relationship)

**
Questions: **

1. Can you explain the fundamental theory behind the
result?

2. How do you prove *Np(all) = p-2* and * Np(oo) =
Np(ee)-1 *?

2. Can you find counter examples?