hleleu

Rewriting your question in a more readable form with some comments:

>>>>>

Define a 2^n-residue modulo m as follows:

An integer a is a 2^n-residue modulo m if and only if it exists an

integer b such as b^(2^n)=a mod m

Obviously if n=1 then we define the quadractic residues.

[not a new definition]

Now, consider the numbers of the form n=k*2^n+1, n and k are

unconstrained. Since Proth number imposes k<2^n, we call these

numbers generalized Proth numbers.

[better be named an arbitrary odd number]

With these definitions we have the following result:

p=k*2^n+1 is prime if and only if there exist exactly k [that is gcd

(p-1,2^n)] integers

comprised between 1 and p-1 who are 2^n-residue modulo p.

<<<<<

> I'm not aware if this result is already known.

It is a special case of a very well known theorem named Euler's

Criterion for solvability of the equation x^n=a(mod p). This

equation is solvable iff a^((p-1)/d)=1(mod p) where d=gcd(p-1,n). If

the equation is solvable then it has exactly d solutions.

Payam