Extended Proth Theorem

[Cyclamen Persicum]

I am writing a Proth prime generator (yet another... ;-)
The Proth theorem says:

"Let n > 1, k < 2^n and N = k*2^n + 1 be a quadratic non-residue (mod a) for some odd prime a. Then the necessary and sufficient condition for N to be a prime is that:
a^((N-1)/2) = (N/a) = -1 (mod N) "

Why the hell a small witness "a" must be obligatory prime?
I believe that every odd number is suit. And I could not find
any quadratic non-residue which gives 1 instead -1 being powered.

[Cyclamen Persicum]

Sorry, I have noticed a pun...
JacobiSymbol = -1 does not mean "quadratic non-residue".
But I'm right, there is a super-extended Proth theorem:
for integer _a_ with Jacobi(a,N)=-1 (or for odd _a_ with Jacobi(N mod a,a)=-1) N=k*2^n+1 (k<2^n) to be prime if and only if a^[(N-1)/2]=-1 (mod N).

So I take a small odd _a_ rather then a small prime _a_ which Ylos Gallot takes in his Proth.exe