Can you prove or disprove the following claim:
Let $N=k \cdot 2^n+1$ be a natural number that is not a perfect square such that $ 2 \nmid k$ , $n>2$ . Let $c$ be the smallest odd prime number such that $\left(\frac{c}{N}\right)=-1$ , where $\left(\frac{}{}\right)$ denotes a Jacobi symbol . Let $z$ be a real number of the form $a+b\sqrt{c}$ equal to the modular $(1-\sqrt{c})^{(N-1)/2} \operatorname{mod} N$ , then $N$ is prime iff $a=b$ .
You can run this test here. I have verified this claim for all $k \in [1,1000]$ with $n \in [3,1000]$ .
This fails for $N=22577=1411\cdot 2^4+1=107\cdot 211$. The value of $c$ that is taken is $3$, but $$(1-\sqrt3)^{11288}\equiv 11502+11502\sqrt 3\bmod 22577.$$ It is true, however, if $N$ is prime. Setting $x=\sqrt c$, we have that $x^N$ is the Galois conjugate of $x$, so $x^N=-x$, whence $$(1+x)^{N+1}+(1+x)^N(1+x)=(1+x)(1+x^N)=(1+x)(1-x)=1-x^2=1-c;$$ this is a product of $-1$ and primes strictly less than $c$ (including $2$) and thus a quadratic residue (using that $c\equiv 1\bmod 8$), so $$(1+x)^{\frac{N+1}{2}}\in\mathbb F_N;$$ this implies that $$(1+x)^{\frac{N+1}{2}}=(1-x)^{\frac{N+1}{2}},$$ which gives that, if $(1-x)^{\frac{N-1}{2}}=a+bx$ with $a,b\in\mathbb F_N$, $$(a-bx)(1+x)=(a+bx)(1-x)\implies x(a-b)=x(b-a)\implies a=b.$$