Conjectured compositeness tests for $N=k \cdot 2^n \pm c$ How to prove that these conjectures are true ?

Definition : $\text{Let}~ P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)~ , \text{where}~ m ~\text{and}~ x ~\text{are nonnegative integers} .$

Conjecture 1 : $ \text{Let} ~N=k\cdot 2^n-c ~\text{such that}~ n&gt;2c ,  k&gt;0 , c&gt;0 ~\text{and}~ c\equiv 3,5 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv -P_{\lfloor c/2 \rfloor}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n-c;
my(s=Mod(2*polchebyshev(k,1,a/2),N)); 
for(i=1,n-1, s=s^2-2); 
if(!(s==N-2*polchebyshev(floor(c/2),1,a/2)) &amp;&amp; isprime(N),print(n)))
}


Conjecture 2 : $ \text{Let} ~N=k\cdot 2^n-c ~\text{such that}~ n&gt;2c ,  k&gt;0 , c&gt;0 ~\text{and}~ c\equiv 1,7 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv P_{\lceil c/2 \rceil}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n-c;
my(s=Mod(2*polchebyshev(k,1,a/2),N)); 
for(i=1,n-1, s=s^2-2); 
if(!(s==2*polchebyshev(ceil(c/2),1,a/2)) &amp;&amp; isprime(N),print(n)))
}


Conjecture 3 : $ \text{Let} ~N=k\cdot 2^n+c ~\text{such that}~ n&gt;2c ,  k&gt;0 , c&gt;0 ~\text{and}~ c\equiv 3,5 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv -P_{\lceil c/2 \rceil}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n+c;
my(s=Mod(2*polchebyshev(k,1,a/2),N)); 
for(i=1,n-1, s=s^2-2); 
if(!(s==N-2*polchebyshev(ceil(c/2),1,a/2)) &amp;&amp; isprime(N),print(n)))
}


Conjecture 4 : $ \text{Let} ~N=k\cdot 2^n+c ~\text{such that}~ n&gt;2c ,  k&gt;0 , c&gt;0 ~\text{and}~ c\equiv 1,7 \pmod{8}$

$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6) , ~\text{thus} $$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv P_{\lfloor c/2 \rfloor}(6) \pmod{N}$$

Searching for counterexample (Pari GP) :

CEk2c(k,c,g)=
{
for(n=2*c+1,g,a=6;
N=k*2^n+c;
my(s=Mod(2*polchebyshev(k,1,a/2),N)); 
for(i=1,n-1, s=s^2-2); 
if(!(s==2*polchebyshev(floor(c/2),1,a/2)) &amp;&amp; isprime(N),print(n)))
}


Any hint is appreciated .
First thing I would do is simplify $P_2(x)$
$$P_2(x)=\dfrac{(x-\sqrt{x^2-4})^2+(x+\sqrt{x^2-4})^2}{4}=\dfrac{x^2-2x\sqrt{x^2-4}+x^2-4+x^2+2\sqrt{x^2-4}+x^2-4}{4}=\dfrac{4x^2-8}{4}=x^2-2$$

How did you come up with these conjectures? And why do you think they would work? Because,if we know that, it's way easier to prove them.
This is not an answer, but this would not enter a comment.
Here is a PARI/gp program that implements a unic formula that manages all 4 conjectures, showing that they are related and can be generalized into only one, showing that it looks to be a deep property.
I've found no counter-example with the two verification loops provided after.
You have found something deep, I think. However, it needs a proof now ! ;)

CEk2c(k,c,g)=
{
a=6;
if(c&gt;0,s=1,s=-1;c=-c);
for(n=2*c+1,g,
N=k*2^n+s*c;
e=c%4;if(e!=1,e=-1);
d=((c-e)%8)/4;
A=(-1)^d;
B=(c-A*s)/2;
s0=Mod(2*polchebyshev(k,1,a/2),N);
sn=Mod(A*2*polchebyshev(B,1,a/2),N);
my(s=s0);
for(i=1,n-1,s=Mod(s^2-2,N));
if(s!=sn &amp;&amp; isprime(N),print("k: ",k," c: ",c," n: ",n))
)
}

for(k=1,100,for(c=0,50,CEk2c(k,2*c+1,100)))
for(k=1,100,for(c=0,50,CEk2c(k,-(2*c+1),100)))


New conjecture summarizing the 4 conjectures:

$$\text{Let } ~N=k\cdot 2^n+ s \cdot c ~\text{ such that: }$$

$$ n&gt;2c ,  k&gt;0 , c&gt;0 ~\text{ and }~ s = \pm 1 ~\text{ and }~ c \equiv 4 \cdot d ~ \pm 1 \pmod 8 ~\text{ and }~ d=0,1$$

$$\text{Let }~ S_i \equiv P_2(S_{i-1}) \pmod{N} ~ \text{ with } ~ S_0=P_k(6) ~ \text{ , thus: }$$

$$\text{If } ~ N ~\text{ is prime then }~ S_{n-1} \equiv {(-1)}^d \cdot P_{(c - {(-1)^d \cdot s})/2} ~ (6) \pmod{N}$$