TITLE: Conjectured compositeness tests for $N=k\cdot 2^n \pm 1$ and $N=k\cdot 2^n \pm 3$
QUESTION [5 upvotes]: How to prove these conjectures ?
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 : $\text{Let} ~N=k\cdot 2^n-1 ~\text{such that}~ n>2 ~\text{and}~ k>0$ .
$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6)~ , \text{thus}$$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv 6 \pmod{N}$$
Conjecture 2 : $\text{Let} ~N=k\cdot 2^n+1 ~\text{such that}~ n>2 ~\text{and}~ k>0$ .
$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6)~ , \text{thus}$$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv 2 \pmod{N}$$
Conjecture 3 : $\text{Let} ~N=k\cdot 2^n-3 ~\text{such that}~ n>6 ~\text{and}~ k>0$ .
$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6)~ , \text{thus}$$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv -6 \pmod{N}$$
Conjecture 4 : $\text{Let} ~N=k\cdot 2^n+3 ~\text{such that}~ n>6 ~\text{and}~ k>0$ .
$$\text{Let}~ S_i=P_2(S_{i-1})~ \text{with}~ S_0=P_k(6)~ , \text{thus}$$
$$\text{If}~ N ~\text{is prime then}~ S_{n-1} \equiv -34 \pmod{N}$$
Any hint would be welcomed . Thanks .

REPLY [6 votes]: Your four conjectures are true.
First of all, we have $$\begin{align}S_0&=P_k(6)\\&=2^{-k}((6-4\sqrt 2)^k+(6+4\sqrt 2)^k)\\&=(3-2\sqrt 2)^k+(3+2\sqrt 2)^k\\&=(\sqrt 2-1)^{2k}+(\sqrt 2+1)^{2k}\end{align}$$
Now since $P_2(x)=2^{-2}(x^2+x^2-4+x^2+x^2-4)=x^2-2$, we have
$$S_1=P_2(S_0)=(\sqrt 2-1)^{4k}+(\sqrt 2+1)^{4k}$$
So, it follows that 
$$S_i=(\sqrt 2-1)^{k\cdot 2^{i+1}}+(\sqrt 2+1)^{k\cdot 2^{i+1}}$$
(if you want to prove this strictly, you can use induction.)
Hence, we have
$$S_{n-1}=(\sqrt 2-1)^{k\cdot 2^{n}}+(\sqrt 2+1)^{k\cdot 2^{n}}$$
Conjecture 1 is true because we have 
$$\begin{align}S_{n-1}&=(\sqrt 2-1)^{k\cdot 2^{n}}+(\sqrt 2+1)^{k\cdot 2^{n}}\\&=(\sqrt 2-1)^{N+1}+(\sqrt 2+1)^{N+1}\\&=\sum_{i=0}^{N+1}\binom{N+1}{i}(\sqrt 2)^i((-1)^{N+1-i}+1^{N+1-i})\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}2^{j+1}\\&=\binom{N+1}{0}2^{1}+\binom{N+1}{2}2^{2}+\cdots+\binom{N+1}{N-1}2^{\frac{N-1}{2}+1}+\binom{N+1}{N+1}2^{\frac{N+1}{2}+1}\\&\equiv \binom{N+1}{0}2^{1}+\binom{N+1}{N+1}2^{\frac{N+1}{2}+1}\pmod N\\&\equiv 2+2\times 2\times 2^{\frac{N-1}{2}}\pmod N\\&\equiv 6\pmod N\end{align}$$
(This is because $2^{(N-1)/2}\equiv 1\pmod N$ for $N\equiv -1\pmod 8$.)
Conjecture 2 is true because we have 
$$\begin{align}S_{n-1}&=(\sqrt 2-1)^{k\cdot 2^{n}}+(\sqrt 2+1)^{k\cdot 2^{n}}\\&=(\sqrt 2-1)^{N-1}+(\sqrt 2+1)^{N-1}\\&=(\sqrt 2-1)(\sqrt 2+1)\left((\sqrt 2-1)^{N-1}+(\sqrt 2+1)^{N-1}\right)\\&=(\sqrt 2+1)(\sqrt 2-1)^{N}+(\sqrt 2-1)(\sqrt 2+1)^N\\&=\sqrt 2\left((\sqrt 2-1)^N+(\sqrt 2+1)^N\right)+\left((\sqrt 2-1)^N-(\sqrt 2+1)^N\right)\\&=\sqrt 2\sum_{i=0}^{N}\binom{N}{i}(\sqrt 2)^i((-1)^{N-i}+1^{N-i})+\sum_{i=0}^{N}\binom{N}{i}(\sqrt 2)^i((-1)^{N-i}-1^{N-i})\\&=\sqrt 2\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(\sqrt 2)^{2j-1}\cdot 2+\sum_{j=0}^{(N-1)/2}\binom{N}{2j}2^{j}\cdot(-2)\\&=\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}2^{j+1}-\sum_{j=0}^{(N-1)/2}\binom{N}{2j}2^{j+1}\\&\equiv 2^{\frac{N+1}{2}+1}-2\pmod N\\&\equiv 2\cdot 2\cdot 1-2\pmod N\\&\equiv 2\pmod N\end{align}$$
(This is because $2^{(N-1)/2}\equiv 1\pmod N$ for $N\equiv 1\pmod 8$.)
Conjecture 3 is true because we have 
$$\begin{align}S_{n-1}&=(\sqrt 2-1)^{k\cdot 2^{n}}+(\sqrt 2+1)^{k\cdot 2^{n}}\\&=(\sqrt 2-1)^{N+3}+(\sqrt 2+1)^{N+3}\\&=\sum_{i=0}^{N+3}\binom{N+3}{i}(\sqrt 2)^i((-1)^{N+3-i}+1^{N+3-i})\\&=\sum_{j=0}^{(N+3)/2}\binom{N+3}{2j}2^{j+1}\\&\equiv 2+2^2\binom{N+3}{2}+2^{\frac{N+1}{2}+1}\binom{N+3}{N+1}+2^{\frac{N+3}{2}+1}\pmod N\\&\equiv 2+2\times 3\times 2+(-1)\times 2\times 3\times 2+(-1)\times 2^3\pmod N\\&\equiv -6\pmod N\end{align}$$
(This is because $2^{(N-1)/2}\equiv -1\pmod N$ for $N\equiv -3\pmod 8$.)
Conjecture 4 is true because we have
$$\begin{align}S_{n-1}&=(\sqrt 2-1)^{k\cdot 2^{n}}+(\sqrt 2+1)^{k\cdot 2^{n}}\\&=(\sqrt 2-1)^{N-3}+(\sqrt 2+1)^{N-3}\\&=(\sqrt 2-1)^3(\sqrt 2+1)^3\left((\sqrt 2-1)^{N-3}+(\sqrt 2+1)^{N-3}\right)\\&=(7+5\sqrt 2)(\sqrt 2-1)^{N}+(-7+5\sqrt 2)(\sqrt 2+1)^N\\&=5\sqrt 2\left((\sqrt 2-1)^N+(\sqrt 2+1)^N\right)+7\left((\sqrt 2-1)^N-(\sqrt 2+1)^N\right)\\&=5\sqrt 2\sum_{i=0}^{N}\binom{N}{i}(\sqrt 2)^i((-1)^{N-i}+1^{N-i})+7\sum_{i=0}^{N}\binom{N}{i}(\sqrt 2)^i((-1)^{N-i}-1^{N-i})\\&=5\sqrt 2\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(\sqrt 2)^{2j-1}\cdot 2+7\sum_{j=0}^{(N-1)/2}\binom{N}{2j}2^{j}\cdot(-2)\\&=5\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}2^{j+1}-7\sum_{j=0}^{(N-1)/2}\binom{N}{2j}2^{j+1}\\&\equiv 5\cdot 2^{\frac{N+1}{2}+1}-7\cdot 2\pmod N\\&\equiv 5\cdot 2\cdot 2\cdot (-1)-7\cdot 2\pmod N\\&\equiv -34\pmod N\end{align}$$
(This is because $2^{(N-1)/2}\equiv -1\pmod N$ for $N\equiv 3\pmod 8$.)