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How to prove that these conjectures are true?

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

Conjecture 1: Let $N= b^n-b-1$ such that $n>2$, $b \equiv 0,6 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv P_{(b+2)/2}(6) \pmod{N}$.

Conjecture 2: Let $N= b^n-b-1$ such that $n>2$, $b \equiv 2,4 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv -P_{b/2}(6) \pmod{N}$.

Conjecture 3: Let $N= b^n+b+1$ such that $n>2$, $b \equiv 0,6 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv P_{b/2}(6) \pmod{N}$.

Conjecture 4: Let $N= b^n+b+1$ such that $n>2$, $b \equiv 2,4 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv -P_{(b+2)/2}(6) \pmod{N}$.

Conjecture 5: Let $N= b^n-b+1$ such that $n>3$, $b \equiv 0,2 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv P_{b/2}(6) \pmod{N}$.

Conjecture 6: Let $N=b^n-b+1$ such that $n>3$, $b \equiv 4,6 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv -P_{(b-2)/2}(6) \pmod{N}$.

Conjecture 7: Let $N= b^n+b-1$ such that $n>3$, $b \equiv 0,2 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv P_{(b-2)/2}(6) \pmod{N}$.

Conjecture 8: Let $N= b^n+b-1$ such that $n>3$, $b \equiv 4,6 \pmod{8}$. Let $S_i=P_b(S_{i-1})$ with $S_0=P_{b/2}(6)$, thus if $N$ is prime, then $S_{n-1} \equiv -P_{b/2}(6) \pmod{N}$.

Any hint is appreciated.

user642796
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Pedja
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  • That's an interesting question, I have two notes in the formulation: I would prefer to include only one conjecture and see if people have any ideas , hints (if you have the proof for the first conjecture then you can prove others yourself by the same methods), the second observation would be a question: Why do you think these conjectures are true? – Elaqqad Aug 05 '15 at 15:51
  • @Elaqqad Because I couldn't find any counterexample . – Pedja Aug 05 '15 at 16:06
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    That's not enough there are other questions that I may ask: how did you start thinking about these conjecture how did you test them.. this information is very important for the readers, if you did not edit your question to include more information (just for the first conjecture, you can even remove the others) then very likely your question will not get attention, when I look at the question I see 8 conjectures, this is a lot and an answer will be very large, so I become unmotivated and I don't want to think about the problem because it will take me a lot of time ... – Elaqqad Aug 05 '15 at 18:07
  • I'm not surprised you couldn't find a counterexample - even with b=14, n=3, the numbers involved are approximately of the order 10^1000. – Patrick Stevens Aug 07 '15 at 15:42

1 Answers1

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Your eight conjectures are true.

First of all, $$\begin{align}S_0=P_{b/2}(6)&=2^{-\frac{b}{2}}\cdot\left(\left(6-4\sqrt{2}\right)^{\frac{b}{2}}+\left(6+4\sqrt{2}\right)^{\frac b2}\right)\\&=\left(3-2\sqrt 2\right)^{\frac b2}+\left(3+2\sqrt 2\right)^{\frac b2}\\&=\left(\sqrt 2-1\right)^b+\left(\sqrt 2+1\right)^b\\&=p^b+q^b\end{align}$$ where $p=\sqrt 2-1,q=\sqrt 2+1$ with $pq=1$.

Now, we can prove by induction that $$S_i=p^{b^{i+1}}+q^{b^{i+1}}.$$

By the way, $$\begin{align}p^{N+1}+q^{N+1}&=\sum_{i=0}^{N+1}\binom{N+1}{i}(\sqrt 2)^{i}\left((-1)^{N+1-i}+1\right)\\&=\sum_{j=0}^{(N+1)/2}\binom{N+1}{2j}2^{j+1}\\&\equiv 2+2^{(N+3)/2}\pmod N\\&\equiv 2+4\cdot 2^{\frac{N-1}{2}}\pmod N\tag1\end{align}$$ Also, $$\begin{align}p^{N+3}+q^{N+3}&=\sum_{i=0}^{N+3}\binom{N+3}{i}(\sqrt 2)^{i}\left((-1)^{N+3-i}+1\right)\\&=\sum_{j=0}^{(N+3)/2}\binom{N+3}{2j}2^{j+1}\\&\equiv 2+\binom{N+3}{2}\cdot 2^2+\binom{N+3}{N+1}\cdot 2^{\frac{N+3}{2}}+2^{\frac{N+5}{2}}\pmod N\\&\equiv 14+12\cdot 2^{\frac{N-1}{2}}+8\cdot 2^{\frac{N-1}{2}}\pmod N\tag2\end{align}$$

Here, for $N\equiv\pm 1\pmod 8$, since $2^{\frac{N-1}{2}}\equiv 1\pmod N$, from $(1)(2)$, we can prove by induction that $$p^{N+2i-1}+q^{N+2i-1}\equiv p^{2i}+q^{2i}\pmod N\tag 3$$

For $N\equiv 3,5\pmod 8$, since $2^{\frac{N-1}{2}}\equiv -1\pmod N$, from $(1)(2)$, we can prove by induction that $$p^{N+2i-1}+q^{N+2i-1}\equiv -\left(p^{2i-2}+q^{2i-2}\right)\pmod N\tag 4$$

To prove $(3)(4)$, we can use $$p^{N+2(i+1)-1}+q^{N+2(i+1)-1}\equiv \left(p^{N+2i-1}+q^{N+2i-1}\right)\left(p^2+q^2\right)-\left(p^{N+2(i-1)-1}+q^{N+2(i-1)-1}\right)\pmod N$$and $$p^{N+2(i-1)-1}+q^{N+2(i-1)-1}\equiv \left(p^{N+2i-1}+q^{N+2i-1}\right)\left(p^{-2}+q^{-2}\right)-\left(p^{N+2(i+1)-1}+q^{N+2(i+1)-1}\right)\pmod N$$

(Note that $(3)(4)$ holds for every integer $i$ (not necessarily positive) because of $pq=1$.)

Conjecture 1 is true because from $(3) $$$\begin{align}S_{n-1}&=p^{N+b+1}+q^{N+b+1}\\&\equiv p^{b+2}+q^{b+2}\pmod N\\&\equiv P_{(b+2)/2}(6)\pmod N\end{align}$$

Conjecture 2 is true because from $(4)$ $$\begin{align}S_{n-1}&=p^{N+b+1}+q^{N+b+1}\\&\equiv -\left(p^{b}+q^{b}\right)\pmod N\\&\equiv -P_{b/2}(6)\pmod N\end{align}$$

Conjecture 3 is true because from $(3) $$$\begin{align}S_{n-1}&=p^{N-b-1}+q^{N-b-1}\\&\equiv p^{-b}+q^{-b}\pmod N\\&\equiv q^b+p^b\pmod N\\&\equiv P_{b/2}(6)\pmod N\end{align}$$

Conjecture 4 is true because from $(4)$ $$\begin{align}S_{n-1}&=p^{N-b-1}+q^{N-b-1}\\&\equiv -\left(p^{-b-2}+q^{-b-2}\right)\pmod N\\&\equiv -\left(q^{b+2}+p^{b+2}\right)\pmod N\\&\equiv -P_{(b+2)/2}(6)\pmod N\end{align}$$

Conjecture 5 is true because from $(3) $$$\begin{align}S_{n-1}&=p^{N+b-1}+q^{N+b-1}\\&\equiv p^{b}+q^{b}\pmod N\\&\equiv P_{b/2}(6)\pmod N\end{align}$$

Conjecture 6 is true because from $(4)$ $$\begin{align}S_{n-1}&=p^{N+b-1}+q^{N+b-1}\\&\equiv -\left(p^{b-2}+q^{b-2}\right)\pmod N\\&\equiv -P_{(b-2)/2}(6)\pmod N\end{align}$$

Conjecture 7 is true because from $(3) $$$\begin{align}S_{n-1}&=p^{N-b+1}+q^{N-b+1}\\&\equiv p^{-b+2}+q^{-b+2}\pmod N\\&\equiv q^{b-2}+p^{b-2}\pmod N\\&\equiv P_{(b-2)/2}(6)\pmod N\end{align}$$

Conjecture 8 is true because from $(4)$ $$\begin{align}S_{n-1}&=p^{N-b+1}+q^{N-b+1}\\&\equiv -\left(p^{-b}+q^{-b}\right)\pmod N\\&\equiv -\left(q^b+p^b\right)\pmod N\\&\equiv -P_{b/2}(6)\pmod N\end{align}$$

mathlove
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  • Thank you for answering my questions . Maybe you should collect all these proofs (your answers on my questions) and publish it in Arxiv. – Pedja Sep 04 '15 at 15:17
  • @MathBot: You are welcome and thank you for your suggestion. I've really enjoyed solving your interesting questions. By the way, don't you ask if this conjecture is true? :) – mathlove Sep 04 '15 at 15:51