How to prove that following conjecture is true ?
Def 1.
Let $R_p(3)=\frac{3^p-1}{2}$ such that $p \in \mathbb{P}$ and $p > 2$ .
Def 2.
Let $S_i=S_{i-1}^3-3S_{i-1}$ with $S_0=52$ , then
Conjecture
If $R_p(3)$ is prime then $S_{p-1} \equiv 52 \pmod{R_p(3)}$
You can run this test here .
For $p=3$, since $R_3(3)=13$, we get $$S_{2}=R_3(3)\times 213127961508000+52\equiv 52\pmod{R_3(3)}$$
In the following, we consider the case for $p\gt 3$.
First of all, let us prove by induction that $$S_i=a^{3^{i+1}}+b^{3^{i+1}}\tag1$$ where $a=2-\sqrt 3,b=2+\sqrt 3$ with $ab=1$.
$$S_0=a^3+b^3=(2-\sqrt 3)^3+(2+\sqrt 3)^3=(8+18)\times 2=52$$
Supposing that $(1)$ holds for some $i\ (\ge 0)$ gives $$\begin{align}S_{i+1}&=S_{i}^3-3S_i\\\\&=(a^{3^{i+1}}+b^{3^{i+1}})^3-3(a^{3^{i+1}}+b^{3^{i+1}})\\\\&=a^{3^{i+2}}+3a^{3^{i+1}}+3b^{3^{i+1}}+b^{3^{i+2}}-3a^{3^{i+1}}-3b^{3^{i+1}}\\\\&=a^{3^{i+2}}+b^{3^{i+2}}\qquad\blacksquare\end{align}$$
Now letting $N:=R_p(3)=\frac 12(3^p-1)$, we get, from $(1)$, $$\begin{align}S_{p-1}&=a^{3^p}+b^{3^p}\\\\&=a^{2N+1}+b^{2N+1}\\\\&=a(a^2)^N+b(b^2)^N\\\\&=(2-\sqrt 3)(7-4\sqrt 3)^N+(2+\sqrt 3)(7+4\sqrt 3)^N\\\\&=2((7-4\sqrt 3)^N+(7+4\sqrt 3)^N)+\sqrt 3\ ((7+4\sqrt 3)^N-(7-4\sqrt 3)^N)\end{align}$$ Using the binomial theorem, $$\small S_{p-1}=2\left(\sum_{i=0}^{N}\binom Ni7^i((-4\sqrt 3)^{N-i}+(4\sqrt 3)^{N-i})\right)+\sqrt 3\ \left(\sum_{i=0}^{N}\binom Ni7^i((4\sqrt 3)^{N-i}-(-4\sqrt 3)^{N-i})\right)$$ Since $N$ is odd, $$\small \begin{align}S_{p-1}&=2\left(\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}7^{2j-1}\cdot 2(4\sqrt 3)^{N-(2j-1)}\right)+\sqrt 3\ \left(\sum_{j=0}^{(N-1)/2}\binom{N}{2j}7^{2j}\cdot 2(4\sqrt 3)^{N-2j}\right)\\\\&=4\left(\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}7^{2j-1}\cdot 4^{N-2j+1}\cdot 3^{(N-2j+1)/2}\right)+6\left(\sum_{j=0}^{(N-1)/2}\binom{N}{2j}7^{2j}\cdot 4^{N-2j}\cdot 3^{(N-2j-1)/2}\right)\end{align}$$ Since $\binom Nr$ is divisible by the prime $N$ for $1\le r\le N-1$, $$S_{p-1}\equiv 4\cdot 7^{N}+6\cdot 4^N\cdot 3^{(N-1)/2}\pmod N\tag2$$
Now we have
$\frac 32(N-1)=\frac 34(3^p-3)$ is even since $3^p-3$ is divisible by $8$
Every prime number larger than $3$ is either of the form $6k+1$ or of the form $6k+5$
$\frac 12(3^{6k+1}-1)\equiv 1\pmod 7$ since $3^{6k+1}-1\equiv 3\cdot (3^6)^k-1\equiv 3\cdot 1-1\equiv 2\pmod 7$
$\frac 12(3^{6k+5}-1)\equiv 2\pmod 7$ since $3^{6k+5}-1\equiv 3^5\cdot (3^6)^k-1\equiv 3^5\cdot 1-1\equiv 4\pmod 7$
So, we get $$7^{(N-1)/2}\equiv \left(\frac 7N\right)=\frac{(-1)^{\frac{7-1}{2}\cdot\frac{N-1}{2}}}{\left(\frac N7\right)}=\frac{(-1)^{\frac{3(N-1)}{2}}}{\left(\frac N7\right)}=\frac 11=1\pmod N$$ from which we have $$7^N\equiv 7\pmod N\tag3$$ Next, $$4^N=4\cdot(2^{(N-1)/2})^4\equiv 4\cdot ((-1)^{(N^2-1)/8})^4=4\cdot (-1)^{(N^2-1)/2}=4\cdot 1=4\pmod N\tag4$$
Also, we have
$\frac{N-1}{2}$ is even since $3^{2k+1}-1=3\cdot 9^k-1\equiv 3\cdot 1-1=2\pmod 8$
$N=\frac 12(3^p-1)=3^{p-1}+3^{p-2}+\cdots +3+1\equiv 1\pmod 3$
So, we get $$3^{(N-1)/2}\equiv \left(\frac 3N\right)=\frac{(-1)^{\frac{3-1}{2}\cdot\frac{N-1}{2}}}{\left(\frac N3\right)}=\frac{(-1)^{(N-1)/2}}{\left(\frac N3\right)}=\frac 11=1\pmod N\tag5$$
It follows from $(2)(3)(4)(5)$ that $$S_{p-1}\equiv 4\cdot 7^{N}+6\cdot 4^N\cdot 3^{(N-1)/2}\equiv 4\cdot 7+6\cdot 4\cdot 1\equiv 52\pmod N$$ as conjectured.