Can you provide a proof or a counterexample to the following claim :
Let $p$ be a prime greater than three , let $R_p(+3)=\frac{3^p-1}{2}$ and $R_p(-3)=\frac{3^p+1}{4}$ . Let $S_i=S_{i-1}^3+3S_{i-1}$ with $S_0=36$ , then $R_p(\pm 3)$ is prime iff $S_{p-1} \equiv 36 \pmod{R_p(\pm 3)}$ .
You can run this test here .
I have tested this claim up to $10000$ .
This is a partial answer.
This answer proves that if $R_p(\pm 3)$ is prime, then $S_{p-1} \equiv 36 \pmod{R_p(\pm 3)}$.
Let us prove first by induction that $$S_i=a^{3^{i+1}}-b^{3^{i+1}}\tag1$$ where $a=\frac{\sqrt{13}+3}{2},b=\frac{\sqrt{13}-3}{2}$ with $ab=1$.
$(1)$ holds for $i=0$ since $a^3-b^3=(a-b)((a+b)^2-1)=3(13-1)=36=S_0$.
Suppose that $(1)$ holds for some $i$.
Then, $$\begin{align}S_{i+1}&=S_{i}^3+3S_{i}=(a^{3^{i+1}}-b^{3^{i+1}})^3+3(a^{3^{i+1}}-b^{3^{i+1}})\\\\&=3^{3^{i+2}}-b^{3^{i+2}}+3a^{3^{i+1}}b^{3^{i+1}}(-a^{3^{i+1}}+b^{3^{i+1}})+3(a^{3^{i+1}}-b^{3^{i+1}})\\\\&=3^{3^{i+2}}-b^{3^{i+2}}+3(-a^{3^{i+1}}+b^{3^{i+1}})+3(a^{3^{i+1}}-b^{3^{i+1}})\\\\&=3^{3^{i+2}}-b^{3^{i+2}}\qquad\blacksquare\end{align}$$
Let $N_1:=R_p(+3)=\frac{3^p-1}{2}$ where $N_1$ is odd since $$3^{2k+1}-1\equiv (-1)^{2k+1}-1\equiv 2\pmod 4$$
Then, from $(1)$, $$\begin{align}2^{N_1}\cdot S_{p-1}&=2^{N_1}(a^{3^{p}}-b^{3^{p}})=2^{N_1}(a^{2N_1+1}-b^{2N_1+1})=a(2a^2)^{N_1}-b(2b^2)^{N_1}\\\\&=\frac{\sqrt{13}+3}{2}\left(11+3\sqrt{13}\right)^{N_1}-\frac{\sqrt{13}-3}{2}\left(11-3\sqrt{13}\right)^{N_1}\\\\&=\frac{\sqrt{13}}{2}(\left(11+3\sqrt{13}\right)^{N_1}-\left(11-3\sqrt{13}\right)^{N_1})\\\\&\qquad\qquad +\frac 32(\left(11+3\sqrt{13}\right)^{N_1}+\left(11-3\sqrt{13}\right)^{N_1})\end{align}$$
By the binomial theorem, $$\begin{align}2^{N_1}\cdot S_{p-1}&=\frac{\sqrt{13}}{2}\ \sum_{i=0}^{N_1}\binom{N_1}{i}11^{N-i}\cdot ((3\sqrt{13})^i-(-3\sqrt{13})^i)\\\\&\qquad\qquad +\frac 32\sum_{i=0}^{N_1}\binom{N_1}{i}11^{N-i}\cdot ((3\sqrt{13})^i+(-3\sqrt{13})^i)\\\\&=\frac{\sqrt{13}}{2}\ \sum_{j=1}^{(N_1+1)/2}\binom{N_1}{2j-1}11^{N_1-(2j-1)}\cdot 2(3\sqrt{13})^{2j-1}\\\\&\qquad\qquad +\frac 32\sum_{j=0}^{(N_1-1)/2}\binom{N_1}{2j}11^{N_1-2j}\cdot 2(3\sqrt{13})^{2j}\\\\&=\sum_{j=1}^{(N_1+1)/2}\binom{N_1}{2j-1}11^{N_1-(2j-1)}\cdot 3^{2j-1}\cdot 13^{j}\\\\&\qquad\qquad +\sum_{j=0}^{(N_1-1)/2}\binom{N_1}{2j}11^{N_1-2j}\cdot 3^{2j+1}\cdot 13^j\end{align}$$
Since $\binom{N_1}{j}\equiv 0\pmod{N_1}$ for $1\le j\le N_1-1$, we have $$2^{N_1}\cdot S_{p-1}\equiv 3^{N_1}\cdot 13^{(N_1+1)/2}+11^{N_1}\cdot 3$$
Since $k^{N_1-1}\equiv 1\pmod{N_1}$ for $k$ such that $\gcd(k,N_1)=1$ by Fermat's little theorem, we get $$2S_{p-1}\equiv 3\cdot 13\cdot \frac{(-1)^{\frac{13-1}{2}\cdot\frac{{N_1}-1}{2}}}{\left(\frac{N_1}{13}\right)}+11\cdot 3\equiv \frac{39}{\left(\frac{N_1}{13}\right)}+33\equiv \frac{39}{1}+33\equiv 72\pmod{N_1}\tag2$$ where $\left(\frac{q}{p}\right)$ denotes the Legendre symbol and $\left(\frac{N_1}{13}\right)\equiv 1\pmod{N_1}$ since $$\begin{align}3^{6k+1}-1&\equiv 3\cdot 27^{2k}-1\equiv 3\cdot 1-1\equiv 2\pmod{13}\implies\frac{3^{6k+1}-1}{2}\equiv 1\pmod{13}\\3^{6k+5}-1&\equiv 9\cdot 27^{2k+1}-1\equiv 9\cdot 1-1\equiv 8\pmod{13}\implies\frac{3^{6k+5}-1}{2}\equiv 4\pmod{13}\end{align}$$ It follows from $(2)$ and $\gcd(N_1,2)=1$ that $$S_{p-1}\equiv 36\pmod{R_p(+3)}$$
Let $N_2:=R_p(-3)=\frac{3^p+1}{4}$ where $N_2$ is odd since $$3^{2k+1}+1\equiv 3\cdot 9^k+1\equiv 3\cdot 1+1\equiv 4\pmod 8$$
Then, from $(1)$, $$\begin{align}2^{N_2}\cdot S_{p-1}&=2^{N_2}(a^{3^{p}}-b^{3^{p}})=2^{N_2}(a^{4N_2-1}-b^{4N_2-1})=2^{N_2}\cdot ab\cdot (a^{4N_2-1}-b^{4N_2-1})\\\\&=2^{N_2}(ba^{4N_2}-ab^{4N_2})=b(2a^4)^{N_2}-a(2b^4)^{N_2}\\\\&=\frac{\sqrt{13}-3}{2}\cdot \left(119+33\sqrt{13}\right)^{N_2}-\frac{\sqrt{13}+3}{2}\cdot \left(119-33\sqrt{13}\right)^{N_2}\\\\&=\frac{\sqrt{13}}{2}(\left(119+33\sqrt{13}\right)^{N_2}-\left(119-33\sqrt{13}\right)^{N_2})\\\\&\qquad\qquad -\frac 32(\left(119+33\sqrt{13}\right)^{N_2}+\left(119-33\sqrt{13}\right)^{N_2})\end{align}$$
By the binomial theorem, $$\begin{align}2^{N_2}\cdot S_{p-1}&=\frac{\sqrt{13}}{2}\sum_{i=0}^{N_2}\binom{N_2}{i}119^{N_2-i}((33\sqrt{13})^i-(-33\sqrt{13})^i)\\\\&\qquad\qquad -\frac 32\sum_{i=0}^{N_2}\binom{N_2}{i}119^{N_2-i}((33\sqrt{13})^i+(-33\sqrt{13})^i)\\\\&=\frac{\sqrt{13}}{2}\sum_{j=1}^{(N_2+1)/2}\binom{N_2}{2j-1}119^{N_2-(2j-1)}\cdot 2(33\sqrt{13})^{2j-1}\\\\&\qquad\qquad -\frac 32\sum_{j=0}^{(N_2-1)/2}\binom{N_2}{2j}119^{N_2-2j}\cdot 2(33\sqrt{13})^{2j}\\\\&=\sum_{j=1}^{(N_2+1)/2}\binom{N_2}{2j-1}119^{N_2-(2j-1)}\cdot 33^{2j-1}\cdot 13^j\\\\&\qquad\qquad -\sum_{j=0}^{(N_2-1)/2}\binom{N_2}{2j}119^{N_2-2j}\cdot 3^{2j+1}\cdot 11^{2j}\cdot 13^j\end{align}$$
Since $\binom{N_2}{j}\equiv 0\pmod{N_2}$ for $1\le j\le N_2-1$, we have $$2^{N_2}\cdot S_{p-1}\equiv 3^{N_2}\cdot 11^{N_2}\cdot 13^{(N_2+1)/2}-7^{N_2}\cdot 17^{N_2}\cdot 3\pmod{N_2}$$
By Fermat's little theorem, we get $$2S_{p-1}\equiv 3\cdot 11\cdot 13\cdot \frac{(-1)^{\frac{13-1}{2}\cdot\frac{N_2-1}{2}}}{\left(\frac{N_2}{13}\right)}-7\cdot 17\cdot 3\equiv 429\cdot\frac{1}{1}-357\equiv 72\pmod{N_2}\tag3$$ where $\left(\frac{N_2}{13}\right)\equiv 1\pmod{N_2}$ since $$\begin{align}3^{6k+1}+1&\equiv 3\cdot 27^{2k}+1\equiv 3\cdot 1+1\equiv 4\pmod{13}\implies\frac{3^{6k+1}+1}{4}\equiv 1\pmod{13}\\3^{6k+5}+1&\equiv 9\cdot 27^{2k+1}+1\equiv 9\cdot 1+1\equiv 10\pmod{13}\implies \frac{3^{6k+5}+1}{4}\equiv 9\pmod{13}\end{align}$$ It follows from $(3)$ and $\gcd(N_2,2)=1$ that $$S_{p-1}\equiv 36\pmod{R_p(-3)}$$