Can you provide a proof or a counterexample for the following claim :
Let $S_{i}=S_{i-1}^3-3S_{i-1}$ with $S_0=52$ . Let $p$ be an odd prime and let $R_p(-3)=\frac{3^p+1}{4}$ . Suppose $R_p(-3)$ is prime , then :
$\bullet$ $S_{p-1} \equiv 52 \pmod{R_p(-3)}$ if $p \equiv 1 \pmod 4$
$\bullet$ $S_{p-1} \equiv 724 \pmod{R_p(-3)}$ if $p \equiv 3 \pmod 4$
You can run this test here .
I have tested this claim for all $p$ up to $152287$ .
The claim is true.
For $p=3$, since $R_3(-3)=7$, we get $$S_2=R_3(-3)\times 395809071371904+724\equiv 724\pmod{R_3(-3)}$$
In the following, $p$ is prime greater than $3$.
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 $ab=1$.
For $i=0$, $(1)$ holds since $$a^3+b^3=(a+b)((a+b)^2-3)=4\times 13=52=S_0$$
Supposing that $(1)$ holds for some $i$ gives $$\begin{align}S_{i+1}&=S_{i}^3-3S_i=(a^{3^{i+1}}+b^{3^{i+1}})^3-3(a^{3^{i+1}}+b^{3^{i+1}})\\\\&=a^{3^{i+2}}+b^{3^{i+2}}+3(ab)^{3^{i+1}}(a^{3^{i+1}}+b^{3^{i+1}})-3(a^{3^{i+1}}+b^{3^{i+1}})\\\\&=a^{3^{i+2}}+b^{3^{i+2}}+3(a^{3^{i+1}}+b^{3^{i+1}})-3(a^{3^{i+1}}+b^{3^{i+1}})\\\\&=a^{3^{i+2}}+b^{3^{i+2}}\qquad\blacksquare\end{align}$$
Now letting $N:=R_p(-3)=\frac{3^p+1}{4}$ where $N$ is odd, we get, from $(1)$, $$\begin{align}S_{p-1}&=a^{3^{p}}+b^{3^{p}}=a^{4N-1}+b^{4N-1}=ab(a^{4N-1}+b^{4N-1})=b(a^4)^{N}+a(b^4)^N\\\\&=(2+\sqrt 3)(97-56\sqrt 3)^N+(2-\sqrt 3)(97+56\sqrt 3)^N\\\\&=2((97-56\sqrt 3)^N+(97+56\sqrt 3)^N))-\sqrt 3\ ((97+56\sqrt 3)^N-(97-56\sqrt 3)^N)\end{align}$$
By the binomial theorem, $$\begin{align}S_{p-1}&=2\sum_{i=0}^{N}\binom{N}{i}\cdot 97^{N-i}\cdot ((-56\sqrt 3)^i+(56\sqrt 3)^i)\\\\&\qquad\quad -\sqrt 3\ \sum_{i=0}^{N}\binom Ni\cdot 97^{N-i}\cdot ((56\sqrt 3)^i-(-56\sqrt 3)^i)\\\\&=2\sum_{j=0}^{(N-1)/2}\binom{N}{2j}\cdot 97^{N-2j}\cdot 2(56\sqrt 3)^{2j}\\\\&\qquad\quad -\sqrt 3\ \sum_{j=1}^{(N+1)/2}\binom N{2j-1}\cdot 97^{N-(2j-1)}\cdot 2(56\sqrt 3)^{2j-1}\\\\&=4\sum_{j=0}^{(N-1)/2}\binom{N}{2j}\cdot 97^{N-2j}\cdot 56^{2j}\cdot 3^j\\\\&\qquad\quad -2\sum_{j=1}^{(N+1)/2}\binom N{2j-1}\cdot 97^{N-(2j-1)}\cdot 56^{2j-1}\cdot 3^j\end{align}$$
Since $\binom Nj\equiv 0\pmod N$ for $1\le j\le N-1$, we get, by Fermat's little theorem, $$\begin{align}S_{p-1}&\equiv 4\cdot 97^{N}-2\cdot 7^{N}\cdot 8^{N}\cdot 3^{(N+1)/2}\pmod N\\\\&\equiv 4\cdot 97-2\cdot 7\cdot 8\cdot 3\cdot \left(\frac 3N\right)\pmod N\\\\&\equiv 388-336\cdot \frac{(-1)^{\frac{N-1}{2}}}{\left(\frac N3\right)}\pmod N\tag3\end{align}$$ where $\left(\frac{q}{p}\right)$ denotes the Legendre symbol.
Here, let us separate it into cases :
If $p\equiv 1\pmod 4$, then, writing $p=4k+1$, since we get$$\small\begin{align}N&=\frac{3^{4k+1}+1}{4}=3^{4k}-3^{4k-1}+\cdots -3+1\equiv 1+1+\cdots +1+1\equiv 4k+1\equiv 1\pmod 4\\\\N&=\frac{3^{4k+1}+1}{4}=3^{4k}-3^{4k-1}+\cdots -3+1\equiv 1\pmod 3\end{align}$$we have, from $(3)$, $$S_{p-1}\equiv 388-336\cdot \frac{(-1)^{\frac{N-1}{2}}}{\left(\frac N3\right)}\equiv 388-336\cdot \frac{1}{1}\equiv 52\pmod{R_p(-3)}$$
If $p\equiv 3\pmod 4$, then, writing $p=4k+3$, since we get$$\small\begin{align}N&=\frac{3^{4k+3}+1}{4}=3^{4k+2}-3^{4k+1}+\cdots -3+1\equiv 1+1+\cdots +1+1\equiv 4k+3\equiv 3\pmod 4\\\\N&=\frac{3^{4k+3}+1}{4}=3^{4k+2}-3^{4k+1}+\cdots -3+1\equiv 1\pmod 3\end{align}$$we have, from $(3)$, $$S_{p-1}\equiv 388-336\cdot \frac{(-1)^{\frac{N-1}{2}}}{\left(\frac N3\right)}\equiv 388-336\cdot \frac{-1}{1}\equiv 724\pmod{R_p(-3)}$$