ARTICLE AD BOX
I'll give you some context — I was trying to implement this algorithm to break RSA in Python, but it's a headache to get an LLM to implement it, for obvious reasons. That's why I came here, to ask for help on how I could implement this.
Below I'm sharing the theory behind exactly what I want to implement in code.
Paper PDF in Zenodo DOI: https://doi.org/10.5281/zenodo.19075019 Paper PDF in IPFS: https://dweb.link/ipfs/bafkreibs3f2mhefuwyhkmyamyae74n7wfijmxuqdg5tusy3ofeplsuectu
Explanation in LaTeX (source code):
\documentclass[12pt]{article} \usepackage{amsmath, amssymb, amsthm} \usepackage[margin=1in]{geometry} \newtheorem{theorem}{Theorem} \title{\textbf{Breaking RSA via Transcendent Reduction}} \author{Kaoru Aguilera Katayama} \date{\today} \begin{document} \maketitle \begin{theorem} The RSA cryptosystem is broken by embedding its ciphertext into a transcendent structure and applying an analytical reduction through the Lambert $W$ function. \end{theorem} \begin{proof} \subsection*{Construction} Let $(e, n)$ be an RSA public key and $M$ the plaintext. Define the \textbf{transcendent ciphertext}: \[ \mathcal{C}(M, e, n) \;=\; \underbrace{M^e \pmod{n}}_{A} \;+\; M^{\,M^e \!\!\pmod{n}} \] This embeds the classical RSA output $A = M^e \pmod{n}$ into the transcendent form: \[ C = A + M^A \] \subsection*{Analytical Recovery of the RSA Ciphertext} Starting from $C = A + M^A$, rewrite as: \[ C - A = e^{\,A \ln M} \] Let $u = C - A$. Then: \[ (C - A)\, e^{(C - A)\ln M} = M^C \ln M \cdot \frac{1}{\ln M} \cdot \ln M \] which simplifies to: \[ (C - A)\ln M \;\cdot\; e^{(C-A)\ln M} \;=\; M^C \ln M \] Applying the Lambert $W$ function to both sides yields: \[ (C - A)\ln M = W\!\bigl(M^C \ln M\bigr) \] and therefore: \[ \boxed{\displaystyle A \;=\; M^e \pmod{n} \;=\; C \;-\; \frac{W\!\bigl(\ln(M)\;\cdot\; M^{\,C}\bigr)}{\ln(M)}} \] \subsection*{Why This Breaks RSA} The value $A = M^e \pmod{n}$ is the standard RSA ciphertext. The formula above recovers $A$ from the transcendent ciphertext $C$ \textbf{without requiring}: \begin{itemize} \item The private key $d$ \item The factorization of $n$ \item Any brute-force or subexponential algorithm \end{itemize} The entire security of RSA is bypassed by a \textbf{single analytical evaluation} of the Lambert $W$ function. The modular trapdoor, which is the foundation of RSA's computational hardness, becomes irrelevant once the problem is lifted into the transcendent domain and solved in closed form. \textbf{RSA is therefore broken under this reduction.} \end{proof} \end{document}It would help me a lot if someone could help me create this code!
