Kryptos - The Cipher (Part 6-b) - A hidden word in K2

It is possible to extract a 'hidden' word from K2 that Sanborn may have inserted intentionally. What could we learn about K4 from this? Or is it just a coincidence?

In Part 6-b, i want to show another finding from years ago, which probably many people didnt know. It is about K2. Below you see on the left the ciphertext (I) as it can be found on the Kryptos statue. Decoding it with "ABSCISSA", yields the plaintext (I) on the right.

Kryptos - The Cipher (Part 6-a) - The Kryptos Mini Sculpture - Sanborns' Proof-Of-Concept

During the Christmas period, I spent many hours reading news articles and forum discussions about Kryptos. I reread some of the old topics on the Yahoo Kryptos Group. It is always surprising to see the findings from many years ago that have somehow been forgotten. I don't mean they're forgotten, just not on the current radar. Especially three topics raised my attention:

1. The Pre-K, i.e. Mini Kryptos Sculpture

2. The reveal of the word ASTATOS in K2

3. Methods to generate the first key PALIMPSEST for K1

In this post, I want to show how these topics might be connected. This post covers the first of the three points, Parts 6-b and Part 6-c follow the next days.

The Diffie-Hellman Oracle

One of the most charming things about modern cryptography is that it often rests on questions we still don’t fully understand. Even after decades of research, some of the most central assumptions — the kind that protect millions of TLS handshakes every second — remain only partially mapped. A perfect example is the relationship between two cornerstone problems in cyclic groups: the Discrete Logarithm (DL) problem and the Computational Diffie–Hellman (CDH) problem.

At first glance, DL seems clearly stronger than CDH. If you can solve discrete logs, you can obviously solve CDH: given $g^x$ and $g^y$, you recover $x$ and $y$, and then compute $g^{xy}$. The reverse implication is where the mystery begins. Suppose you had access to a machine — an oracle — that, given $(g^x,g^y)$, instantly returns $g^{xy}$. Could you then use this ability to compute discrete logarithms efficiently?

Jim Sanborn - Open Letter

A few days ago Jim Sanborn made an announcement - he wrote an open letter to the community, which is shown below and be accessed at [2]:

Nov 12, 2025

Another open letter to the kryptos community:

K4 has not been solved or decrypted. The scrambled plain text was found, but without the coding method or key, this is a very important distinction. The method of decryption also known as the Key is in this auction and is still very much a secret. As is the discovered plain text which is still unpublished, In fact several people have seen the K4 plain text over the years but have chosen to not release it for the greater good and have long since forgotten what it says.

Todays K4 clues:

"Mighty" Euler Quotients - One query to rule them all

Fermat quotients have been mentioned several times in this blog and they are fascinating peace of math. The definition is

\begin{equation}q_p(2) = \frac{2^{p-1}-1}{p},\;\;p \in \mathbb{P} \end{equation} One can extend this definition to composite integers $n$, which are called Euler quotients. \begin{equation} q^*_n(2) = \frac{2^{\varphi(n)}-1}{n},\;\;n \in \mathbb{N} \end{equation} In this short post, i would like to demonstrate the power of these Euler Quotients.

Suppose you have an Oracle $\mathcal{O}$ that, given an integer $n$, returns $q^*_n(2)$. Only one such query is allowed. We treat $q^*_n(2)$ simply as a binary string and show that this string contains enough information to compute things that are otherwise hard to compute.