EASTNORTHEAST - This is not exactly the hint Jim Sanborn (JS) gave for K4 on the 29th of January this year. He only gave NORTHEAST - which refers to the positions 26-34 of K4's plaintext. Beside BERLIN and CLOCK it is the third revealed plaintext word of K4. However, also this hint does not seem to help much. However, it just so happened, that a member in the yahoo kryptos group had a conversation with Jim Sanborn due to a submitted solution. Sandborn's answer to the question contained again the last clue which surprisingly was EASTNORTHEAST at position 22-34. Jim Sanborns compass rose at CIA There is disagreement if Jim revealed this on purpose or he did it accidentially, but the new extended clue seem to be serious and valid.Interestingly, EASTNORTHEAST is exactly the direction which is illustrated on the compass rose on one of the stones around kryptos, also created by Jim Sanborn. Actually, i dont really kn...
In one of my previous posts , i talked about the possibilities to insert backdoors in prime numbers (beside the property that $p-1$ is smooth). One of the potential ideas for a backdoor was to generate primes that allow an easy lift from $\mathbb{Z}/p$ to $\mathbb{Z}/{p^2}\mathbb{Z}$. But under the consideration of the $l$-Hensel Discrete Logarithm Assumption , this should be actually hard in the general case. Because if you could successfully lift the element $e\equiv g^x\pmod{p}$ to $E\equiv g^x\pmod{p^2}$, then you could, as long as $x < \mathsf{ord}_p(g) = r$, compute \begin{align*} \left(g^x\right)^{r} \equiv (g^{r})^x \equiv (1+p\lambda)^x \pmod{p^2} \end{align*} Next, the Binomial Theorem gives \begin{align*} (1+p\lambda)^x \equiv \sum^x_{i=0}\binom{x}{i}p^i\lambda^i \equiv 1 + xp\lambda\pmod{p^2} \end{align*} and hence \begin{align*} \frac{E^{r} - 1}{p\lambda} \equiv x \pmod{p} \end{align*}