We got strong numerical evidence that primes of the form $p=27a^2+27a+7$
are unsafe for cryptographic purposes since they can be found in the factorization.
Consider the following generic factoring algorithm for factoring
$n$ which is divisible by $p$. Suppose you known an elliptic curve
in Weierstrass form with known multiple $m$ of the order of $E$ modulo
$p$. Let $\psi_n$ denote the $n$-th division polynomial of $E$.
Choose random integer $X$. If $X$ is the $x$ coordinate on $E$ modulo $p$
then $\psi_m(X) \equiv 0 \pmod{p}$. If it is not, $\psi_m(X)$ need not
vanish modulo $p$. By selecting few random $X$ we probabilistically find
$p$ from $\gcd(n,\psi_m(X))$.
This question conjectures
that if $p=27a^2+27a+7$ and $\textrm{kronecker}(2k^3,p)= -1$ then the order of
$y^2=x^3+2k^3$ is $p$. In this case $n$ is multiple of the order, so we
can use the above algorithm by trying in addition random $k$.
We found $200$ bit factor with pari/gp in just few seconds.
Are these conjecturally unsafe RSA primes known?
Best Answer
This appears special case of "A New Special-Purpose Factorization Algorithm":
https://pdfs.semanticscholar.org/1843/73605e846f90b0a9d7252931bab4c47a1ec7.pdf
From the abstract: a new factorization algorithm is presented, which finds a prime factor $p$ of an integer $n$ in time $(D \log{n})^{O(1)}$, if $4p - 1 = Db^2$
If $p=27a^2+27a+7$ we have $4p-1=3\square$ which is covered by the paper.