Security of RSA with $\gcd(pq, (p-1)(q-1))\ne 1$ The Congress of ThéOùÇa enacted that civilian use of any public-key encryption system is authorized on the territory of ThéOùÇa, subject to meeting certain security requirements published by the Ministry of Defense. The essence of these is:


  ..must operate per the internationally recognized RSAES-OAEP algorithm and relevant requirements in PKCS#1 V2.2, with a 2048-bit public modulus $n$ product of two primes $p$ and $q$. In order to ensure separation of key domains with the Pailler cryptosystem used for national defense purposes, the key generator shall be such that $p$ and $q$ also meet the requirement $\gcd(pq, (p-1)(q-1))\ne 1$. Demonstration of conformance to that requirement shall be by submitting the mathematical description of the key generation method to the Ministry of Defense of ThéOùÇa.


Can a public-key encryption system lawfully usable by civilians on the territory of ThéOùÇa be secure? If yes, give a possible submission to the Ministry of Defense (English is spoken in ThéOùÇa). In either case, make a cryptographically convincing argument.

Thanks to Poncho for spotting a big mistake in my transcription of the Official Journal of ThéOùÇa, now fixed.

This is not homework. If you wonder, inspiration was that recent question.
Okay, I came up with this, it's not a complete answer and the attack presented is pretty weak without a follow-up algorithm for breaking a somewhat unbalanced modulus but let me know if you spot any flaws or have any ideas to improve it...



If $n = pq$ with $p, q$ prime and $\gcd(pq, (p - 1)(q - 1)) = r &gt; 1$ then clearly $r = p$ or $r = q$. Now obviously $q - 1$ cannot divide $p$ and so we must have $p$ divides $q - 1$, without loss of generality. Thus:
$$q - 1 = kp ~ ~ ~ \text{for some} ~ k &gt; 1$$
Now suppose that $k$ is $B$-powersmooth for small $B$, then we have $M = \alpha k$ for $M$ a suitable product of prime powers up to $B$, some $\alpha$ and so for any $a$ we have:
$$a^{Mn} \equiv a^{Mpq} \equiv a^{(\alpha k p)q} \equiv a^{\alpha(q - 1)q} \equiv 1 \pmod{q}$$
And so in general we find that:
$$\gcd(a^{Mn} - 1, n) = q$$
Therefore in order for $n$ to be usable in the context of RSA, $k$ should not be smooth. Realistically we will be able to try $B$ up to $2^{80}$ or so (you won't be factoring the 2048-bit modulus through general purpose algorithms anyway) which means $k \approx 2^{81}$ and a special key generation technique needs to be used, probably involving letting $2rp = q - 1$ for large prime $r$ with $k = 2r$.

And since $k$ is large, that means $q/p \approx 2^{81}$ so the modulus is unbalanced. Perhaps this enables polynomial-time factorization of the 2048-bit modulus via existing methods which rely on particularly poor selection of $p$ and $q$, or perhaps the scheme is secure with careful selection of $k$. Further work is needed to determine which conclusion is the right one (but I am less and less confident in this scheme).