Which impact on security (factorization) has a common prime factor among the prime factors $P$,$Q$ of a number $N$ $$N=P\cdot Q$$ $$P=2\cdot F\cdot p+1$$ $$Q=2\cdot F\cdot q+1$$ with $F,q,p$ different primes and $F$ the biggest prime factor of $P$ and $Q$ with $$F\gg p,q$$

A potential adversary who want to factorize $N$ does know about the internal structure but does not know $F,p,q,P,Q$

For example $N$ is a $1024$-bit number with $P,Q \approx$ $512$-bit each.
$F \approx 461$-bit and $p,q \approx 50$-bit each.
Would security significantly change for larger $N,F$ but constant size $p,q$?
Or how would security change for larger/smaller $p,q$ but constant size of $N$?


Edit Update: It turned out a common factor is not necessary. I did some more detailed question.


1 Answer 1


There is a large weakness in 1024 bit products created according to the method described if you reuse F.

If N1 and N2 are both created with the same F, F can be calculated immediately:

G = gcd(N1-1,N2-1) = 2Fk.  

Factor G to get F, which is easy to spot because it is length 461 bits.

Additionally, security can be significantly weakened if the difficulty of factoring N-1 is significantly easier than factoring N.

N-1 is a composite which can be significantly easier to factor than a 1024 bit product of two 512 bit primes.

Based on the definitions and limitations of F, p and q in the question: N = (2Fp+1)(2Fq+1)

Expanding N = 4*F^2pq+2Fp+2Fq+1

Rearranging N = 2F(2Fpq+p+q)+1

N-1 = 2F(2Fpq+p+q)

After factoring N-1 you have F, an approx. 461 bit prime.

Let u be (N-1)/2F = (2Fpq+p+q)

u = (2Fpq+p+q)

Then calculate s = mod(u,2F) = p+q,

q = s-p

Substitute s-p for q and s for p+q

u = (2Fp(s-p)+s)
u = 2Fps-2Fp^2+s

This results in a quadratic in p

-2Fp^2+2Fsp+s-u = 0

p = (Fs - sqrt(F(Fs^2 + 2s - 2u)))/(2F)

q = s-p

The two approx 512 bit primes can now be calculated.

N = (2Fp+1)(2Fq+1)

Note that factoring N-1 may still take a significant amount of time requiring GNFS or CADO-NFS, but still significantly easier to factor than a 1024 bit product of two 512 bit primes.

  • $\begingroup$ is factoring '$N-1 = 2F(2Fpq+p+q)$' that easy? It's still an 922-bit number. How much easier is it compared to a regular used 922-bit number? Current record of a hard number is 829-bit. If it is easy. Does it significantly depend at the size of $F$? We could scale it as big as we want as long $p,q$ have a constant size. $\endgroup$
    – J. Doe
    Mar 25, 2022 at 14:46
  • $\begingroup$ ah ok, '$(2Fpq+p+q)$' may be factorized easily. How about we take care about this and set it to a small number times a $500$-bit prime? Would it be still easy? $\endgroup$
    – J. Doe
    Mar 25, 2022 at 14:56

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