It's my understanding that there's no longer a requisite of safe primes for $q$ and $p$ when choosing a RSA modulus. How is it that this does not change the hardness of factoring $N$?

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    $\begingroup$ "There's no longer a requisite of safe primes", said who? $\endgroup$
    – fkraiem
    May 26, 2017 at 11:54
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    $\begingroup$ Because the probability of randomly picking a non-safe prime is negliglible at current, standard sizes, but the risk was found to be too big back in the day when smaller key sizes were used. $\endgroup$
    – SEJPM
    May 26, 2017 at 11:54
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    $\begingroup$ @fkraiem The HAC said so back in the nineties: "These attacks are described at length in Note 8.8(iii); as noted there, it is now believed that strong primes offer little protection beyond that offered by random primes, since randomly selected primes of the sizes typically used in RSA moduli today will satisfy the constraints with high probability." (section 4.4.2 on page 149) And I think Boran is asking for why we don't explicitely check / search for strong / safe primes anymore. $\endgroup$
    – SEJPM
    May 26, 2017 at 11:58
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    $\begingroup$ Boran: Are you asking about safe primes (ie primes $p$ where $(p-1)/2$ is also prime) or strong primes? $\endgroup$
    – SEJPM
    May 26, 2017 at 12:00
  • $\begingroup$ the former, your answer was great though, thank you $\endgroup$
    – boran
    May 26, 2017 at 12:04

2 Answers 2


First we may want RSA primes to be something like a safe prime, ie a prime $p$ where $(p-1)/2$ is prime as well.

Back in 1974 Pollard found an algorithm to factor moduli whereby you can factor $N=pq$ if $p-1$ or $q-1$ are smooth, that is all prime-factors of $p-1$ or $q-1$ are smaller than a bound $B$. The algorithm will then factor $N$ in time $\mathcal O(B\cdot \log B\cdot \log^2N)$. This implies that you want at least one "large" prime factor for both $p-1$ and $q-1$ where large is probably something like $>2^{80}$ or a similar value to match your key strength.

The most "effective" solution is to pick $p,q$ as safe primes of course. In practice however it suffices if they are strong primes because:

  1. you can find those much quicker because the constraints are less severe and;
  2. there are other attacks against which a strong prime will protect you, such as p+1 factoring.

Now the argument is that the Elliptic-Curve Method is a strict generalization / improvement over Pollard's $p-1$ method in finding small factors as it allows one to "retry" if a "bad" group order (that is $p-1$) was encountered, this is the quote from Rivest's paper (see below):

Thus we see that it is useless to protect against factoring attacks by building in large prime factors into $p-1$ or $p+1$, since the enemy cryptanalyst can instead attempt to find an elliptic curve $E_{a,b}$ such that $|E_{a,b}|$ is smooth. The cryptographer has no control over whether $|E_{a,b}|$ is smooth or not, since this is essentially a random number approximately equal to $p$. Only by making $p$ sufficiently large can he adequately protect against such an attack.

If you want to dive further into the history of strong primes, I suggest you read "Are 'Strong' Primes Needed for RSA" by Rivest and Silverman from 1999.

I have used material from section 8.3 of said paper in my above answer. They go on after the above quote and actually list numbers for the probability of breaking random prime factors for 1024 bit moduli and find the chances to be negligible for ECM. As ECM is strictly better than $p-1$ factoring, random primes suffice.

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    $\begingroup$ As far as I know, there has been no recommendation to use safe primes (that is primes $p$ with $(p-1)/2$ also primes) in RSA. Definitely there was no such recommendation in the original RSA article. Safe primes are used in some algorithms of the Discrete-Log family; but if they have been used as secret factors of a public modulus in a factorization-based algorithm, I want to know! $\endgroup$
    – fgrieu
    May 28, 2017 at 8:40
  • $\begingroup$ I just re-invented a context where RSA with safe primes makes some beginning of a sense; see the last paragraph of this. $\endgroup$
    – fgrieu
    May 28, 2017 at 11:12
  • $\begingroup$ Suppose p is a safe prime. So, p = 2p' + 1, where p' is a prime. Now, p+1 is 2p'+2. How can we guarentee that p+1 is smooth? I am just trying to understand, why safe primes eliminate p+1 factoring attack. $\endgroup$
    – satya
    Dec 17, 2018 at 14:27
  • $\begingroup$ @satya you may want to ask this as a new question so it gets more exposure (also I don't think I know the answer to this question right now). $\endgroup$
    – SEJPM
    Dec 17, 2018 at 14:35

just on a side-note: it can have other benefits to use safe primes as RSA factors, namely, you can then use any odd integer as "public" exponent $e$ and derive a corresponding "private" exponent $d$ (disclaimer: I'm not saying this is secure of even recommended to do so!). If $p$ and $q$ are safe primes, then the Euler's totient is $\phi(N) = (p-1)(q-1) = 2p' * 2q' = 2²\ p'q'$ with $p', q'$ prime. Hence, for any* odd $e$ the $\gcd(e, \phi(N)) = 1$ , meaning, a multiplicative inverse $d$ of $e$ modulus $\phi(N)$ exists (for RSA to work you need $e*d\equiv 1 \mod \phi(N)$ ).
* any odd $e \notin \{1, p', q'\} $


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