I found this recent paper The Wiener Attack on RSA Revisited: A Quest for the Exact Bound, which reported a new bound $d\le \frac 1 {\sqrt[4]{18}} N^\frac 1 4$. Is this well accepted in the cryptanalysis research community?

With a Python script, I personally verified the d revealed from the given (N,e) in Section 4 (page 392~395) of this paper, it worked indeed. The interesting thing is that even though this d was generated with $\lambda(N)=LCM(p-1,q-1)$, the Wiener Attack function could successfully reveal this d with the assumption of $ed = 1 \pmod {\phi(N)}$. How does this work?

import gmpy2

def solve_rsa_primes(s: int, m: int) -> tuple:
    """ Solve RSA prime numbers (p, q) from the quadratic equation
    p^2 - s * p + m = 0 with the formula p = s/2 +/- sqrt((s/2)^2 - m)

        s - sum of primes (p + q)
        m - product of primes (p * q)
    Return: (p, q)
    half_s = s >> 1
    tmp = gmpy2.isqrt(half_s ** 2 - m)
    return int(half_s + tmp), int(half_s - tmp)

def wiener_attack(n: int, e: int) -> (int, int, int):
    """ Wiener's Attack on RSA public key cryptosystem
        N - RSA modulus N = p*q
        e - RSA public exponent
        d - RSA private exponent

    cfe = cf_expansion(e, n) # Convert e/n into a continued fraction
    cvg = cf_convergent(cfe) # Get all of its convergents

    for k, d in cvg:
        # Check if k and d meet the requirements
        if k == 0 or d % 2 == 0 or (e * d) % k != 1:

        # assume ed ≡ 1 (mod ϕ(n))
        phi = (e * d - 1) // k 
        p, q = solve_rsa_primes(n - phi + 1, n)
        if n == p * q:
            return d

    return None

N = int(

e = int(

d = wiener_attack(N, e)
assert not d is None, "Wiener's attack failed!"
print("d =", d)

new_b = int(gmpy2.root(N, 4)/gmpy2.root(18, 4))
print("new_b =", new_b)
assert d <= new_b

old_b = int(gmpy2.root(N, 4)/3)
print("old_b =", old_b)
assert d > old_b

1 Answer 1


Is this $\frac 1{\sqrt[4]{18}}N^\frac1 4$ bound well accepted in the cryptanalysis research community?

I see no reason why there would be a doubt. However, the exact bound for Wiener's attack is not the subject of much scrutiny by the cryptanalysis research community, because:

  • Dan Boneh & Glenn Durfee's Cryptanalysis of RSA with Private Key $d$ Less than $N^{0.292}$ (in proceedings of Eurocrypt 1999) factors $N$ for $d$ significantly larger that Wiener's attack can, making Wiener's attack interesting only by it's relative simplicity.
  • Practice must use $d$ comfortably above these thresholds for security reasons. Thus it can only give a modest speed advantage compared to using the CRT method for the RSA private key operation, and then only if using this method, where essentially it lowers $d_p$ and $d_q$ (and, for two moduli of about equal size, makes $d_p=d_q$, which is far from reassuring from a security standpoint).
  • Practice uses small $e$ because it is widely considered safe and leads to fast public key operation, which is desirable. It follows $d$ is with practical certainty far above Wiener's $N^{0.25}$ and Boneh & Durfee's $N^{0.292}$ bound (unless we use a modulus with at least 4 prime factors and have these factors exactly one above some multiple of a large prime, with the later at least worrying from a security standpoint).

This does not answer the "how does this work" part of the Q.

  • $\begingroup$ Do you have answer to my 2nd question regarding using $\phi(N)$ to solve it but d was actually generated by $\lambda(N)$ ? $\endgroup$
    – Zixi Sean
    Commented Nov 9, 2023 at 19:03
  • $\begingroup$ If it was used a small $d$ (which again, is never the case in practice), it would be $e$ that's generated from $d$ as $d^{-1}\bmodλ(N)$ or $d^{-1}\bmodϕ(N)$, not $d$ generated from $e$. With probability like $1/3$, $ϕ$ and $λ$ yield the same $e$, which would be a first reason why Wiener's attack works. That occurs because $ϕ(N)=gλ(N)$ with $g=\gcd(p-1,q-1)$ typically small (often $2$, $4$, $6$). I'm totally ready to believe Wiener's attack works more often than that, but sorry, I have no clear explanation of why. $\endgroup$
    – fgrieu
    Commented Nov 9, 2023 at 22:00
  • $\begingroup$ "With probability like 1/3, ϕand λ yield the same 𝑒" - so in this case, we would get ed = 1 (mod ϕ) and ed = 1 (mod λ) , right? - where did this 1/3 come from? Any mathmatical proof? $\endgroup$
    – Zixi Sean
    Commented Nov 9, 2023 at 22:50
  • 1
    $\begingroup$ For the lowest positive such $e$, yes. My $1/3$ estimate is derived from simulation over 1000 tests. We could find it mathematically. 1) Estimate the distribution of $g=\gcd(p-1,q-1)$ for large random primes $p$ and $q$ about the same size with $\gcd(p-1,d)=1=\gcd(q-1,d)$ for predetermined random odd $d$. $g=2$ is most common, followed by $4$ and $6$. 2) For given $g$, the probability that the values derived using $ϕ$ and $λ$ are equal is next to $1/g$. We sum these over the aforementioned distribution to find (a good estimate of) the probability that $ϕ$ and $λ$ yield the same $𝑒$. $\endgroup$
    – fgrieu
    Commented Nov 10, 2023 at 8:01

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