According to this and this paper from Alexander May is possible to factor given the knowledge of the RSA private key. This is possible via Coppersmith and LLL reduction. Now I am trying to implement the code using the reduction in 2 (I prefer this to 1 given the fact []1] uses bivariate integer polynomials).

Now I struggle to understand the parameter of the reduction in section 4.6 of 2. In particular it refers to and M to be applied to Theorem 1. Giving a look to Theorem 1 though it doesn't mention M. The problem is that M is the only place where d (the secret exponent d is mentioned) hence why I am lost....

  • $\begingroup$ Actually, May's paper would appear to be incorrect; he specifically assumed that, for RSA public/private exponents, we always have $ed \equiv 1 \pmod {\phi(N)}$; that is not always the case. We always have $ed \equiv 1 \pmod {\phi(N)/gcd(p-1, q-1)}$, but that doesn't imply the relation he assumed. It's possible that his results could be adjusted to account for this; I haven't dug through his results thoroughly enough to be sure... $\endgroup$
    – poncho
    Commented Nov 1, 2017 at 13:34
  • $\begingroup$ thanks @poncho. But let assume he was correct. How about the M in the Coppersmith formula in Theorem 1. How is it connected ? $\endgroup$ Commented Nov 1, 2017 at 13:36
  • $\begingroup$ $M=ed-1$ is the known modulus with the unknown factorization and called $N$ in Theorem 1. $\endgroup$
    – j.p.
    Commented Nov 2, 2017 at 7:30

1 Answer 1


May and Coron's result is a deterministic reduction from knowing a multiple of $\varphi(n)$, in this case $de-1$, to the factorization of $n$. This method involves lattice reduction and Coppersmith's algorithm.

But there is a much simpler probabilistic method to factor $n$ discovered by Miller, of the famed Miller-Rabin primality test. Suppose you are able to find a nontrivial square root of $1$, i.e., $$ x^2 \equiv 1 \pmod{n}, \quad x \neq \pm 1\,. $$ Then $x^2 - 1 \equiv (x+1)(x-1) \pmod{n}$, and we find a factor by computing $\gcd(x \pm 1, n).$ But how do we find such a root? By exploiting the relation $$ a^{\varphi(n)} \equiv 1 \pmod{n}\,. $$ Let $t = \varphi(n) / \gcd\left(\varphi(n), 2^{\log_2(n)}\right)$ be $\varphi(n)$ with every power of 2 removed from it. If we knew $\varphi(n)$ exactly, all we had to do was compute $a^{t} \bmod n$ for a random base $a$, and repeatedly square until we obtained a $1$—that value would be our $x$ as above. This will yield a factor depending on the order of $a$ modulo $p-1$ and $q-1$. This process naturally extends to any multiple of $\varphi(n)$, and in practice only a couple of bases are necessary to factor $n$.

Here's some Sage code that implements this:

def factor_from_d(n, d, e):
  kphi = d*e - 1
  # remove powers of 2 from phi multiple
  kphi = kphi // gcd(kphi, 2^int(log(n, 2)))
  while True:
    # random base
    b = randint(0, n)
    x = power_mod(b, kphi, n)
    # try to find a nontrivial square root of 1
    while x != 1 and x != n - 1:
      # found one!
      # x^2 = 1 (mod n)
      # x^2 - 1 = (x + 1)*(x - 1) (mod n)
      if x^2 % n == 1:
        return gcd(n, x + 1), gcd(n, x - 1)
      x = x^2 % n

p = random_prime(2^512)
q = random_prime(2^512)
n = p*q
e = 2^16 + 1
d = inverse_mod(e, (p-1)*(q-1))
factor_from_d(n, d, e)

Coming back to May and Coron, their algorithm is conceptually simple, but more complex to implement from scratch. Here we want to find a small root of $$ f(x) = n - x \pmod{de - 1}\,, $$ which is guaranteed to exist because $n$ and $\varphi(n)$ are relatively close together. This root is precisely $p+q-1 = pq - (p-1)(q-1)$. Having $p+q$, we can solve the polynomial $x^2 - (p+q)x + n$, which has $p$ and $q$ as its roots.

Using Sage's small_roots function, we can do this easily:

p = random_prime(2^512)
q = random_prime(2^512)
n = p*q
e = 2^16 + 1
d = inverse_mod(e, (p-1)*(q-1))

P.<X> = Zmod(d*e-1)[]
f = X - n
pq = ZZ(f.small_roots(X=2^513, beta=0.9, epsilon=0.1)[0] + 1)
disc = pq^2 - 4*n
(pq + isqrt(disc))/2, (pq - isqrt(disc))/2

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