In an RSA-encryption scenario, Bob's public key pair $(n, e)$ is $(143, 43)$. An attacker Mallory tries brute-force and comes to $d = 7$ as the private key.

The value of $φ(143) = 120$ is not known to Mallory.

However from $43 \cdot d \equiv 1 \bmod 120)$, one can calculate the first positive element $d = 67$ from multiplicative group $d = 120 + 67n$ and $n \in \mathbb{Z}$

$d = 7$ clearly doesn't fit in that multiplicative group, so how come it can successfully decrypt the encryption?

  • $\begingroup$ I was doing some RSA math trying different numbers. I noticed that $d = 7$ can decrypt at least for messages in $1 ≤ x ≤ 142$ and $n ∈ ℤ$ $\endgroup$ – user86295 Jan 17 at 17:53

This question can be summarized: the attacker found a $d$ that did not satisfy $e \cdot d \equiv 1 \pmod{ \phi(n) }$, but it works; what's going on.

It turns out that $e \cdot d \equiv 1 \pmod{ \phi(n) }$ is not necessary (it is sufficient).

The necessary and sufficient conditions are:

$$e \cdot d \equiv 1 \pmod{p-1}$$ $$e \cdot d \equiv 1 \pmod{q-1}$$

If both of these hold, then $d$ will always work [1]; conversely if $d$ always works, then both of these hold.

These two conditions can be summarized as a single relation:

$$e \cdot d \equiv 1 \pmod{\text{lcm}(p-1, q-1)}$$

This $\text{lcm}(p-1, q-1)$ modulus is known as the Carmichael function of $n$.

In the specific example you have, $\text{lcm}(p-1, q-1) = 60$, and we have $7 \cdot 43 \equiv 1 \pmod{60}$, and hence $d = 7$ works

[1]: Assuming $p, q$ are distinct primes.

  • $\begingroup$ Consequently, we want to avoid simple relations between $p$ and $q$. For example if $q=3p-2$ then $\operatorname{lcm}(p-1,q-1)=p-1\ll \phi(n)$. (And apart from that, such bad choices of primes are also easily factored wiht standard algorithms) $\endgroup$ – Hagen von Eitzen Jan 18 at 12:20

RSA private key can be found in two ways with $n = p\cdot q$, $p = 11$ and $q = 13$

  1. if Euler's totient function is used as in RSA paper: $$\varphi(n)= (p-1)(q-1) = 120$$ is used then
    • $d = 67 = e^{-1} \bmod 120$
  2. If Carmichael Function used as requried in FIPS 180.4and allowed in PKCS#1 v2.2 standards: $$\lambda(n) = \text{LCM}(p-1,q-1) = 60$$ is used then
    • $d= 7 = e^{-1} \bmod 60$

Both are valid and Carmichael Function provides always the smallest $d$. The easy relation between both of them is that $\lambda(n)| \varphi(n)$. Therefore this indicates that in some setups we can have more than one valid private key where each of them $\leq \varphi(n)$. Actually, in the two distinct prime case we have the relation;

$$\varphi(n) = \lambda(n) \cdot \gcd(p-1,q-1).$$ This is due to the fact that $$a \cdot b = \operatorname{lcm}(a,b) \times \gcd(a,b) $$

Since RSA primes are distinct odd primes $p$ and $q$, then $\gcd(p-1,q-1) \geq 2$ and this implies that there is always at least two $d$ in the range $[1,\varphi(n)]$

This is your case and you have the $\varphi(n)$ and the attacker has $\lambda(n)$.

The PKCS#1 standard requires the Carmichael Function to be used for the calculation of $d$. Original RSA paper used Euler's totient function. Using shorter $d$ will decrease the signature time and less used decryption time.

Carmichael Function: For a positive integer $n$, $\lambda(n)$ is defined to be the smallest positive integer $k$ such that $$a^k \equiv 1 \pmod n$$ for all $a$ such that $\gcd(a,n)=1$

Little proof of $\lambda(n)| \varphi(n)$:

The proof relies on the exponent definition of group theory.

Let $G$ be group then the non-negative generator of the ideal $\{z \in \mathbb{Z}: \forall g \in G (g^z=1)\}$ is called the exponent of the group $G$. For finite groups like RSA groups, it is finite and positive, and then it is the smallest positive natural number $z$ such that $g^z=1$ for all $g \in G$.

The exponent of any finite group must divide the order of the group. $\lambda(n)$ is the exponent by the definition and the order of the group is $\varphi(n)$ also by definition. This clearly implies $\lambda(n)| \varphi(n)$.

More than 2 private key example;

  • $n = 6901$
  • factors $6901 = 103 \cdot 67$, $p=103,q=67$
  • $\varphi(n) = 6732$
  • $\lambda(n) = 1122$
  • $e = 43$
  • $g =\gcd(p-1,q-1)=6$
  • inverse of $e$ by $d = \varphi(n) = 5323$
  • inverse of $e$ by $d' =\lambda(n) = 835$

Now all $d+k\cdot \lambda(n)$ are valid private key where $k \in [0,g]$, listing;

  1. 835
  2. 1957
  3. 3079
  4. 4201
  5. 5323
  6. 6445

SageMath code to find the above example;

p = random_prime(200, 400) #upper and lower range
q = random_prime(200, 400)
n = p*q
e = 43
print("n = ",n)
print("factors %s = " % n, factor(n))

phi = (p-1)*(q-1) # or call euler_phi(n)
print("phi    = ",phi)

if gcd(e,phi) != 1:
    print( gcd(e,phi))
lmd = lcm(p-1,q-1) #or call carmichael_lambda(n)

print("lambda = ",lmd)

print("gcd(%s,%s) = " % (p-1,q-1), gcd(p-1,q-1))

print("inverse of %s by phi   " %e, inverse_mod(43,phi))
print("inverse of %s by lambda" %e, inverse_mod(43,lmd))

d = inverse_mod(43,lmd)

for k in range(gcd(p-1,q-1)):
  • $\begingroup$ The PKCS#1 standard does not require that the Carmichael function $\lambda$ be used for the calculation of $d$. It allows it, and allows any $d$ with $e\,d\equiv1\pmod{\lambda(N)}$ and $0<d<p$. AFAIK, only FIPS 186-4 requires use of $\lambda$, leading to a single $d$. $\endgroup$ – fgrieu Jan 18 at 11:17
  • $\begingroup$ @fgrieu both FIPS 186-4 and PKCS#1 v2.2 only mentions $\lambda$. $\endgroup$ – kelalaka Jan 18 at 11:28
  • $\begingroup$ @kelaka. Yes. But FIPS 186-4 (B.3.1. step 3.a) tequires $d=e^{-1}\bmod{\lambda(N)}$ thus $d<\lambda(N)$ (and $2^{\left\lceil\log_2(N)\right\rceil/2}<d<\lambda(N)$ in step 3.b), thus uniquely defines $d$, and requires computing $\lambda(N)$. While PKCS#1 requires $d\equiv e^{-1}\pmod{\lambda(N)}$ and $0<d<n$, thus does not require computing $\lambda(N)$ since we can get away with $d\gets d=e^{-1}\bmod{\varphi(N)}$. That's a common way to compute $d$, is PKCS#1 conformant since $d\equiv e^{-1}\bmod{\varphi(N)}\implies d\equiv e^{-1}\bmod{\lambda(N)}$, but often is not FIPS 186-4 conformant. $\endgroup$ – fgrieu Jan 18 at 14:27
  • 1
    $\begingroup$ @fgrieu I see, the boundary makes the allowance. Thanks, $\endgroup$ – kelalaka Jan 18 at 14:34

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