# Need help to understand RSA common modulus attack to get private key

I'm learning about common modulus attack and learned that public modulus attacks can find out the private key. Assume there are 2 users with public and private keys $$(e_1, d_1)$$ and $$(e_2, d_2)$$. Scenario is attacker has his public and private keys $$(e_2, d_2)$$ and victim's public key $$e_1$$ Here are the steps to get the secret key:

1. $$t= e_2\cdot d_2-1$$
2. Attacker uses the extended Euclidean algorithm to find $$f=\gcd(t,e_1)$$ whose pairs of numbers $$(r,s)$$ satisfy the equation $$r\cdot t + s\cdot e_1 = f$$
3. If $$f=1$$ set $$d_1' = s$$ and return private key $$d_1'$$
4. If $$f\neq 1$$ set $$t=\frac t s$$ and return step 2

I have a example with $$d_1=17, M = 25$$ is two value we should find. n = 253 (common modulus) $$e_1 = 13$$ (victim's public key) $$e_2 = 23$$ (attacker's public key ) $$d_2 = 67$$ (attacker's private key) $$C = 27$$ (Ciphertext). Attacker will find $$d_1'$$ according there steps:

1. $$t= e_2\cdot d_2-1 = 23\cdot 67 = 1540$$
2. $$\gcd(t,e_1) = 1 \implies r = -2, s = 237$$
3. So $$d_1' = s = 237$$
4. Verify $$M = C^{d_1} \bmod n = 27^{237} \bmod 253 = 25$$

Problem is I don't understand why with such steps we will find the secret key. Can someone explain me pls ?

• multiple questions and answers exist on this. see the questions that come up onthe related tab Oct 28, 2021 at 3:54
• Can you provide me a specify link no . I have searched but couldn't find a suitable answer . . . Oct 28, 2021 at 4:07
• This does not answer the question, but: with $(n,e_2,d_2)$ the attacker can factor $n$ and then find a working $d_1$. This is known since the original RSA article: “knowledge of $d$ enables $n$ to be factored…”. See these related questions.
– fgrieu
Oct 28, 2021 at 5:56

We'll heavilly use the following fact: for a fixed public modulus $$n$$ product of distinct primes, a pair of integers $$(e,d)$$ forms a matching pair of RSA exponents [that is, with $$c\mapsto c^d\bmod n\,=\,m$$ capable of reliably deciphering any plaintext $$m$$ in $$[0,n)$$ enciphered per $$m\mapsto m^e\bmod n\,=\,c$$ ] if and only if$$e\cdot d\equiv1\pmod{\lambda(n)}$$where $$\lambda$$ is the Carmichael function. That can be shown to follow from the definition of $$\lambda(n)$$ as the smallest strictly positive integer $$y$$ such that $$m^y\equiv 1\pmod n$$ for all $$m\in\mathbb Z^*$$. This holds regardless of the sign of $$d$$.
It follows that $$t$$ of step 1 of the question's algorithm is such that exists $$k\in \mathbb Z$$ with $$t=k\cdot\lambda(n)$$.
If the algorithm finds $$f=1$$ at the first execution of step 2, it thus holds $$r\cdot k\cdot\lambda(n)+s\cdot e_1=1$$, hence $$s\cdot e_1=1+(-r\cdot k)\cdot \lambda(n)$$, hence when the algorithm sets $$d'_1=s$$ in step 3 it holds $$e_1\cdot d'_1\equiv1\pmod{\lambda(n)}$$. Applying the first fact, $$(e_1,d'_1)$$ is a matching pair of RSA exponents for public modulus $$n$$. If we want $$d'_1$$ to be non-negative we can make $$d'_1=s\bmod t$$, which by definition is in range $$[0,t)$$ and also such that $$e_1\cdot d'_1\equiv1\pmod{\lambda(n)}$$.
Things get wrong when $$f\ne1$$ at the first execution of step 2. Often $$s$$ won't divides $$t$$ in step 4, preventing application of the algorithm as is. Example: $$p=13$$, $$q=19$$, $$n=247$$, $$\varphi(n)=216$$, $$\lambda(n)=36$$, $$e_1=91$$, $$e_2=25$$, $$d_1=19$$, $$d_2=121$$, $$t=3024$$, $$r=-4$$, $$s=133$$, $$f=7$$, $$t/s=432/19\not\in\mathbb Z$$.
Changing $$t:=\frac t s$$ to $$t:=\frac t f$$ in step 4 insures divisibility, and leaves the algorithm working. Argument: $$f$$ divides $$e_1$$, and $$\gcd(e_1,\lambda(n))=1$$, thus $$f$$ is coprime with $$\lambda(n)$$, thus we restart steps 2 with $$t$$ still a multiple of $$\lambda(n)$$.
Alternatively: given $$(n,e_2,d_2)$$ the adversary can factor $$n$$ (see this) and from that get $$\hat{d_1}=e^{-1}\bmod\lambda(n)$$ matching $$(n,e_1)$$, typically with $$\hat{d_1}$$ smaller/faster than $$d'_1$$; or get a working private key in the form allowing CRT operation thus even faster decryption or signature.