- #1

mathmari

Gold Member

MHB

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1. Construct a pair of private/public key RSA, where the prime numbers that we use are $p=11, q=13$.

2. Describe how we can calculate a RSA signature at the message $m=2$ without using a hash function.

3. Show that, given the above signature, we can calculate a valid signature at the message $m'=8$ without using the private key.

I have done the following:

1. $n=p \cdot q=11 \cdot 13$

$\phi(n)=(p-1)(q-1)=10 \cdot 12=120$

We choose a $e$ such that $(e,\phi(n))=1$. We take for example, $e=7$.

Then we calculate $d$ such that $ed \equiv 1 \pmod {\phi(n)}$. So, $d=13$.

The private key is $d=13$ and the public key is $(e, n)=(7, n)$.

2. The signature is $c=m^d \pmod {\phi(n)}$.

3. There is a $m_1$ such that $m=m'm_1$.

$c=m^d \pmod {\phi(n)} \Rightarrow c=m'^dm_1^d \pmod {\phi(n)} \\ \Rightarrow c(m_1^d)^{-1}=m'^d \pmod {\phi(n)} \Rightarrow cm_1^{-d}=m'^d \pmod {\phi(n)} \\ \Rightarrow ((cm_1^{-d})^{-e})^{-\frac{1}{e}}=m'^d \pmod {\phi(n)} \Rightarrow (c^{-e}m_1^{ed})^{-\frac{1}{e}}=m'^d \pmod {\phi(n)} \\ \Rightarrow (c^{-e}m_1)^{-\frac{1}{e}}=m'^d \pmod {\phi(n)}$

That means that the signature of the message $m'$ is $(c^{-e}m_1)^{-\frac{1}{e}}$.

Could you tell me if it is correct what I have done?? (Wondering)