Take the 2-minute tour ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

Given $n=pq$ for $p,q$ known, I can calculate $\phi(n)$.

$e$ is selected such that $\gcd (e,\phi(n)) = 1$.

Using this, how do I calculate the RSA private key?


Example:

I have $n = 35$, with $(p,q)=(5,7)$. I have also computed $\phi(n)=24$, and selected $e$ such that $\gcd (e,\phi(n)) = 1$ by taking $e=23$.

Calculate the private key.

share|improve this question
    
Take a look at the extended Euclidean algorithm –  DrLecter Dec 6 '13 at 3:21
    
I figured it out and I'll answer my question soon –  Ali Gajani Dec 6 '13 at 3:25
2  
This question appears to be off-topic because its scope is too local. It's unlikely that anyone else will need to calculate a key with these exact parameters. Maybe if you edited the question to make it more general... –  rath Dec 6 '13 at 6:24
    
I recommend you reading Conrado's post in here, it's easier to understand than that on the Wiki –  Alex Dec 6 '13 at 10:42

2 Answers 2

The private key $d$ of RSA algorithm with public parameters $(N,e)$ is such that:

$ed \equiv 1\mod{\phi(N)}$. Since by definition $e$ and $\phi(N)$ are coprime then with extended euclidean algorithm you can find such $d$: $ed +k\phi(N)=1$

Consider that to compute $\phi(N)$ you should know how to factor $N$ since $\phi(N)=\phi(p)\phi(q)=(p-1)(q-1)$

To see why this is correct imagine an encryption of the message $m$ to be $c=m^e\mod{N}$. Then to decrypt you compute $c^d=m^{ed}\mod{N}=m \mod{N}$

share|improve this answer

I figured out the decent way of solving for $d$ (the private key).

I have $n=35$, with $(p,q)=(5,7)$. I have also computed $\phi(n)=24$, and selected $e$ such that $\gcd(e,\phi(n))=1$ by taking $e=23$. To calculate the private key, we need to use the formula:

$$d = e^{-1} \mod \phi(n)$$

This gives us $d = 23$, which happens to be the same as $e$, a coincidence.

share|improve this answer
1  
These are standard techniques you can find in all books.We say the same thing.In order to compute the inverse you can use the extended euclidean algorithm –  curious Dec 7 '13 at 12:09
2  
Also, this isn't a much of a coincidence, because $e=23=-1\pmod{24}$ and so $e^2=(-1)^2=1\pmod {24}$. –  figlesquidge Dec 7 '13 at 15:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.