1
vote
2answers
82 views

Proof for exponentiation in modular arithemtic

If $e$ is a natural number, then this is true: $$m^e \bmod\ n = (m\bmod\ n)^e\bmod\ n$$ This is often used when encrypting, especially with RSA, since one can avoid directly calculating $m^e$, ...
3
votes
1answer
323 views

How do institutions like banks do RSA with big primes?

When encrypting with RSA it is often infeasible to decrypt by just doing c^d mod n, because for example when using the primes $(p,q)=(12553,1233)$, which are small ...
3
votes
2answers
176 views

Avoiding overflow when encrypting with RSA

When encrypting with RSA one calculates $ m^e \pmod n $ by doing the following: m^e % n Where $m$ is what we encrypt. Often $e$ is a very big number to make it ...
2
votes
1answer
142 views

Why encrypting with private and public keys produce the same result?

Let say $e = 5$, $n = 119$ and $d = 77$. If I encrypt, for example, $m=15$ I get: $$m_1 = m^e=15 ^ 5 \mod 119 = 36\qquad\text{and}\qquad m_2 =m^d= 15 ^ {77} \mod 119 = 36$$. Why? Is it always like ...
4
votes
4answers
381 views

Where does the $\varphi(n)$ part of RSA come from?

$e d \equiv 1 \pmod{\varphi(n)}$ Where does the $\varphi(n)$ part come from? How did the inventors of RSA arrive at $\varphi(n)$?
6
votes
2answers
319 views

How does Clifford Cocks 'Non-Secret Encryption' work?

I have read Clifford Cocks "A Note on 'Non-secret Encryption'" and thought I would try to implement this, but I don't seem to be able to get it to work. I'm obviously missing something. From the ...
4
votes
1answer
141 views

Solve a Modular Exponentiation

It might be common, but if we had to solve an equation like this $m=s^{e}$ mod $n$ where $m,e,n$ are known. How can we find $s$. What optimisations could be applied? And what would the complexity of ...
3
votes
2answers
355 views

In RSA, why is it important choosing e so that it is coprime to φ(n)?

When choosing the public exponent e, it is stressed that $e$ must be coprime to $\phi(n)$, i.e. $\gcd(\phi(n), e) = 1$. I know that a common choice is to have $e = 3$ (which requires a good padding ...
2
votes
2answers
193 views

How does this happen in RSA malleability?

I don't understand how the $E(m)$ turns into $E(mt)$. I mean, I don't know how does that transformation happen and how does the equation occur. $$E(m) \cdot t^e \bmod n = (mt)^e \bmod n = E(mt)$$ ...
3
votes
1answer
209 views

How difficult is it to brute force d in RSA: d = (1/e) mod φ in a CPT attack?

Given that RSA key generation works by computing: n = pq φ = (p-1)(q-1) d = (1/e) mod φ If I was an attacker who wanted to brute force d, could I brute force d given just the public key, the ...
1
vote
1answer
52 views

RSA Proof - di-mgt - modulo properties

I´m trying to follow one of the very detailed RSA Proofs given by di-mgt: "RSA theory", but unfortunately I stuck at the beginning of solution (chapter 3). I don´t understand where the second part ...
4
votes
2answers
154 views

how to iteratively calculate a^emod n with modulus n sized 4096 bits

In most sites the exponent of the RSA public key is 24 bits. But the modulus can get to 4096 bits size. I have an accelerator that can get max. 2112 bit size modulus. It calculates ...
0
votes
1answer
212 views

RSA smaller number work-out-by-hand not working - I think I made a mistake

I tried out the paper/pencil explanation @ http://sergematovic.tripod.com/rsa1.html, and it seemed to make sense just fine until I came to decryption. Here is what I worked out: Key Creation: Choose ...
4
votes
3answers
405 views

Is sharing the modulus for multiple RSA key pairs secure?

In the public-key system RSA scheme, each user holds beyond a public modulus $m$ a public exponent, $e$, and a private exponent, $d$. Suppose that Bob's private exponent is learned by other users. ...
7
votes
1answer
3k views

Calculating RSA private exponent when given public exponent and the modulus factors using extended euclid

When given $p = 5, q = 11, N = 55$ and $e = 17$, I'm trying to compute the RSA private key $d$. I can calculate $\varphi(N) = 40$, but my lecturer then says to use the extended Euclidean algorithm to ...