# Questions tagged [rsa]

An asymmetric (e.g. public-key) cryptosystem, based on modular exponentiation with big exponents and modulus. RSA can be used both for signature and encryption with proper paddings.

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### Prove RSA scheme is insecure

Im reviewing my crypto and I would like to prove the following: RSA experiment where the adversary is given (N, e, y) where N = pq and p, q are two primes. The adversary knows p, q. How can I build an ...
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### Can we use several times the same RSA and ECC keys?

For RSA or elliptic cryptography, can we use the same public/private keys for several exchanges, or is that unsafe? Does it lower the security of the protocol? To me it doesn't matter, as receiving ...
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### Clarifying terms for cipher, algorithm, rsa, asymmetric, openssl

I am a self taught front-end developer and I am just trying to get my cryptography terminologies straight. I will make some statements. Can someone correct my mistakes? Question 1 I read that cipher ...
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### How could a 1024‒bits RSA modulus be most economically factored within months today?

Of course this is a question with an answer that is due to evolve. A 2002 paper about TWIRL stated that the cost would be around 10M\$and an other 10M\$ to manufacture the device. A later 2007 paper ...
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### RSA given 30% MSB of p and 30%MSB of q

is factoring RSA given 30% MSB of p and 30%MSB of q possible in polynomial time? Notice that it is known that given 50% MSB of p or q it is doable in polynomial time using Coppersmith's theorem
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### Is it possible to perform reverse rsa encryption ? (using the public key for decrypting and the private key for encrypting)

Simple question, but I couldn’t find an answer as I know about signing but don’t want it. The aim would be to encrypt someone’s bet parameter as a short message using the private key and use the ...
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### What exactly is RSA_WITH_AES256_CBC_SHA256?

Quick disclaimer, I am pretty new to cryptography. I only understand the basics of asymmetric and symmetric encryption. In particular, RSA and AES. I have a project where my service is receiving an ...
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### Implications of switching RSA padding schemes

In my company we want to switch from PKCSv1.5 to PSS for RSA signing. This is because of the marvin attack vector. My question is, if an adversary can see many similar messages that were signed with ...
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### What happens if we know that for an RSA key pair, the equation $d^e \equiv c \pmod{n}$ holds?

If we possess the knowledge that the expression $d^e \equiv c \pmod{n}$ holds for an RSA key pair, what would be the implication or consequence of this? Here, $e$, $n$, and $d$ represent the public ...
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### Chinese Remainder Theorem in a strange configuration

Normally, the chinese theorem looks like this:$$x = a_1 \bmod n_1\\ x = a_2 \bmod n_2\\ x = a_3 \bmod n_3$$ but what to do if in my situation it looks like this: c_1 = x^e \bmod n_1\\ c_2 = x^e \...
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### Are Safe and Sophie Germain primes evenly distributed?

Do Safe and Sophie Germain primes maintain a relatively stable distribution as numbers get larger, or do they become rarified beyond a predictable value? This is important in one area of triangular ...
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### Textbook RSA on random numbers mod a smaller prime

I have a protocol where I have a random one-time number message $m \in \mathbb{Z}_p$, where $p$ is a 256 bit prime from an Elliptic Curve (EDIT: The order of an elliptic curve prime order group). I ...
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### Is there an algebra group (or ring) in which computing the inverse element is hard without some trapdoor information?

Specifically, I want an algebra group $G$ (or ring $R$) features: Given elements $g,h\in G$ (or $R$ ), computing $g\cdot h \in G$ (or $R$ ) is easy. Given an element $g \in G$ (or $R$ ), finding the ...
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### How to create RSASSA-PSS keys using openssl?

I am trying to understand the RSASSA-PSS and process of generating keys with rsassaPss. I am using below ...
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The encryption oracle only allows you to encrypt a custom message $m$ or the secret message $m_s$. In both cases it asks you to choose a public exponent $e>16$. Of course, $n$, is not available. If ...