# Tag Info

## Hot answers tagged rsa

32

Surprisingly, very basic algorithms which the children learn at the basic schools are used. For instance: http://www.wikihow.com/Do-Long-Multiplication You can find a similar algorithm for sum, sub and division. Try to ask google for: "division on paper" The "power of" is little tricky. In cryptography you don't really need the "real power of". Instead ...

23

There are two reasons by which such "huge" numbers can be computed in reasonable time. The first one is that we do not raise one integer x to some big exponent d. What we do is that we compute x raised to power d modulo an integer n. The modulo means that we are not interested in the final integer xd but only in the remainder of the Euclidian division of xd ...

4

I want to show that low values of the public key exponent can make it easy to 'invert' the function so that the encrypted message can be recovered. That is not known to be true; as long as the modulus is large enough to make factorization infeasible, there is no known way to compute e-th roots in general. Now, if the plaintext $p$ is small enough that ...

3

I'll use these common definitions and notations: $a\equiv b\pmod c$ means that $c>0$ and $c$ divides $b-a$ $a\equiv b^{-1}\pmod c$ means that $a\cdot b\equiv 1\bmod c$ $a=b\bmod c$ means that $a\equiv b\pmod c$ and $0\le a<c$ $a=b^{-1}\bmod c$ means that $a\equiv b^{-1}\pmod c$ and $0\le a<c$ $\varphi$ is the Euler totient function (also noted ...

3

The security of $\varphi$ and $\lambda$ should be equivalent since they are mathematically equivalent in the context in which they are used. (That is: the $d'$th power in $(\mathbb Z/pq\mathbb Z)^\times$ is exactly the same operation as the $d$th power.) However, the mathematically "right" modulus for computing $d$ is $\lambda(pq)$: it is precisely the ...

2

If Jeff has the public key and an encrypted message, why can't he guess the message, encrypt it with the public key, and see if he gets the encrypted message. Well, that particular attack is foiled by the RSA padding method. All RSA encryption padding methods include randomness in the padding (specifically to foil this attack). Whenever John encrypts ...

1

We assume an RSA signature scheme with appendix where the signature of message $M$ is $S=\left(\operatorname{MD5}(M)\right)^d\bmod N$, and the verification procedure checks that $0\le S<N$ and $\left(S^e\bmod N\right)=\operatorname{MD5}(M)$, with $e=3$ (or other relatively small odd $e\ge3$). Eve somewhat got $k$ rightful signatures $S_i$ and perhaps the ...

1

You forgot to mention one additional advantage of elliptic curves: the generation of keys is much faster than with RSA. In europe, many government smart card solutions are now based on ECC: The european electronic pass ports The Austrian card The German ID card The new German health insurance card

1

The premise that you don't need to split the messages before encryption is incorrect. Given that there are schemes giving (partial) message recovery for signing also means that that premise is incorrect as well. The message doesn't need to be split up before signing (which includes hashing), so that premise, finally, is also incorrect. Encryption operation, ...

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