# Tag Info

67

The $GF$ in $GF(p^n)$ is not a function — it just stands for "Galois field (of $p^n$ elements)". As for what a Galois field is, it's a finite set of things (which we might represent e.g. with the numbers from $0$ to $p^n-1$), with some mathematical operations (specifically, addition and multiplication, and their inverses) defined on them that let us ...

37

The extended Euclidean algorithm is essentially the Euclidean algorithm (for GCD's) ran backwards. Your goal is to find $d$ such that $ed \equiv 1 \pmod{\varphi{(n)}}$. Recall the EED calculates $x$ and $y$ such that $ax + by = \gcd{(a, b)}$. Now let $a = e$, $b = \varphi{(n)}$, and thus $\gcd{(e, \varphi{(n)})} = 1$ by definition (they need to be coprime ...

32

There are some widely used cryptographic algorithms which need a finite, cyclic group (a finite set of element with a composition law which fulfils a few characteristics), e.g. DSA or Diffie-Hellman. The group must have the following characteristics: Group elements must be representable with relatively little memory. The group size must be known and be a ...

21

First, we are talking about multiplications, so we work in $\mathbb{Z}_p^*$, not $\mathbb{Z}_p$. By definition, any integer $g \in \mathbb{Z}_p^*$ is the generator for... the subgroup generated by $g$, i.e. the set of $g^k \mod p$ for all integer values $k$. The order of $g$ is the smallest $k \geq 1$ such that $g^k = 1 \mod p$. For soundness (Alice and ...

19

How could this allow for a backdoor? Well, if you do DH modulo a composite, an attacker can recover the shared secret if they can solve the DH problem (or the DLog problem) modulo each of the primes that make up the composite. There are a couple of ways that could be used by someone who knows the factorization to solve the DLog problem easier than expected....

17

The quoted recommendations do little to prevent fields that are subject to the recent developments. Take the $\mathbb{F}_{2^{6120}}$ example: it clearly passes the field size criterion, but also the subgroup rule, as the group order $2^{6120} - 1$ has one $1536$-bit prime factor. Not all binary fields are affected equally, however. Both Göloğlu et al and ...

16

Not only does g not need to be a generator for the entire group, general practice is that it is not. As Thomas has mentioned, the order of $g$ is the smallest $k \ge 1$ such that $g^k = 1 \mod p$. Let $q$ be the order of the value $g$ we use. If $g$ is a generator for the entire group, then $q = p-1$, if not, it is some proper divisor of $p-1$. Now, if $... 16 This procedure is known as incremental search and his described in the Handbook of Applied Cryptography (note 4.51, page 148). Although some primes are being selected with higher probability than others, this allows no known attacks on RSA; roughly speaking, incremental search selects primes which could have been selected anyway and there are still ... 14 I think there are some gaps and some misunderstandings in what you say. A finite field or Galois field$GF(p^n)$is a collection of$p^nn$-dimensional vectors. Here,$p$is a prime, and each coordinate in a vector is an integer in the range$[0,p-1]$; that is, an element of$GF(p)$. Thus, $$\mathbf A = (a_0, a_1, \ldots, a_{n-1}), ~~ a_i \in GF(p)$$ is ... 13 There is a reduction from DL to RSA if the DL oracle accepts composite modulus. For prime modulus, a reduction is not known. I copied the following from this wikipedia page with minor edits. Let$n = p\,q$be an RSA modulus. Generate random integer$a$co-prime to$n$and a random integer$x$taken in an interval much larger than$n$, say$[1,1000n]$. ... 13$\phi(n)$is the order of the multiplicative group of the numbers in$\mathbb{Z}_n$.$\phi$is known as Euler's totient function. A consequence Lagrange's theorem is that any element of a group, raised to the order of the group is equal to the identity element. So, using$\phi(n)$ensures that decryption works. Since$ed\equiv 1\bmod{\phi(n)}$, we can say ... 13 It is indeed a hard problem - in fact, it is at least as hard as the square Diffie-Hellman problem (SDH), which states that given$(g,g^a)$, it is infeasible to compute$g^{a^2}$. It is a standard and well-studied assumption, and it can be reduced do CDH (correcting a previous version of this answer where I said it does not - I was confusing with the ... 12 No, there is no known test that we can run on a 2048 bit composite number that would indicate whether it was the product of two primes, or whether it was the product of more than two primes. About the closest we can get is a zero knowledge proof; we know how someone (who does know the factorization) can run an interactive proof with us that can demonstrate ... 11 There are two approaches to such a validation: Test: you can look at the number and decide without involving the person who gave it to you. Proof: The person who generated the number can also give you additional information that will convince you it is a correct RSA number. There are no tests for RSA numbers. There are proofs for RSA numbers, including "... 11 In RSA, the public key is$e$and private key is$d$, if:$ed=1 \mod{\phi (n)} $To rearrange:$d=e^{-1} \mod{\phi (n)}$In an public key system, it should be the case that one cannot compute the private key from the public key. Therefore, at least one of the variables should be kept private. In the above equation, everyone knows$e$, everyone can ... 11 Where does the$\phi(n)$part come from? Well, the actual requirement is that, if$n = pq$and both$p$and$q$are prime, we have:$de \equiv 1 \mod p-1de \equiv 1 \mod q-1$The first ensures that RSA encryption, followed by RSA decryption, will obtain the original value modulo$p$. The second ensures that RSA encryption, followed by RSA decryption, ... 11 I’m trying to understand which properties of a group are used in DHKE at each step. Actually, you can implement a DH-style operation in any semigroup; you need closure, and you need associativity (so$A^3 = A\times A \times A = (A \times A) \times A = A \times (A \times A)$is well defined), but other than that, you really don't need anything. You don't ... 10 According to this: To summarize: solving the discrete logarithm problem for a composite modulus is exactly as hard as factoring and solving it modulo primes. So, given your question "Would the ability to efficiently find Discrete Logs have any impact on the security of RSA?" the answer would be yes. Furthermore, if you can solve DLP for composite moduli, ... 10 Can an attacker learn some bits of a using this information? No. In the case of multiplication modulo a prime, we have, for any possible value of$a$, there is a unique value of$b$that makes$a \cdot b \bmod p$give any particular value of$c$in the range$(1, p-1)$. That is, even if we knew all the bits of$c$, no particular value of$a$are any more ... 9 There are no known implications of the ABC Conjecture to RSA. The ABC problem doesn't have even a superficial resemblance to the security of RSA. (The only point of connection is the fact that they both relate to prime numbers, but that is extremely thin. Much of number theory can say it is somehow related to prime numbers. It'd be like assuming that ... 9 What is Rijndael's finite field? Rijndaels finite field is$F=\mathrm{GF}(2^8)$with minimal polynomial$f(x)=x^8 + x^4 + x^3 + x + 1$. Formally, we have$F=\mathbb F_2[x] / (f)$but don't worry about that. So what does this mean? Well, elements of$F$should be thought of as polynomials over$\mathbb{F}_2$, with the added fact that the minimal polynomial ... 9 A trapdoor in a discrete log group was first suggested in 1992 by Daniel M. Gordon[1] in response to the recently proposal by NIST for the Digital Signature Standard (among hundreds of other responses[2] including an objection to the now-infamous random generation of the per-signature secret). Though the computational cost was too high for an academic ... 9 If you want to end up in the industry, I strongly doubt a PhD is a good investment of your time, regardless of the rest of this discussion. I believe a general purpose quantum computer, the kind that will send us all to relearn all our algorithm theory, is far from certain in 20 years. I won't even be surprised if someone publishes tomorrow a serious proof ... 9 I am not sure if this question should be considered on topic here, but I will answer anyway. Theorem: All prime numbers larger than$3$can be written as$6k+1$or$6k-1$for some natural number$k$. Proof: The remainder of a number modulo$6$is between$0$and$5$. If it is$1$or$5$, the above criterion holds. It remains to show that, if it is$0$,$2$,... 8 Checking for smoothness can be computationally expensive, depending on the size of the "small" primes (there is no "natural" definition of "small", one has to define an arbitrary limit). Also, it is not really useful. The need for non-smooth integers comes from the$p-1$factorization method. Let$n = pq\$ be a RSA modulus that we wish to factor. Now suppose ...

8

Yes, using Miller-Rabin with a random witness does give a practical factoring method. When you run the Miller-Rabin algorithm, it can end in one of three ways: The final value is not 1; this case causes Miller-Rabin to output "Composite" An intermediate value was not 1 or N-1, but the next value was 1; this causes Miller-Rabin to output "Composite" The ...

8

Let me try a simple explanation of NFS. I will necessarily skip lots of details, but I hope you will get the main ideas. The number field sieve algorithm (NFS) is a member of a large family: index calculus algorithms. All algorithms in the family, which can be used for factoring and discrete logarithms in finite fields, share a common structure: ...

8

The zerocoin paper mentions such a technique: implementers can use the technique of Sander for generating so-called RSA UFOs for accumulator parameters without a trapdoor and refers to: T. Sander, “Efficient accumulators without trapdoor extended abstract,” in Information and Communication Security, vol. 1726 of LNCS, 1999, pp. 252–262. I can'...

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