[PDF] RSA and ASYMMETRIC (PUBLIC-KEY) ENCRYPTION





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RSA Cryptosystem The RSA cryptosystem is a example of a “public

At the center of the RSA cryptosystem is the RSA modulus N. It is a positive and then Alice picks p and q so that equation (1) holds.



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Ø Permutation: RSA(M) = Me (mod N) where M?Z N Ø Trapdoor: d– decryption exponent Where e?d = 1 (mod ?(N) ) Ø Inversion: RSA(M) d = Me k??(N) +1 = M (mod N) Ø “Assumption”: no efficient alg can invert RSA without trapdoor Page 2 Textbook RSA is insecure Ø Textbook RSA encryption: • public key: (Ne) Encrypt: C = Me (mod N)



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RSA and ASYMMETRIC (PUBLIC-KEY) ENCRYPTION

RSA: what to remember The RSA function f(x) = xe mod N is a trapdoor one way permutation: Easy forward: given N;e;x it is easy to compute f(x) Easy back with trapdoor: Given N;d and y = f(x) it is easy to compute x = f 1(y) = yd mod N Hard back without trapdoor: Given N;e and y = f(x) it is hard to compute x = f 1(y) Nadia Heninger UCSD 21



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RSA With Low public exponent Ø To speed up RSA encryption (and sig verify) use a small e C = Me (mod N) Ø Minimal value: e=3 ( gcd(e ?(N) ) = 1) Ø Recommended value: e=65537=216+1 Encryption: 17 mod multiplies Ø Several weak attacks Non known on RSA-OAEP Ø Asymmetry of RSA: fast enc / slow dec • ElGamal: approx same time for both

How do you calculate RSA key?

    In fact that's just what happens during normal RSA key generation. You use that e ? d = 1 mod ?(n), and solve for d using the extended Euclidian algorithm. i.e., d is the multiplicative inverse of e mod ?(n). This is often computed using the extended Euclidean algorithm.

How do you factor n in RSA?

    Given ?(n) and n it's easy to factor n by solving the equations n = p ? q and ?(n) = (p ? 1) ? (q ? 1) for p and q. Remember that with RSA the number N is the product of two large secret primes. Let's call them P and Q. We will treat them as our unknowns: Now N is known, as part of the public key.

What is RSA and how does it work?

    Clifford Cocks, an English mathematician, had developed an equivalent system in 1973, but it was classified until 1997. A user of RSA creates and then publishes the product of two large prime numbers, along with an auxiliary value, as their public key.

RSA and ASYMMETRIC (PUBLIC-KEY) ENCRYPTION

Nadia Heninger UCSD 1

RSA generators

An RSA generator with security parameterkis an algorithmKrsathat returnsN;p;q;e;dsatisfying p;qare distinct odd primes

N=pq, and is called the (RSA) modulus

jNj=k, meaning 2k1N2k e2Z'(N)is called the encryption exponent d2Z'(N)is called the decryption exponent edmod'(N) = 1

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A formula for Phi

Fact:

Supp oseN=pqfor distinct primespandq. Then

'(N) = (p1)(q1):

Example:LetN= 15 = 35. Then the Fact says that

'(15) = (31)(51) = 8: As a check,Z15=f1;2;4;7;8;11;13;14gindeed has size 8.

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Recall

Given'(N) ande2Z'(N), we can computed2Z'(N)satisfying edmod'(N) = 1 via d MOD-INV(e;'(N)): We have algorithms to eciently test whether a number is prime, and we know that a random number has a pretty good chance of being a prime.

We use these facts to build RSA generators.

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Building RSA generators

Say we wish to havee= 3. (We will see that the smaller ise, the more ecient is encryption.) The generatorK3rsawith (even) security parameter kis as follows: repeat p;q$ f2k=21;:::;2k=21g;N pq;M (p1)(q1) until

N2k1andp;qare primeand gcd(e;M) = 1

d MOD-INV(e;M) returnN;p;q;e;d

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One-wayness of RSA

The following should be hard:

Given:N;e;ywherey=f(x) =xemodN

Find:x

Formalism picksxat random and generatesN;evia an RSA generator.

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One-wayness of RSA, formally

LetKrsabe a RSA generator andIan adversary.

GameOWKrsa

procedure Initialize (N;p;q;e;d)$ Krsa x $ ZN;y xemodN returnN;e;yprocedure Finalize(x0) return(x=x0)The ow-advantage ofIis Adv owKrsa(I) = Prh

OWIKrsa)truei

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Inverting RSA

Inverting RSA : givenN;e;yndxsuch thatxemodN=y

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Inverting RSA

Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodN Knowd

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Inverting RSA

Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodN

KnowdEASY

becaused=MOD-INV(e;'(N))

Know'(N)

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Inverting RSA

Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodN

KnowdEASY

becaused=MOD-INV(e;'(N))

Know'(N)EASY

because'(N) = (p1)(q1)

Knowp;q

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Inverting RSA

Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodN

KnowdEASY

becaused=MOD-INV(e;'(N))

Know'(N)EASY

because'(N) = (p1)(q1)

Knowp;q?

KnowNNadia Heninger UCSD 12

Factoring Problem

Given:NwhereN=pqandp;qare prime

Find:p;q

If we can factor we can invert RSA. We do not know whether the converse is true, meaning whether or not one can invert RSA without factoring.

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A factoring algorithm

AlgFACTOR(N)//N=pqwherep;qare primes

fori= 2;:::;lpN m do ifNmodi= 0then p i;q N=i;returnp;q

This algorithm works but takes time

O(pN) =O(e0:5lnN)

which is prohibitive.

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Factoring algorithms

AlgorithmTime taken to factorNNaiveO(e0:5lnN)Quadratic Sieve (QS)O(ec(lnN)1=2(lnlnN)1=2)Number Field Sieve (NFS)O(e1:92(lnN)1=3(lnlnN)2=3)Nadia Heninger UCSD 15

Factoring records

bit-length of numberWhen factoredAlgorithm used

4001993QS

4281994QS

4311996NFS

4651999NFS

5151999NFS

5762003NFS

7682009NFS

7952019NFS

8292020NFS

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Moduli sizes

We estimate that a 1024-bit RSA modulus provides 80 bits of security, meaning factoring it takes 2

80time.

Factorization of a 1024-bit modulus hasn't been done yet in public, but is within reach of large organizations. Longer moduli, like 2048 bits, have been recommended since around 2010. Just because factoring some large numbers seems to be hard does not mean factoringalllarge numbers is hard. For example, a random integer has probability 1=2 of having 2 as a prime factor. This is why RSA uses moduliNdesigned to resist known factoring algorithms.

Nadia Heninger UCSD 17

Choices of encryption exponent

Common choices aree= 3,e= 17 ande= 65;537. Why these?ebin(e)311

1710001

65,53710000000000000001

Recall that the modular exponentiation algorithm computingx7! x emodNusesc(b) modular multiplications per bitb2 f0;1gin the binary expansion bin(e), wherec(0) = 1 andc(1) = 2. So the fewer the number of 1s in bin(e), the faster is the operation.

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Textbook RSA is insecure

Choosing a good modulus and exponent is not enough to make an RSA implementation secure.

For example:

Lete= 3,Nhave 1024 bits, and encrypty=xemodN.

To avoid this, typical RSA implementationspadmessages before encrypting. But even this is hard to get right: the most common RSA padding scheme in use is broken if it throws an error when it discovers incorrect padding.

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RSA Video

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RSA: what to remember

The RSA functionf(x) =xemodNis a trapdoor one way permutation:

Easy forward: givenN;e;xit is easy to computef(x)

Easy back with trapdoor: GivenN;dandy=f(x) it is easy to computex=f1(y) =ydmodN Hard back without trapdoor: GivenN;eandy=f(x) it is hard to computex=f1(y)

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The quantum threat

On a quantum computer, Shor's algorithm can compute discrete logarithms and factor in polynomial time.

Eorts to build quantum computers are underway.

Eorts are underway to standardize public-key cryptography based on computational problems like nding short vectors in lattices for which there are currently no known ecient quantum algorithms.

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Two settings

Symmetric encryption:

Before Alice and Bob can communicate securely, they need to have a common secret keyKAB. If Alice wishes to also communicate with Charlie then she and Charlie must also have another common secret keyKAC. If Alice generatesKAB;KAC, they must be communicated to her partners over private and authenticated channels.

Asymmetric (public-key) encryption:

Alice has a secret decryption keydkthat is shared with nobody, and an associated public encryption keyekthat is known to everybody. Anyone (Bob, Charlie,:::) can use Alice's encryption keyekto send her an encrypted message which only she can decrypt.

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Syntax of a PKE scheme

A public-key (or asymmetric) encryption schemeAE= (K;E;D) consists of three algorithms that operate as follows: (ek;dk)$ K| generate an encryption keyekand matching decryption keydk C$ Eek(M) | encrypt messageMunder encryption keyekto get a ciphertextC. AlgorithmEmay be randomized. M0 Ddk(C) | decrypt ciphertextCunder decryption keydkto get an outputM02 f0;1g[ f?g.

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Correct decryption requirement

LetAE= (K;E;D) be an asymmetric encryption scheme. The correct decryption requirement is that

Pr[Ddk(Eek(M)) =M] = 1

for all (ek;dk) that may be output byKand all messagesMin the message spaceofAE. The probability is over the random choices ofE. This simply says that decryption correctly reverses encryption to recoverquotesdbs_dbs6.pdfusesText_11
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