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Public‐Key Cryptography and RSA

CSE 651: Introduction to Network Security

Abstract

• We will discuss– The concept of public-key cryptography

– RSA algorithm

– Attacks on RSA

• Suggested reading: – Sections 4.2, 4.3, 8.1, 8.2, 8.4

– Chapter 9

2

Public-Key Cryptography

• Also known as asymmetric-key cryptography.• Each user has a pair of keys: a public key and

a private key.• The public key is used for encryption.

– The key is known to the public.

• The private key is used for decryption.– The key is only known to the owner.

3

4

Bob Alice

Why Public-Key Cryptography?

• Developed to address two main issues:– key distribution– digital signatures

• Invented by Whitfield Diffie & Martin Hellman 1976.

5

6

1

1

trapdoor

Easy:

Hard:

Easy:

Use as the private key.

Many public-key cryptosystems are based on trapdoor one-way fu

trapdoor

One-way function with trapdoorf

f

f

x y

x y

x y

−

−

⎯⎯→

←⎯⎯

←⎯⎯⎯

•nctions.

Modular Arithmetic

Mathematics used in RSA(Sections 4.2, 4.3, 8.1, 8.2, 8.4)

| : divides , is a divisor of . gcd( , ): greatest common divisor of and . Coprime or relatively prime: gcd( , ) 1. Euclid's algorithm: computes gcd( , ). Extented Eucl

Integersa b a b a b

a b a ba ba b

••• =•• id's algorithm: computes integers and such that gcd( , ).x y ax by a b+ =

0

1

1

1 1

Comment: compute gcd( , ), where 1.

:

:

for : 1, 2, until = 0

: mod

return ( )

Euclidean Algorithm

n

i i i

n

a b a b

r a

r b

i r

r r r

r

+

+ −

> >

=

=

=

=

…

1 782 65

1 13

6

299 221221 7878 65

5 5 13 0

gcd(229,221) 13( 2 ) 3

3 (299 221) 221

299 2

78 6578 221 78 78 22

2114

1

3

Extended Euclidean Algorithm:Example

gcd(299,221) ?= ⋅ += ⋅ += ⋅ += ⋅ +

= = −= − − ⋅ = ⋅ −= ⋅ − ⋅ −= ⋅ − ⋅

=

id

A group, denoted by ( , ), is a set with a binary operation : such that 1. ( ) ( ) (associative) 2. s.t. , ( ) 3. , s.t.

entity

GroupG G

G G Ga b c a b c

e G x G x x xx G y G x y y x

e e

•× →

=∃ ∈ ∀ ∈ = =∀ ∈ ∃ ∈ = = ( )

A group ( , ) is if , , .

Examples: ( ,

abel

), ( , ), ( \ {0}, ), ( ,

inver

), ( \ {0}

ian

s

, ).

ee

G x y G x y y x

Z Q Q RR

• ∀ ∈ =

• + + × +×

Let 2 be an integer.

Def: is congruent to modulo , written

mod , if | ( ), i.e., and have the

same remainder when d

m

ivided by .

Note: dod an

Integers modulo n

a b n

a b n n a b a b

n

a b n

n•

•

• ≡

≥

≡ −

{ } are different.

Def: [ ] all integers congruent to modulo .

[ ] is called a residue calss modulo , and is a

representati

mod

ve of that class.

n

n

a b

a a n

a n a

n

•

=

=

•

[ ] [ ] if and only if mod .

There are exactly residue classes modulo :

[0], [1], [2], , [ 1].

If [ ], [ ], then [ ] and [ ].

Define addition and multiplication

n na b a b n

n n

n

x a y b x y a b x y a b

= ≡

−

∈ ∈ + ∈

•

+• ⋅

•

•

∈ ⋅

…

for residue classes:

[ ] [ ] [ ]

[ ] [ ] [ ].

a b a b

a b a b

+ = +

⋅ = ⋅

{ }

{ }

( )

Define [0], [1], ..., [ 1] .

Or, more conveniently, 0, 1, ..., 1 .

, forms an abelian additive group.

For , ,

( ) mod . (Or, [ ] [ ] [ ] [ mod ].)

0 is th

n

n

n

n

Z n

Z n

Z

a b Za b a b n a b a b a b n

= −

= −

+

∈+

•

= + + =

•

•+ =

•

+ii

10

e identity element. The inverse of , denoted by , is .

When doing addition/substraction in , just do the regular addition/substraction and reduce the result modulo .

In , 5

n

a a n a

Zn

Z

•

− −i

i 5 9 4 6 2 8 3 ?+ + + + + + + =

( )

( )

1

1

1

, is not a group, because 0 does not exist.

Even if we exclude 0 and consider only \ {0},

, is not necessarily a group; some may not exist.

For , exists if and on

n

n n

n

n

Z

Z Z

Z a

a Z a

−

+

+ −

−

•

•

•

∗

=

∗

∈ ly if gcd( , ) 1.

gcd( , ) 1 1 for some integers and

[ ] [ ] [ ] [ ] [1] in

[ ] [ ] [1] in

[

n

n

a n

a n ax ny x y

a x n y Z

a x Z

=

= ⇔ + =

⇔ ⋅ + ⋅ =

⇔ =

⇔

•

⋅1] [ ] in na x Z− =

{ }

( )

*

*

1

Let : gcd( , ) 1 .

, is an abelian multiplicative group.

mod .

mod . 1 is the identity element. The inverse of , written , can be computated

n n

n

Z a Z a n

Z

a b ab n

a b ab n

a a −

= ∈ =

∗

∗ =

∗

•

•

=

•

iii

{ }*12

*

by the Extended Euclidean Algorithm.

For example, 1,5,

Q: How many

7,11 . 5

ele

7 35 mod12 11.

ments are t here in ? nZ

Z = ∗•

•

= =

{ }

*

*

( ) : 1 , gcd( ,

Euler's totient function:

Facts:

) 1

1. ( ) ( 1) for prime .

2. ( ) ( ) ( ) if gcd( , ) 1.

How many elements are there in ?

n

n

n Z a a n a n

p p p

pq p q p q

Z

ϕ

ϕ

ϕ ϕ ϕ

•

= = ≤ ≤ =

= −

=

•

=

Let be a (multiplicative) finite group. The order of , written , is the smallest

p ( , identity elos emitiv ent.e integer such that . The order of ,

ord( )

ord( , is the number )

)

t

a

Ge

Ga G

t a eG

•• ∈

=•

*15

* *15 15

2

3 2

of elements in . Example: Consider

ord( ) (15)

ord(8) 4, since 8 64 mod15 4 8 (8 8) mod15 (4 8) mod15 2

GZ

Z Z ϕ

•

= =

= = =

= ⋅ = ⋅ =

i

i

4 2 2 8 (8 8 ) mod15 (4 4) mod15 1= ⋅ = ⋅ =

{ }*15

*15

*15

* (15) 815

= 1, 2, 4, 7, 8, 11, 13, 14

(15) (3) (5) 2 4 8

: 1 2 4 7 8 11 13 14ord( ) : 1 4 2 4 4 2 4 2

For all , we have 1

Example: 15Z

Z

a Za

a Z a a

n

ϕ

ϕ ϕ ϕ

•

•

•

•

= = × = × =

∈

∈ = =

=

( ) 1

ord( )

*

Theorem: For any element , ord( ) | ord( ).

Fermat's little theorem: If ( a prime), th

Corollary: For any element ,

en 1.

Euler's theorem:

.

If

pp

p

G

a G a G

a p a

a G a e

a

a

Z ϕ −

• ∈

•

•

∈ = =

•

∈ =

∈ ( ) *

(

*

)

, then 1. ( ord( ) ( ).)

That is, for any integer relatively prime to , 1 mod .

nnn

n

a Z n

a na

Z

n

ϕ

ϕ

ϕ= =

≡

∵

The Chinese Remainder Problem• A problem described in an ancient Chinese

arithmetic book.• Problem: We have a number of things, but we

do not know exactly how many. If we count them by threes we have two left over. If we count them by fives we have three left over. If we count them by sevens we have two left over. How many things are there?

[ ] [ ] [ ] [ ]3 5 7 105

2mod3, 3mod5, 2mod7? All integers 2 3 2 23 .

x x xx x≡ ≡ ≡

= ∈ =∩ ∩

( )

1

1 1

2 2

10

If integers , , are pairwise coprime, then the system of congruences

modmod

modhas a unique solution in :

mod

Chinese remainder theorem

k

k k

N

i i i i

n n

x a nx a n

x a nZ

x a N N n−

≡⎧⎪ ≡⎪⎨⎪⎪ ≡⎩

=

…

[ ]

1

1 2

0

mod

where , and .

Furthermore, every integer is a solution.

k

i

k i i

N

N

N n n n N N n

x x

=

= =

∈

∑…

1 1 1

1 1 1

Suppose 1 mod 3 6 mod 7 8 mod 10By the Chinese remainer theorem, the solution is:

1 70 (70 mod3) 6 30 (30 mod7) 8 21 (21 mod10) 1 70 (1 mod3) 6 30 (2 mod7) 8 21 (1 mod10)

xxx

x − − −

− − −

≡≡≡

≡ × × + × × + × ×

≡ × × + × × + × × 1 70 1 6 30 4 8 21 1 mod 210

958 mod 210 118 mod 210

≡ × × + × × + × ×≡≡

Example: Chinese remainder theorem

( )( )

1 1 2

1 2

* * *

1

(the numbers are pairwise coprime) There is a one-to-one correspondence :

also,

, , , where

Chinese remainder theorem (another version)

k

k i

N n n N n n

k N

N n n n n

Z Z Z Z Z Z

A a a A Z

• =•

↔ × × ↔ × ×

↔ ∈…

( )1

and mod ? ? , ,

i i

k

a A nA

a a

=

• →

• ← …

( )

( )( )

1

1

1

1

1 1

One-to-one correspondence :

, ,

Operations in can be performed individually in each .

, , If

, ,

then, ,

k

i

N n n

k

N n

k

k

Z Z Z

A a aZ Z

A a aB b b

A B a b

•↔ × ×

↔

•

↔⎧⎨ ↔⎩

± ↔ ±

…

……

…( )( )( )

1 1*

1

1 1

, ,

, , if

mod mo d mod

k k

k k

k k N

k

a bA B a b a bA B a b a

N n n

b B Z

±× ↔ × ×÷ ↔ ÷ ÷↑ ↑

∈↑

……

( )( )( )

15

* * *15 3 5 15 3 5

Suppose we want to compute 8 11 in .

8 (2, 3) 8mod3, 8mod5

11 (2, 1) 11mod3, 11mod5 8 11 (2 2, 3 1) (1, 3). (1, 3)

Example: Chinese remainder theorem

Z

Z Z Z Z Z Z

x

• ×

• ↔ × ↔ ×

↔ =

↔ =

× ↔ × × =• ↔

1mod3 Solve 13

3mod5x

xx≡⎧

• ⇒ =⎨ ≡⎩

1

3

gcd( , ),

mod ,

mod

Can be done in (log ) time.

Important Problems

k

a b

a n

a n

O n

−

•

•

•

•

1

1 *

1

Compute in .

exists if and only if gcd( , ) 1. Use extended Euclidean algorithm to find ,

such that gcd( , ) 1 1 (beca

mod ?How to compute na Z

a a nx y

ax ny a na

a

x

n

−

−

−•

• =•

+ = =⇒ =

1

use 0 in )

. Note: every computation is reduced modulo .

nny Z

a xn

−

=

⇒ =•

1 Compute 15 mod 47. 47 15 3 (divide 47 by 15; remainder 2) 15 2 7 (divide 15 by 2; remainder 1) 1 15 7 (mod 47) 1

21

25 ( ) 7 (mod 47)47 15 3

Example−•

= × + == × + == − ×= − ×− ×

1

15 22 47 7 (mod 47) 15 22 (mod 47) 15 mod 47 22−

= × − ×= ×

=

30

By ivest, hamir & dleman of MIT in 1977. Best known and most widely used public-key scheme. Based on the one-way property

of mo

R S

du

lar powering:

A

assumed

The RSA Cryptosystem•••

1

: mod (easy)

: mod (hard)

e

e

f x x n

f y y n−

→

→

1

RSA

RSA

*

Encryption (easy):

Decryption (hard):

Looking for a trapdoor: ( ) .If is a number such that 1mod ( ), then

( )

It works in group

1

.

Idea behind RSA

e

e

e d

n

x x

x x

x xd ed n

e n

Z

d kϕ

ϕ

−

⎯⎯⎯→

←⎯⎯⎯

=≡

= +

( )( ) 1 ( )

for some , and

( ) 1 .ke ed n nd k

k

x x x x x x xϕ ϕ+= = = ⋅ = ⋅ =

Setting up an RSA Cryptosystem

• A user wishing to set up an RSA cryptosystem will:– Choose a pair of public/private keys: (PU, PR).– Publish the public (encryption) key.– Keep secret the private (decryption) key.

32

*( )

Select two large primes and at random. Compute . Note: ( ) ( 1)( 1). Select an encryption key satisfying 1 ( ) and

gcd( , ( )) 1. (i.e., , 1.)

Co

RSA Key Setup

n

p qn pq n p q

e e ne n e Z eϕ

ϕϕ

ϕ

= = − −< <

∈ ≠

•

•

=

••

1mpute the descryption key: mod ( ). 1 mod ( ). is the inverse of mod ( ). Public key: . Private key: . Important: , , and ( ) must be kep

( , ) ( ,t sec

e .)

r tPU n e P

d e ned nd e n

np

dn

Rq

ϕϕ

ϕ

ϕ

−

••

=

=

≡

=

ii

Alice

Suppose Bob is to send a secret message to Alice. To encrypt, Bob will

obtain Alice's public key { , }.

encrypt as mod . Note:

RSA Encryption and Decryption

e

m

PU e n

m c m nm

••

=

=

iii *

Alice

. To decrypt the ciphertext , Alice will compute

mod , using her private key { , }. What key will Alice use to encrypt

her reply to Bob?

n

d

Zc

m c n PR d n•

•

∈

= =i

( )

{ }

*

*

*

* *

The settig of RSA is the group , :

Plaintexts and ciphertexts are elements in .

Recall: : 0 , gcd( , ) 1 .

has ( ) elements. (The group h a

s

Why RSA Works

n

n

n

n n

Z

Z

Z x x n x n

Z n Zϕ

= <

•

< =

ii

i

i

( )

( )

* * ( )

*

order ( ).)

In group , , for any , we have 1.

We have chosen , such that 1 mod ( ), i.e., ( ) 1 for some positive integer .

For ,

nn n

de edn

n

Z x Z x

e d ed ned k n k

x Z x x

ϕ

ϕ

ϕϕ

∈ =

≡= +

∈ =

ii

i

i ( )( ) 1 ( ) .kk n nx x x xϕ ϕ+= = =

Select two primes: 17, 11. Compute the modulus 187. Compute ( ) ( 1)( 1) 160. Select between 0 and 160 such that gcd( ,160) 1.

Let 7. Compute

RSA Example: Key Setupp q

n pqn p q

e ee

d

ϕ

= == =

••• = −

•

==

•− =

1 1mod ( ) 7 mod160 23 (using extended Euclid's algorithm). Public key: . Private k

( ,ey

) (7, 187)( , ) (23: ., 7 18 )

PU e n

e

P n

n

R d

ϕ− −

= == =

=

•

= =

•

7

23

23

23

Suppose 88. Encryption: mod 88 mod187 11. Decryption: mod 11 mod187 88. When computing 11 mod187, we first

compute 11 andd

theo

not

n

RSA Example: Encryption & Decryption

e

d

mc m nm c n

•

•

•

•

=

= = =

= = =

23

reduce it modulo 187. Rather, when conmputing 11 , reduce the intermediate

results modulo 187 whenever they get bigger than 187.•

( )

1 0

2

Comment: compute mod , where in binary.

1 for downto 0 do mod

if 1

then mod

Algorithm: Square-and-Multiply( , , )c

k k

i

x n c c c c

zi k

z z nc

z z x

x c n

−=

←←

←

=

← ×

…

( )

...Note: At

i.e.,

the e

mod

nd of

retu

iteratio

rn

n , .

( )

k

i

i

c

c ci

z z x n

z

z

n

x=

⎫⎪ ← ×⎬⎪⎭

2

2

2

2

3

2

23 10111

1 11 mod 187 11 (square and multiply) mod 187 121 (square) 11 mod 187 44 (square and multiply) 11 mod 187 165 (square and

11 mod187

mu

Example:

b

zz zz zz zz z

=

←

← ⋅ =

← =

← ⋅ =

← ⋅ =2

ltiply) 11 mod 187 88 (square and multiply)z z← ⋅ =

4 16

To speed up encryption, small values are usually used for .

Popular choices are 3, 17 2 1, 65537 2 1. These values have only two 1's in their binary representation.

Encryption Key

e

e

= + = +

•

•

• There is an interesting attack on small .e

A message sent to users who employ the same encryption expo

nent is not protected by RSA.

Say, 3, and Bob sends a message to three receipients encry

Low encryption exponent attackm e

ee m

•

• =

1 2 3

1 3

1 2 3

3 3 31 2 2 3

31

3 3 3

2 3

pted as: mod , mod , mod .

Eve intercepts the three ciphertexts, and recovers : mod , mod , mod .

By CRT, mod for som

m mc n c n c n

mm c n m c n m c n

m c n

m

n n

= = =

≡ ≡ ≡

•

≡

i

i 1 2 3

3 3 31 2 3

e .

Also, . So, , and .

c n n n

m n n n m c m c

<

< = =i

1/4

One may be tempted to use a small to speed up decryption.

Unfortunately, that is risky.

Wiener's attack: If /3 and 2 , then the decryption exponent

Decryption Key d

dd n p q p

d•

•

• < < <can be computed

from ( , ). (The condition 2 often holds in practice.)

CRT can be used to speed up decryption by four times.

n e p q p< <

•

2 Multiplying two numbers of bits takes ( ) time. . 1024-2058 bits. , half the size. Decryption: . Instead of computing

mod direod ctlm

Speeding up Decryption by CRT

d

d

c

k O kn pq n

nc

p

n

q•

= Šť

•

•

( ) ( ) 1 2

1 1 2 2

1

2

y, we compute mod and mod

compute mod and mod

mod recover the plaintext b

y solv

ing o

m d

d d

c c p c c q

m c p m c q

x m px m q

= =

= =

≡⎧⎨ ≡⎩

i

i

i

( )

Four categories of attacks on RSA: brute-force key search

infeasible given the large key space math

ematical attacks timing attacks

chosen ciphe r

Security of RSA•i

iii text attacks

1

Then ( ) ( 1)( 1) andmod ( ) can be calculated

Factor into .

Determine ( ) directly

easily.

Equivalent to factoring . Knowing ( ) will enable us to f

.

Mathematical Attacksn p q

d e n

nn

n pq

nϕ

ϕ

ϕ

ϕ

−

= − −

=

•

•actor by solving

( ) ( 1)( 1)

Best known algorithms are not faster than those for factoring . Besides, if is know

Detn,

then can be factored with

erm

hi

ine dir

g

ectly.d

nn pq

n p q

n dn

ϕ=⎧

⎨ = − −

•

⎩

h probability.

A difficult problem, but more and more efficient algorithms have been developed.

In 1977, RSA challenged resea

rchers to decode a ciphertex encrypted with a key ( ) of

Integer Factorization

n

•

•129 digits

(428 bits). Prize: $100. Would take quadrillion years using best algorithms of that time.

In 1991, RSA put forward more challenges, with prizes, to encourage research on f•

actorization.

Each RSA number is a semiprime. (A number is semiprime if it is the product of two primes.) There are two labeling schemes.

by the number of decimal digits: RSA-100, .

RSA Numbers•

•i

.., RSA-500, RSA-617. by the number of bits: RSA-576, 640, 704, 768, 896, 1024, 1536, 2048.i

RSA-100 ( bits), 1991, 7 MIPS-year, Quadratic Sieve. RSA-110 ( bits), 1992, 75 MIPS-year, QS. RSA-120

3323653 ( bits), 1993, 830 MIPS-year, QS.

RSA-129 98

4(

RSA Numbers which have been factored•••• bits), 1994, 5000 MIPS-year, QS. RSA-130 ( bits), 1996, 1000 MIPS-year, GNFS. RSA-140 ( bits), 1999, 2000 MIPS-year, GNFS. RSA-155 ( bits), 1999, 8000 MIPS-year, GNFS.

284

314655

RSA-161

0 (2

530

••••

576 6

bits), 2003, Lattice Sieve. RSA- (174 digits), 2003, Lattice Sieve. RSA- (193 digits), 2005, Lattice Sieve. RSA-200 ( bits), 2005, Lattice

40663 Sieve.

•••

49

RSA-200 =

27,997,833,911,221,327,870,829,467,638,

722,601,621,070,446,786,955,428,537,560,

009,929,326,128,400,107,609,345,671,052,

955,360,856,061,822,351,910,951,365,788,

637,105,954,482,006,576,775,098,580,557,

613,579,098,734,950,144,178,863,178,946,

295,187,237,869,221,823,983.

*

In light of current factorization technoligies, RSA recommends that be of 1024-2048 bits.

Encrypting messages \ is insecure. Such an is not relatively prime to , i. e.,

Remarks

n n

n

m Z Zm n

∈

•

•i

*

gcd( , ) 1. By computing gcd( , ), one will be able to factor . gcd( , ) 1 gcd( , ) 1, where mod

Question: what is the probability that \

?

e

n n

m nm n n pq

m n c n c m n

m Z Z

>=

> ⇒ > =

∈•

ii

Paul Kocher in mid-1990’s demonstrated that a snooper can determine a private key by keeping track of how long a computer takes to decipher messages.

RSA decryption: mod .

Timing Attacks

dc n

•

•

( )

Countermeasures: Use constant decryption time Add a random delay to decryption time modify the ciphertext to

Blin and computeding

mod .

:d

c c

c n

′

′

•iii

1 2 1 2

RSA encryption has a homomorphism property: RSA( ) RSA( ) RSA( ). To decrypt a ciphertext RSA( ):

Generate a random secret mess

ag .

e

Blinding in Some of RSA Products

m

m m m mc m

r

× =•

=•×

i

1

Encrypt as RSA( ). Multiply the two ciphertexts: RSA( ). Decrypting yields a value e

qual to .

Multiplying that value by yields . Note: all calculation

r

mr m r

mr

r c rc c c mr

c mr

r m−

==

•

=iiii

*s are done in (i.e., modulo ).nZ n

RSA's homomorphism property is the basis of a simple chosen-ciphertext attack. The attacker intercepts a ciphertext . An oracle can decrypt ciphertexts, except ,

Chosen-Ciphertext Attacks

m

m

cc

•

•• for you. To launch a chosen-ciphertext attack:

Generate a message, say, 2. Encrypt as RSA( ). Multiply the two ciphertexts: RSA( ). Ask the oracle

to decryp

r

m r

rr c r

c c c mr

=

= =

•

=iiii

1

t , yielding a value equal to . Multiplying that value by yields the plaintext .

c mrr m−i

If the message space is small. The adversary can encrypt all messages and compare them with the intercepted ciphertext.

For inst

ance, if the message is known t

Small message space attack•

• o be a 56-bit DES key, or a social security number.

The RSA we have described is the basic or textbook RSA, susceptible to many attacks.

In real world, RSA is not used that way.

(now in

version 2.1) is a speRSA PK cificati onCS #1

Textbook RSA•

•

• by RSA Labs specifying how to implement RSA.

PKCS: ublic ey ryptography tandard. Let ( , , ) give a pair of RSA keys. Let denote the length of in bytes (e.g., 216). To encrypt a message :

P K C S

Padded RSA as in PKCS #1 v.1.5

n e dk

mk n

••

•=•

i

( ) ( )

pad so that 00 02 00 ( bytes) where 8 or more random bytes 00. original message must be 11 bytes.

the ciphertext is : RSA mo

d . This format makes RSA esis

r

e

r kr

k

c n

m

m

m

mm

m•

==

≤ −

=

′

′

≠

=′

ii

itant to many of the

aforementioned attacks Q: Whic h ones?

A padded message is called PKCS conforming if it has the specified format: 00 02 padding string 00 orig

PKCinal message.

usually send youS #1 implemen

(v.1.5

tati (sons

)RSA PKCS #1•

•1

ender) an error message if RSA ( ) is PKCS conforming. There was a famous chosen-ciphertext attack taking

advantage of these error messages. PKCS #1 (v 2.1) now uses a scheme call

n

e Opt

o

t

d

c−

•

• imal Asymmetric Encryption Padding (OAEP).

A message is called PKCS conforming if it has the specified format: 00 02 padding string 00 original message. PKCS #1 implementations usually

Bleichenbacher's chosen-ciphertext attack•

•1

1

send you (sender) an error message if RSA ( ) is PKCS conforming. It is just like you have an Oracle which, given , answers

whether or not RSA ( ) is PKCS conforming. Bleich

not

enbacher'

cc

c

−

−

•

• s attack takes advange of such an Oracle.

*

Given RSA( ), Eve tries to find . (Assume is PKCS conforming.)

How can Oracle help? Recall that RSA is homomorphic: RSA( ) =RSA( ) RSA( ) (computa

te

d in )

G n

c m mm

a b a b Z

=

⋅

•

•

⋅

i

i

i

*

*iven RSA( ), Eve can compute RSA( ) for any . She can ask the Oracle, Is PKCS conforming? (That is, is PKCS conforming?

)

Whmo

y is this info ed

us fu

n

n

m m s

msn

s

Z

Z

ms

⋅ ∈

∈

•

i

l?

8( 2) 8( 2)

8( 2) 8( 2)

Recall PKCS Format ( bytes): 00 02 padding string 00 original message

Let 00 01 0 2 (as a binary integer) Then,

2 00 02 0 and 3 00 03 0 .

If is PKCS c

o

k k

k k

k

BB B

m

− −

− −

= =

•

•

••

= =

modmo

nforming 2 3 . If, in addition, is PKCS con

dforming

2 3 2 3 for some 2 3 for some

m

ms nms n

msm

B B

B BB tn B tn tB s t n s B s t n s t

⇒ < <

⇒ < <⇒ + < < +⇒ + ⋅

•

⋅ < < +

• • •

• • •

0 n 2n 3n 4n

ns

2B 3B

If is PKCS conforming is in the blue area. If is also PKCS conforming

is in the blue area is in the red areas is in the red lines. Thus, is in the red line

mod

s

m

o

o

d

f t

m mms n

ms nmsm

m

⇒

⇒

•

⇒

•

⇒

•

he blue area.

blue area Let's focus on the blue area, (2B, 3B). If is PKCS conforming is in the . If is also PKCS conforming

mod

red areas/is in If is also PKCS conforming

modline

s

ms

ms nm

m

m

mn

•

•⇒

⇒

•

•⇒

′

purple areas/line is in So, blue red purple

sm• ∈ ∩ ∩

2B 3B

1 2 3

1

So, starting with the fact that is PKCS conforming, Eve finds a sequence of integers , , , ... such that

2 and mod is PKCS conforming.

To find , ra n

i i

i

i

ms s s

s sms n

s

−

•

≤

•

1domly choose an 2 , and ask the oracle whether is PKCS conforming. If not, then try a different .

This way, Eve can repeatedly narrow down the area containing , and even

d

mo

tu

i

mss

m

ns

s

−

•

≥

1 2 3

ally finds . For having 1024 bits, it takes roughly 1 million accesses

to the oracle in order to find , , , ...

mn

s s s•

( ) ( )( )( )

OAEP

RSA

The current version of PKCS #1 (v 2.1) uses a scheme called Optimal Asymmetric Encryption Padding (OAEP).

:

: mod

(v.2

.1

:

)RSA PKCS #1

e

m m m G r r h m G r

c m n

G

′⎯⎯⎯→ = ⊕ ⊕ ⊕

′⎯⎯⎯

•

•

→ =

•

pseudorandom generator : hash function : randomhr

••

Public-Key Applications

• Three categories of applications:– encryption/decryption (provide secrecy)– digital signatures (provide authentication)– key exchange (of session keys)

• Public-key cryptosystems are slower than symmetric-key systems.

• So, mainly used for digital signatures and key exchange.

• RSA: basic RSA, textbook RSA

• Padded-RSA: PKCS #1 v.1.5

• Original message

• Padded message•

66