RSA Use in Encryption

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Lecture 6
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RSA Use in Encryption

• to encrypt a message M the sender
• obtains public key of recipient PUe,n
• computes C Me mod n, where 0Mltn
• to decrypt the ciphertext C the owner
• uses their private key PRd,n
• computes M Cd mod n
• note that the message M must be smaller than the
  modulus n (block if needed)

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RSA Example - En/Decryption

• sample RSA encryption/decryption is
• given message M 88
• Publish public key PU7,187
• Keep secret private key PR23,187
• encryption
• C 887 mod 187 11
• decryption
• M 1123 mod 187 88

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Assignments

• Perform encryption and decryption using RSA
  algorithm, as in Figure 1, for the following
• p 3 q 11, e 7 M 5
• p 5 q 11, e 3 M 9
• In a public-key system using RSA, you intercept
  the ciphertext C 10 sent to a user whose public
  key is e 5, n 35. What is the plaintext M?

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Why RSA Works

• because of Euler's Theorem
• aø(n)mod n 1 where gcd(a,n)1
• in RSA have
• np.q
• ø(n)(p-1)(q-1)
• carefully chose e d to be inverses mod ø(n)
• hence e.d1k.ø(n) for some k
• hence Cd Me.d M1k.ø(n) M1.(Mø(n))k
• M1.(1)k M1 M mod n

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Efficient Encryption

• Encryption uses exponentiation to power e
• Hence if e small, this will be faster
• often choose e65537 (216-1)
• also see choices of e3 or e17
• But if e too small (eg e3) can attack
• using Chinese remainder theorem 3 messages with
  different moduli

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Efficient Decryption

• Decryption uses exponentiation to power d
• this is likely large, insecure if not
• can use the Chinese Remainder Theorem (CRT) to
  compute mod p q separately. then combine to get
  desired answer

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Exponentiation

• can use the Square and Multiply Algorithm
• a fast, efficient algorithm for exponentiation
• concept is based on repeatedly squaring base
• and multiplying in the ones that are needed to
  compute the result
• look at binary representation of exponent
• only takes O(log2 n) multiples for number n
• eg. 75 74.71 3.7 10 mod 11
• eg. 3129 3128.31 5.3 4 mod 11

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Exponentiation

• The algorithm for computing ab mod n
• The b integer is expressed as as binary number
  bk , bk-1 ,, b0

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Example

• Get 7 5 mod 11 ?? ? ab mod n as in th
  algorithm
• So b 5 ? 101
• n 11 a 7
• The final result is 10

1 0 1
b0 b1 b2
d c Ask? b(k) k
1 0
1 0 2
7 1 Yes b2 1
5 2 1
No
3 4 0
10 5
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Example

• Get 10 3 mod 60 ?? ? ab mod n as in th
  algorithm
• So b 3 ? 11
• n 60 a 10
• The final result is 40

1 1
b0 1b2
d c Ask? b(k) k
1 0
1 0 1
10 1 Yes b1 1
40 2 0
40 3 Yes b0 1
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RSA Security

• three approaches to attacking RSA
• brute force key search (infeasible given size of
  numbers)
• mathematical attacks (based on difficulty of
  computing ø(N), by factoring modulus N)
• timing attacks (on running of decryption)

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Factoring Problem

• mathematical approach takes 3 forms
• factor Np.q, hence find ø(N) and then d
• determine ø(N) directly and find d
• find d directly
• currently believe all equivalent to factoring
• have seen slow improvements over the years
• as of Aug-99 best is 130 decimal digits (512) bit
  with GNFS
• biggest improvement comes from improved algorithm
• cf Quadratic Sieve to Generalized Number Field
  Sieve
• barring dramatic breakthrough 1024 bit RSA
  secure
• ensure p, q of similar size and matching other
  constraints

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Summary

• have considered
• prime numbers
• Fermats and Eulers Theorems
• Primality Testing
• Chinese Remainder Theorem
• Discrete Logarithms
• principles of public-key cryptography
• RSA algorithm, implementation, security