EECS 690

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EECS 690

• Weichao Wang

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• Number theory and background
• Modular arithmetic
• The complete set of residues 0 to n-1
• Properties of modular arithmetic
• (ab) mod n ((a mod n) (b mod n)) mod n
• (ab) mod n ((a mod n) (b mod n)) mod n
• a(bc) mod n (((ab) mod n) (ac) mod n))
  mod n
• These features help the cryptographers a lot, for
  example, how to you calculate (a16) mod n
• What about the number that is not 2x

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• Prime numbers
• We use p(n) to represent the number of prime
  numbers that are lt n
• First, the mathematicians find that when n --gt
  infinity, p(n) / n goes to 1 / ln(n), so we have
  enough prime numbers

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• Greatest Common Divisor (gcd)
• Two numbers are relatively prime if gcd(a, b) 1
• If we assume that a gt b,
• If a is bs multiple, gcd(a, b) b
• Otherwise gcd (a, b) gcd (b, a b)
• This is called Euclids algorithm
• Prove that the Euclids algorithm will finish in
  O(log_2(a)) rounds

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• Inverse modular of a number
• If (ax) mod n 1, then x is called as inverse
  modular to n
• General problem, if x is as inverse modular to
  n, can we find a number that ay mod n b?
• Do we have (x mod a) mod b (x mod b) mod a??
• Usually, if gcd(a, n) 1, there exists a unique
  solution of inverse modular
• Extended Euclidean algorithm can be used to
  calculate the inverse modular
• gcd(a, b) min (ax by gt 0)

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• Reduced set of residues and Fermats little
  theorem
• Reduced set of residues the subset of residues
  that is relatively prime to n, for example, n
  12, the reduced set of residues include 1, 5, 7,
  11. Its size is called F(n).
• If n is a prime, F(n)n-1
• If n pq and p, q are prime numbers, then F(n)
  (p-1)(q-1)
• If x mod p 1, and x mod q 1, p and q are two
  different prime numbers, x mod pq ?

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• Fermats Little theorem
• Fermats little theorem if n is a prime number
  and a is not multiple of n, then a(n-1) mod n
  1
• Eulars generalization of Fermats little
  theorem If gcd(a, n) 1, then a(F(n)) mod n
  1.
• Using this method, as modular inverse is
  a(F(n)-1). Pay attention to the requirement of
  gcd(a, n) 1
• An example, what is 5s inverse modular to 7

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• Using Fermats little theorem to examine whether
  a number is a prime number
• For p, select a number a and calculate a(p-1)
  mod p. If it is not 1, we know p is not prime
  number
• What is for all a, a(p-1) mod p 1?
• There exists a set of number called Carmichael
  numbers, they satisfy the above test (to the base
  that they are relatively prime), but they are
  pseudo primes

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• Chinese remainder theorem
• If a number n can be factored as the product of a
  group of primes n p1 p2 --- pt, then a
  number x lt n can be uniquely represented by the
  residues mod these prime numbers
• For example, 22 mod 5 2, 22 mod 7 1, for the
  numbers lt 35, 22 is the only number satisfying
  these conditions. Therefore, 22 can be uniquely
  represented by these residues (2, 1)
• If p and q are different prime numbers, how can
  we find x lt pq and x mod p a, x mod q b, agt
  b
• Since p and q must be relatively prime, we can
  find (uq mod p 1)
• Then x (((a-b) u) mod p) q b

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• How to calculate x
• x mod p1 x1, x mod p2 x2, ---, x mod pt xt
• x (p2p3---pt)y1x1
• (p1p3---pt)y2x2 ----
• (p1p2---p(t-1))ytxt
• x mod p1 (p2p3---pt)y1x1 mod p1
• We know that p1 and p2p3---pt are relatively
  prime, so there exists an inverse modular for
  p2p3---pt, let that be y1
• Then x mod p1 x1
• Similarly, we can find out y2, y3, ---, yt

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• How does RSA works?
• Select 2 large prime numbers p and q, n pq
• Now select a public key e where gcd(e,
  (p-1)(q-1)) 1
• Since e and (p-1)(q-1) are relatively prime, we
  can find out ed mod (p-1)(q-1) 1. d is the
  modular inverse of e. d is the private key, e is
  the public key
• For a message m, me mod n c, cd mod n m
• Now we prove m(ed) mod n m
• Question Why both e and d are odd numbers?

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• RSA
• Make e and n public, so everyone can encrypt
• What if the plaintext m is multiple of p or q?
  The secret key d will be compromised
• Avoid such attacks Make m shorter than both p
  and q
• In RSA, E(m1) E(m2) E(m1 m2)
• Now we can explain why RSA is not used for
  digital signature?

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• Common modulus attack to RSA
• We assume that two parties choose the same npq,
  but different public keys e1 and e2
• If we ever encrypt the same message m with both
  public keys and gcd(e1, e2) 1, we are in
  trouble
• Malicious node can find e1x e2y 1, so
  (me1)x (me2)y m (e1xe2y) mod n m
• What if xlt0? Calculate the inverse modular
• Lesson we learn do not share n with among users