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Question 13 page 160

Question 13 page 160. n = p x q. a. m = k x Φ (n) = k x ( p – 1) x (q – 1). gcd (a, n ) = 1  a m ≡ a m (mod p – 1) (mod p). ≡ a k x (p – 1) x (q – 1) (mod p – 1) (mod p). ≡ a 0 ≡ 1 (mod p). n = p x q. m = k x Φ (n) = k x ( p – 1) x (q – 1).

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Question 13 page 160

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  1. Question 13 page 160 n = p x q a. m = k x Φ(n) = k x (p – 1) x (q – 1) gcd (a, n ) = 1  am≡ a m (mod p – 1) (mod p) ≡ a k x (p – 1) x (q – 1) (mod p – 1) (mod p) ≡ a0 ≡ 1 (mod p)

  2. n = p x q m = k x Φ(n) = k x (p – 1) x (q – 1) gcd (a, n ) = 1  am≡ a m (mod p – 1) (mod q) ≡ a k x (p – 1) x (q – 1) (mod q – 1) (mod q) ≡ a0 ≡ 1 (mod q)

  3. First, we prove that: If n = p x q, p and q are primes Then a Φ(n) ≡ 1 (mod p) if a and p are relatively prime ≡ 0 (mod p) if a and p are not relatively prime And a Φ(n) ≡ 1 (mod q) if a and q are relatively prime ≡ 0 (mod q) if a and q are not relatively prime • Proof: • If a and n are relatively prime • a Φ(n) ≡ a Φ(n) (mod p – 1) (mod p) ≡ a (p – 1) x (q – 1) (mod p – 1) (mod p) ≡ a0 ≡ 1 (mod p) Similar proof for q If gcd(a,p) = b ≠ 1 As p is prime  a = c x p and b = p So a Φ(n)≡ (cp) Φ(n)≡ 0 (mod p) and a ≡ cp ≡ 0 (mod p) Similar proof for q

  4. b. Prove that : a m + 1≡ a (mod p) for all a • a m + 1≡ a kΦ(n) + 1 (mod p) • ≡ (a kΦ(n) (mod p) x a (mod p)) (mod p) • ≡ (a k ) Φ(n) (mod p) x a (mod p)) (mod p) • ≡ 1 x a (mod p) if ak and p are relatively prime • ≡ a (mod p) • If ak and p are not relatively prime • (ak)Φ(n)≡ 0 (mod p) (proven above) And a ≡ 0 (mod p) (proven above) • a m + 1 ≡ 0 x 0 ≡ 0 (mod p) • a m + 1 ≡ a ≡ 0 (mod p)

  5. Prove that: a m + 1≡ a (mod q) for all a • a m + 1≡ a kΦ(n) + 1 (mod q) • ≡ (a kΦ(n) (mod q) x a (mod q)) (mod q) • ≡ (a k ) Φ(n) (mod q) x a (mod q)) (mod q) • ≡ 1 x a (mod q) if ak and q are relatively prime • ≡ a (mod q) • If ak and q are not relatively prime • (ak)Φ(n)≡ 0 (mod q) And a ≡ 0 (mod q) • a m + 1 ≡ 0 x 0 ≡ 0 (mod q) • a m + 1 ≡ a ≡ 0 (mod q)

  6. c. Prove that: a ed≡ a (mod n) for all a Recall that d is chosen such that ed ≡ 1 (mod (p – 1) x ( q – 1)) ≡ 1 (mod Φ(n))  a ed≡ a kΦ(n) + 1 (mod n)

  7. a ed≡ a kΦ(n) + 1≡ a (mod p) (proven before) (1) a ed≡ a kΦ(n) + 1≡ a (mod q) (proven) (2) (1)  a ed= u x p + a • a ed – a = u x p , u is integer However, (2)  a ed = v x q + a  a ed – a = v x q , v is integer • u x p = v x q  u = (v x q) / p As p, q are primes, so q/p cannot be integer, and u is an integer  v/p is integer  v = t x p • a ed = v x q + a = t x p x q + a  a ed≡ a (mod p x q)

  8. d. gcd(a, n) = 1, n = p x q p, q are large primes assume gcd(a,n) = b  a = u x b = p x q  in order that b ≠ 1: b = p or b = q or b = p x q  for gcd(a, n) ≠ 1 : a = k x p or a = k x q Large p p 2p 3p 4p 5p Small p

  9. As p and q are large primes, within a certain bound of values, the number of values that makes gcd(a, n) ≠ 1 is small i.e. the number of values between 2 consecutive numbers that make gcd(a, n) ≠ 1 is large • With large primes p and q, and we randomly choose a, then gcd(a, n) is likely to be 1

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