Question 13 page 160

1
Question 13 page 160
n p x q
a.
m k x F(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 F(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)
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First, we prove that If n p x q, p and q are
primes Then a F(n) 1 (mod p) if a and p are
relatively prime 0 (mod p)
if a and p are not relatively prime And a F(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 F(n) a F(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 F(n) (cp) F(n) 0 (mod p)
• and a cp 0 (mod p)

Similar proof for q
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b.
Prove that a m 1 a (mod p) for all a

• a m 1 a kF(n) 1 (mod p)
• (a kF(n) (mod p) x a (mod p))
  (mod p)
• (a k ) F(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)F(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)

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Prove that a m 1 a (mod q) for all a

• a m 1 a kF(n) 1 (mod q)
• (a kF(n) (mod q) x a (mod q))
  (mod q)
• (a k ) F(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)F(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)

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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 F(n))
? a ed a kF(n) 1 (mod n)
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a ed a kF(n) 1 a (mod p) (proven before)
(1)
a ed a kF(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)

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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