Applications of Number Theory

1
Applications of Number Theory
2
Learning Objectives

• Further results in number theory.
• Some applications of number theory
• computer arithmetic with large numbers
• public key cryptography

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Number Theory

• Theorem 1 gcd(a,b) sa yb (This is called
  the Extended Euclidean Algorithm).
• Examplegcd(57, 23)
• Execute the Euclidean algorithm keeping track of
  the quotients and remainders
• r0 57, r1 23
• 57 23 . 2 11 r2 11 q1 2
• 23 11 . 2 1 r3 1 q2 2
• 11 11x1 0 r4 0 q3 1
• gcd (57,23) 1.
• To obtain the desired linear combination
• 23 11.2 1 gt 1 23 2. 11
• 57 23.2 11 gt 11 57 2. 23
• Thus gcd(57,23) 1 23 2. 11 23 2.(57
  2.23) -2. 57 5.23

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Number Theory

• Definition if ab mod c 1 we say that a is the
  inverse of b mod c. Can also be written ab ? 1
  (mod c).
• Note If a has an inverse mod c, then it has an
  inverse which is lt c. Indeed if b is an inverse
  of a MOD c then so are all the integers b nc.
• A 1,2,4,5,8,10,11,13,16,17,19,20 are all the
  integers lt 21 that are relatively prime to 21.
  That is ?n?A (gcd(21,n) 1).
• Note 2x11 mod 21 1 8x8 1 mod 21 5x17
  1 mod 21. A little more effort and you can
  quickly check that ?n ? A ?m (nm mod 21 1). (In
  words every member of A has an inverse mod 21
  belonging to A).

5
Number Theory

• Lemma 1 a,b and c are positive integers.
  gcd(a,b) 1 (p) and a bc (q) then a c (r).
• (We have here 3 propositions, p, q and r. The
  lemma states that the compound proposition p ? q
  ? r is TRUE).
• Proof
• p ? 1 sa tb (Theorem 1).
• c sac tbc (multiplication by c)
• since a bc (q) a sac ? a tbc
• thus a sac tbc c. QED.
• Remark if a c ? b c MOD e then a ? b MOD e.
  That is in modular arithmetic a cancellation rule
  for addition holds. On the other hand, if ac ? bc
  MOD e there is no general cancellation rule.
• For example 14 ? 8 MOD 6 but 14 / 2 ? 1 (mod 6)
  while 8/2 ? 4 (mod 6).

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Number Theory

• Theorem 2 Let m, a, b, c be positive integers.
  Let ac ? bc (mod m) and gcd(c,m) 1 then a ? b
  (mod m). (the compound proposition p ? q ? r is
  TRUE where p ac ? bc (mod m), q gcd(c,m) 1
  and r a ? b (mod m) ).
• Proof
• p ? m ac bc c(a b) (m divides ac-bc)
• q ? m (a b) (by Lemma 1).
• But this means that a ? b (mod m) QED.
• Observations
• 1. If p is prime and p a1a2 . . . an then ?j
  such that p aj.2. Every integer n has a unique
  factorization into a product of primes.

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Number Theory

• Theorem 3 if a and b are relatively prime
  integers (b gt 1), then an inverse of a modulo b
  exists. Furthermore, this inverse is unique
  modulo b.
• This is a direct consequence of the Extended GCD.
  Indeed if GCD(a,b) 1 then there are integers n1
  and n2 such that n1a n2b 1. Or n1a MOD b 1
  because n1b ? 0 (mod b). So n1 is an inverse of a
  modulo b.
• In the example above gcd(57,23) 1 thus 57 has
  an inverse MOD 23.
• We have gcd(57,23) 1 -2x57 5x23.
• Thus 2 is an inverse of 57 MOD 23 and so is 2
  23 21. Indeed 5721 1197 2352 1.

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Number Theory

• Linear Congruences ax ? b (mod m)
• Example Solve 35x ? 36 (mod 41)
• Answer x 35.
• Verify 3535 1225 4129 36
• 3535 ? 36 (mod 41)
• How do we solve it?
• Recall a is the inverse of b (mod m) if ab ? 1
  (mod m)
• How to solve ax ? b (mod m)?
• Let y be the inverse of a (mod m).
• x by (mod m)

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Number Theory

• Example solve 72x ? 47 (mod 133)
• Step 1 find 72-1 (mod 133)
• 133 721 61
• 72 611 11
• 61 115 6
• 11 61 5
• 6 51 1
• gcd(133,72) 1.
• 1 6 5 26 11 261 1111
• 1361 1172 13133 2472 ( 1729 1728)
• so 72-1 (mod 133) 133 24 109.
• x 10947 (mod 133) 13338 69 (mod 133)
  69
• Verify 6972 (mod 133) 4968 (mod 133)
• 37133 47 (mod 133) 47.

10
Number Theory

• Theorem 1 gcd(a,b) sa yb (This is called
  the Extended Euclidean Algorithm).
• Lemma 1 a,b and c are positive integers.
  GCD(a,b) 1 (p) and a bc (q) then a c (r).
• Theorem 2 Let m, a, b, c be positive integers.
  Let ac ? bc (mod m) and gcd(c,m) 1 then a ? b
  (mod m). (the compound proposition p ? q ? r is
  TRUE where p ac ? bc (mod m), q gcd(c,m) 1
  and r a ? b (mod m) ).
• Theorem 3 if a and b are relatively prime
  integers (b gt 1), then an inverse of a modulo b
  exists. Furthermore, this inverse is unique
  modulo b.

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Number Theory

• Examplesgcd(35,78) a . 35 b . 78 29 . 35
  - 13 . 7878 2 . 35 835 4 . 8 38 2 .
  3 23 1 . 2 1gcd(35, 78) 11 3 -
  1.22 8 - 2 . 3 1 3 - 8 2 .3 3 . 3 - 83
  35 - 4 . 81 3 . 35 - 12 . 8 - 8 3 . 35 -
  13 .88 78 - 2 . 351 3 . 35 - 13 . 78 26 .
  35 29 . 35 - 13 . 78

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Number Theory

• Examples937 is an inverse of 13 modulo
  2436937 . 13 ? 1 (mod 2436) 937 . 13 12181
  5 . 2436 1

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Number Theory

• Examplesfind an inverse of 19 modulo
  141gcd(19, 141) 1, so there is an inverse of
  19 modulo 141.141 7 . 19 819 2 . 8
  38 2 .3 23 1 . 2 11 3 - 1 . 2 3
  - 8 2 . 3 - 8 3 . 31 -8 3 . (19 - 2 .
  8) -7 . 8 3. 191 -7. 141 49 . 19 3 .
  19 -7 . 141 52 . 19inverse of 19 (mod 141)
  52

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Number Theory

• Examplessolve the congruence 4x ? 5 (mod 9)x
  4 -1 . 5 an inverse of 4 (mod 9) is -2
  because 9 2. 4 1, which means 1 -2.4
  9x -2 . 5 -10 ? 8 (mod 9)

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Number Theory
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Number Theory
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Number Theory
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Number Theory

• Some of The Mathematics used in Encryption.
• Private key encryption both sender and receiver
  share the same secret key. For instance the key
  might be a 128 bit long binary sequence B. To
  encode break your message into 128 bit long
  chunks. For each chunk C send the 128 bit
  sequence B ? C. To decode, the receiver will
  retrieve B by calculating B ? (B ? C) C.
• There are two problems with this scheme
• Each pair of (sender, receiver) must have their
  own private keys.
• The key must be somehow sent to the sender (or
  receiver).
• To overcome this difficulty, Diffie and Hellman
  developed the idea of breaking each key into two
  parts a public part which every one can see and
  use and a private part which only the intended
  party will have. Thus, if a sender wishes to send
  a message to a receiver, hell use the receivers
  public key to encode the message. Only the person
  knowing the private key will be able to decode
  the message.
• A mathematical implementation of this idea was
  developed by Rivest, Shamir and Adelman (3
  mathematicians at MIT).

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Number Theory

• The RSA (Rivest, Shamir, Adelman) encryption
• Select two large primes p and q. Let m pq.
• Select a number e such that gcd(e, (p-1)(q-1))
  1.
• Tell the world that if anyone wants to send you
  a message R, send f(R) Re mod m. (if R gt m,
  then break R into chunks, each smaller than m,
  encrypt each chunk separately).
• How do we decrypt?
• Calculate d e-1 mod (p-1)(q-1).
• Decrypt assume k Re mod m.
• Calculate kd mod m. This is R.

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Number Theory

• Why?
• kd mod m Red mod m Ra(p-1)(q-1) 1 mod m
  
• R(R(p-1)(q-1))a mod m
• R(p-1) mod p 1 (Fermat)
• R(q-1) mod q 1 (Fermat)
• Therefore R(p-1)(q-1) mod pq 1
• And (R(p-1)(q-1))a mod pq 1
• So R(R(p-1)(q-1))a mod pq R
• Can anyone else retrieve R? Currently, the
  only way we know how to calculate R efficiently
  is by factoring m pq, retrieving p and q,
  calculate (p-1)(q-1). Calculate e-1, the inverse
  of e mod (p-1)(q-1).
• As of today, no efficient factoring algorithm
  is known. It is a hotly researched subject!

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RSA Encryption

• Caesar cipher f(p) (p 3) mod
  26Decryption f-1(p) (p - 3) mod 26
• RSA (Rivest /Shamir / Adleman) system for public
  key cryptography
• encryption key n pq (p, q large primes)
• exponent e relatively prime to (p-1)(q-1)

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RSA Encryption

• RSA encryption plaintext --gt integer M --gt
  integer C Me mod n
• RSA decryptiond decryption key an inverse
  of e mod (p-1)(q-1)integer P Cd mod n