The Mathematics of RSA Encryption

THE MATHEMATICS OF
RSA ENCRYPTION
Casco Bay .NET User Group
Nov. 2016

Goals
 Make encryption less mysterious

Outline
 What is RSA Encryption?
 Math Refresher
 How does it work?
 Math Deep Dive
 Where do keys come from?
 Why does it work?
 Why is it secure?

What is RSA Encryption?
 Public Key Cryptography Scheme

Math Refresher - Modulus
 %
 “The Remainder” operator
 546 % 31 = 19
 546 = 17*31 + 19

Math Refresher - Primes
 A prime can only be divided by 1 and itself
 Every number can be factored into a list of
primes
 360 = 2 * 2 * 2 * 3 * 3 * 5
 11 = 11
 Two numbers are coprime if they have no
common prime factors
 6 = 2 * 3, 35 = 5 * 7, so 6 and 35 are coprime
 26 = 2 * 13, 4 = 2 * 2, so 26 and 4 are not
coprime

How Does it Work?
 Public key: (e, n)
 Private key: (d, n)
 Message: M
 Encrypted Message: EM = M**e % n
 Decrypted Message: DM = EM**d % n

Example
 Public key: (e, n) = (3593, 150349)
 Private key: (d, n) = (957, 150349)
 Message: M = 90001
 EM = 90001**3593 % 150349
 131425
 DM = 131425**957 % 150349
 90001

Example
 Public key: (e, n) = (3593, 150349)
 Private key: (d, n) = (957, 150349)
 Message: M = 22621
 EM = 22621**3593 % 150349
 62033
 DM = 62033**957 % 150349
 22621

Why did that work?
 Math
 Cleverly chosen keys

Why did that work?
 Math
 Cleverly chosen keys
 Euler’s Theorem

φ, the totient function
 φ(n) is called the totient of n
 Number of integers less than n, coprime with n
 n = 15 = 5 * 3
 φ(15) = Number of integers coprime with 15
 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
 1, 2, 4, 7, 8, 11, 13, 14
 φ(15) = 8

Calculating φ(n)
 φ(15) = 8
 Notice: φ(5 * 3) = (5-1) * (3-1)
 n = p * q
 n is the product of 2 different primes, p and q
 There are p multiples of q
 There are q multiples of p
 0 is counted twice
 φ(n) = p*q – p – q + 1 = (p-1) * (q-1)

Euler’s Theorem
 x**φ(n) % n = 1, where x is coprime with n
 Euler’s Theorem predicts x**8 % 15 = 1
 For x coprime with 15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 1 1 6 1 10 6 1 1 6 10 1 6 1 1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 1 1 6 1 10 6 1 1 6 10 1 6 1 1
1 2 4 7 8 11 13 14
1 1 1 1 1 1 1 1

Proof of Euler’s Theorem
X 1 2 4 7 8 11 13 14
1 1 2 4 7 8 11 13 14
2 2 4 8 14 1 7 11 13
4 4 8 1 13 2 14 7 11
7 7 14 13 4 11 2 1 8
8 8 1 2 11 4 13 14 7
11 11 7 14 2 13 1 8 4
13 13 11 7 1 14 8 4 2
14 14 13 11 8 7 4 2 1

Proof of Euler’s Theorem
 Consider the product of each number in the
first row
 1*2*4*7*8*11*13*14 % 15
 What if we multiply this value by 7**8?
 7**8 * (1*2*4*7*8*11*13*14) % 15
 (7*1)*(7*2)*(7*4)*(7*7)*(7*8)*(7*11)*(7*13)*(7*14)
% 15
 7*14*13*4*11*2*1*8 % 15
 1*2*4*7*8*11*13*14 % 15
 It didn’t change the value, so 7**8 % 15 = 1

Key Generation
 How did we get our keys from the example?
 Public key: (e, n) = (3593, 150349)
 Private key: (d, n) = (957, 150349)

Key Generation
 p, q = 251, 599
 n = p * q
 150349
 e = 3593
 φ(n) = (p-1) * (q-1)
 149500
 d*e % φ(n) = 1 solve for d
 d*3593 % 149500 = 1
 d = 957 is the only solution

What makes those keys work?
 EM = M**e % n
 DM = EM**d % n
 DM = (M**e % n)**d % n
 DM = M**(e * d) % n

Why does it work?
 DM = M**(e*d) % n
 e*d % φ(n) = 1
 e*d = 1 + k*φ(n)
 DM = M**(1 + k*φ(n)) % n
 = (M**1) * (M**φ(n))**k % n
 = M * (1**k) % n
 = M

Why is it secure?
 Can we get the private key from the public
key?
 e*d % φ(n) = 1
 3593*d % φ(150349) = 1

Why is it secure?
key?
 e*d % φ(n) = 1
 3593*d % φ(150349) = 1
 No, because factoring appears to be difficult

Why is it secure?
key?
 e*d % φ(n) = 1
 3593*d % φ(150349) = 1
 No, because factoring appears to be difficult
 RSA-200
 279978339112213278708294676387226016210
704467869554285375600099293261284001076
093456710529553608560618223519109513657
886371059544820065767750985805576135790
987349501441788631789462951872378692218

Why is it secure?
 Can we solve for M given the encrypted
message?
 EM = M**e % n
 131435 = M**3593 % 150349

Why is it secure?
 Can we solve for M given the encrypted
message?
 EM = M**e % n
 131435 = M**3593 % 150349
 No, because taking the eth root (The RSA
Problem) appears to be difficult

Why is it secure?
 Can we solve for d given a decrypted
(authenticated) message?
 EM = M**d % n
 131435 = 90001**d % 150349

Why is it secure?
 Can we solve for d given a decrypted
(authenticated) message?
 EM = M**d % n
 131435 = 90001**d % 150349
 No, because the discrete logarithm appears to
be difficult

Why is it secure?
 Can we take a guess at M, given the
encrypted message?
 EM = M**e % n
 131435 = M**3593 % 150349

Why is it secure?
encrypted message?
 EM = M**e % n
 131435 = M**3593 % 150349
 90001**3593 % 150329 = 131435 !!

Why is it secure?
encrypted message?
 EM = M**e % n
 131435 = M**3593 % 150349
 90001**3593 % 150329 = 131435 !!
 Yes, that is a “chosen plaintext attack”, and
that is why you must pad your messages

More References
 Video demonstrating Public Key Cryptography
 https://www.youtube.com/watch?v=GSIDS_lvRv4
 Wikipedia Page
 https://en.wikipedia.org/wiki/RSA_(cryptosystem)

Appendix: Properties of %
 A + B % n = (A % n) + (B % n) % n
 517 + 878 % 10 = 7 + 8 % 10
 A * B % n = (A % n) * (B % n) % n
 318 * 73 % 10 = 8 * 3 % 10
 A ** B % n = (A % n) ** B % n ≠ (A % n) ** (B
% n)
 93 ** 57 % 10 = 3 ** 57 % 10 ≠ 3 ** 7 % 10
 A ** B % n = A ** (B % phi(n)) % n
 (For A and n coprime)
 93 ** 57 % 10 = 93 ** (57 % 4) % 10
 A % n = A’ => A = A’ + k*n

The Mathematics of RSA Encryption

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