More Modular Arithmetic: Multiplicative Inverses and Afine
Ciphers
The integers
2, 3, 5, 7, 11, 13, 17, 19, 23, . . .
are primes, that is, each has no proper divisors other than 1 and
itself.
Fundamental Theorem of Arithmetic Every positive integer
is either a prime or product of powers of primes. Moreover, there
is only one such factorization, up to the order of the factors.
Example 588 = 2 · 294 = 2 · 2 · 147 = 2 · 2 · 3 · 49 = 2 · 2 · 3 · 7 · 7 =
22 · 3 · 72. This is the prime factorization of 588.
1
Two positive integer integers a and b are relatively prime if
they have no common prime factors. Equivalently, a and b are
relatively prime if their greatest common divisor is 1: a and b
are relatively prime if
gcd(a, b) = 1.
.
Example 588 and 605 are relatively prime because 588 = 22 ·3·72
and 605 = 5 · 112 have no common prime factors.
2
Solving Congruences ax ≡ 1 (mod m)
Multiplication Tables can be helpful for this process
Multiplication Tables
· (mod 2) 0 1
0 0 0
1 0 1
· (mod 4)
0
1
2
3
0
0
0
0
0
1
0
1
2
3
2
0
2
0
2
3
0
3
2
1
· (mod 3)
0
1
2
0
0
0
0
1
0
1
2
2
0
2
1
· (mod 5)
0
1
2
3
4
0
0
0
0
0
0
1
0
1
2
3
4
2
0
2
4
1
3
3
0
3
1
4
2
4
0
4
3
2
1
3
Example: Solve 3x ≡ 1 (mod 5).
Solution: By the mod-5 multiplication table x = 2. Also x = 7,
12, . . . are solutions.
!
Example: Solve 5x ≡ 1
(mod 6).
Solution: Only need to look at possible values x = 0, 1, 2, 3,
4, 5.
5 · 0 ≡ 0,
5 · 1 ≡ 5,
5 · 2 ≡ 10 ≡ 4
5 · 3 ≡ 15 ≡ 3, 5 · 4 ≡ 20 ≡ 2, 5 · 5 ≡ 25 ≡ 1
So x ≡ 5 is a solution. Also x ≡ 11, 17, . . ..
!
4
Example: Solve 4x ≡ 1
(mod 6).
Solution: 4 · 0 ≡ 0, 4 · 1 ≡ 4, 4 · 2 ≡ 8 ≡ 2, 4 · 3 ≡ 12 ≡ 0,
4 · 4 ≡ 16 ≡ 4, 4 · 5 ≡ 20 ≡ 2.
The product 4x is never 1; there is no solution.
!
5
When is there a solution x of the congruence
a · x ≡ 1 (mod m)?
I.e. What is the relationship between a and m?
Example: Is there a solution of
13x ≡ 1 (mod 64)?
Solution: Yes. Look at
13 · 1
13 · 2
13 · 3
13 · 63
MOD 64
MOD 64
...
MOD 64
MOD 64
All of these numbers are in the range 1, . . ., 63.
6
If 1 is not in this list, then two numbers must be the same. That
is,
13 · i ≡ 13 · j (mod 64)
for some i "= j. Then
13(j − i) ≡ 0 (mod 64),
so 13(j − i) is a multiple of 64. Because 13 and 64 are relatively
prime, j − i is a multiple of 64, which is not possible. Thus 1 is
in the list:
13 · x ≡ 1 (mod 64)
for some x among 1, 2, . . ., 63.
!
7
Note: The proof type we use is by contradiction or reductio
ad absurdum. The same line of reasoning proves the following
theorem.
Theorem: The congruence
ax ≡ 1 (mod m)
has a solution x if and only if a and m are relatively prime. The
solution in the range 1 to m − 1 is denoted a−1 or a−1 (mod m),
and called the multiplicative inverse of a modulo m.
Property: a · a−1 ≡ 1 (mod m).
8
Example: Find 5−1 (mod 14).
Solution: Solve 5x ≡ 1 (mod 14). 5 · 1 ≡ 5, 5 · 2 ≡ 10, 5 · 3 ≡
15 ≡ 1 (mod 14), so the solution is x = 3 and thus
5−1 ≡ 3 (mod 14).
(Check: 3 · 5 ≡ 15 ≡ 1 (mod 14).)
9
Property: If
and
then
a ≡ b (mod m)
c ≡ d (mod m)
ac ≡ bd (mod m)
Example: Solve 5x ≡ 12 (mod 14).
Solution: Because 5−1 ≡ 3 (mod 14), multiply the congruences:
5−1 · 5x ≡ 3 · 12 (mod 14)
1 · x ≡ 36 (mod 14)
x ≡ 8 (mod 14)
10
Example: Solve 4x + 2 ≡ 1 (mod 9).
Solution: Strategy: substract 2 from both sides of the congruence and multiply both sides by the multiplicative inverse of 4
modulo 9. The first yields
4x ≡ 1 − 2 ≡ −1
Also 4 · 7 ≡ 28 ≡ 1 (mod 9) so 4−1 ≡ 7 (mod 9). Then
4−1 · 4 · x ≡ (−1) · 7 (mod 9)
x ≡ −7 ≡ 2 (mod 9).
!
11
Cryptography with Modular Arithmetic
Key Questions: Which numbers a in the range 0 to 25 have
multiplicative inverses modulo 26? What are those inverses?
Partial Answer: By the Theorem, the numbers with multiplicative inverses are those in the range 0 to 25 relatively prime to
26.
So 1, 3, 5, . . . have multiplicative inverses.
12
Table of Multiplicative inverses Modulo 26
a
a−1 (mod 26)
1
1
3
9
5
21
7
15
9
3
11
19
15
7
17
23
19
11
21
5
23
17
25
25
Idea: Use an affine cipher
E(x) = (ax + b) MOD 26
to encipher plaintext letter numerical equivalent x, where a is
relatively prime to 26.
13
To find the decipherment formula solve y ≡ ax + b (mod 26) for
x in terms of y:
y − b ≡ ax
a−1(y − b) ≡ a−1 · a · x
a−1(y − b) ≡ x
x ≡ a−1(y − b) (mod 26)
So decipherment of letter y is
D(y) = a−1(y − b) MOD 26.
14
Example: Encipher ITS COOL with
E(x) = (5x + 8) MOD 26.
Solution:
x
5x + 8
(5x + 8) MOD 26
y
I
8
48
22
W
T
19
103
25
Z
S
18
98
20
U
C
2
18
18
S
O
14
78
0
A
O
14
78
0
A
L
11
63
11
L
15
This can also be done in a spreadsheet, as shown here
A
1
2
3
4
5
6
7
B
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=CHAR(65 + MOD($A$1 * (CODE(B2)-65) + $B$1, 26))
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Example: If H P C C X A Q was enciphered with E(x) = (5x +
8) MOD 26, find the plaintext.
Solution: Find the decipherment formula:
y
y−8
5−1(y − 8)
21(y − 8)
≡
≡
≡
≡
5x + 8 (mod 26)
5x
5−1 · 5 · x
x
Decipherment: D(y) = 21(y − 8) MOD 26:
cipher
y
y−8
21(y − 8)
MOD 26
plain
H
7
-1
-21
5
F
P
15
7
147
17
R
C
2
-6
-126
4
E
C
2
-6
-126
4
E
X
23
15
315
3
D
A
0
-8
-168
14
O
Q
16
8
168
12
M
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The decipherment can also be implemented in a spreadsheet as
shown here
1
2
3
4
5
6
7
8
9
A
21
B
8
H
F
C
P
R
D
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E
E
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M
Q
M
=CHAR(65 + MOD($A$1*(CODE(B3) - 65 - $B$1),26))
Notice that the decryption formula is condensed from earlier
ones.
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Example: (Cryptanalysis) Suppose an affine cipher
E(x) = (ax + b) MOD 26
enciphers H as X and Q as Y. Find the cipher.
Solution:
H → X means E(7) = 23
Q → Y means E(16) = 24
I.e.
a · 7 + b ≡ 23 (mod 26)
(1)
a · 16 + b ≡ 24
(2)
19
Subtract (1) from (2) to eliminate b:
a · 16 − a · 7 ≡ 24 − 23
a(16 − 7) ≡ 1
9a ≡ 1
a ≡ 9−1 ≡ 3
So
b ≡ 23 − 7a
≡ 23 − 7 · 3
≡ 23 − 21
≡ 2 (mod 26)
Thus E(x) = (3x + 2) MOD 26.
20
Brute-force Cryptanalysis of an Affine Cipher
Suppose that Eve has intercepted
WLMHL UPUHI FMIEC LNMWK CVU
which she suspects to have come from an affine cipher E(x) =
(ax + b) MOD 26. What is involved in a brute force cryptanalysis
in which she tries out choices of a and b until she hits one that
yields recognizable plaintext?
21
Solution The decipherment formula is
D(y) = a−1(y − b) MOD 26
= (a−1)y + (a−1b) MOD 26,
so Eve can decipher the message with every possible choice of
a−1 and a−1b and hope that she can recognize plaintext. The
choices of these two numbers will tell her the right values for a
and b.
The spreadsheet on the next slide implements this strategy. By
plugging in values of 1, 3, 5, . . . to the cell marked a−1, she eventually arrives at the page shown. There is recognizable plaintext
in spreadsheet row 25.
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A
B
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Cryptanalysis of an affine encipherment: W L M H L U P U H I F M I E C L N M W K C V U came
from E(x) = (a x + b) MOD 26. Find a and b by brute force.
Decipherment is D(y) = a-1 (y - b) MOD 26 = (a-1 y + (- a-1 b) ) MOD 26. Try deciphering
systematically using every possible b value for each possible a.
a-1-value
2
3
7
4
5
6
7
8
9
10
11
1 2 -1
1 3 a b-values
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=CHAR(65+MOD($A$4*(CODE(C$4)-CODE("A"))+$B5,26))
Fill C5 right and then down, and then change the value in A4 until a plaintext
is visible one on of the lines in the array.
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Thus a−1 = 7, so a = 15; then a−1b = 20, so b ≡ 20a ≡ 300 ≡
14 (mod 26).
24