0. I538/B609: Introduction to Cryptography
Fall · Lecture 16

2. Tuesday’s lecture: Today’s lecture: One-way permutations (OWPs)
PRGs from OWPs
Today’s lecture:
Basic number theory
“secret key”
So far:
cryptography
Going forward:
“public key”
cryptography

3. Assignment 4 is due Tuesday, October 27!
There will be a quiz on Tuesday, October 27!

4. Divisibility The set of integers is ℤ={…,-3,-2,-1, 0, 1, 2, 3,…}
Let a and b be any two integers
Defn: b is called a divisor of a if
We write b | a (read “b divides a”) to indicate that b is a divisor of a and b ∤ a to indicate that b is not a divisor of a
∃c∈ℤ such that a=bc
???
If b | a, then a is called a multiple of b
Defn: a is called a prime if a>1 and
???
b | a implies that b∈{±1,±a}

5. Divisibility Does 6 divide 12? Does 6 divide 6? Does 0 divide 6?
YES! (Because 6·2=12)
every integer divides itself
YES! (Because 6·1=6)
only 0 divides 0
No! (Because 0·c=0 for all c∈ℤ)
every integer divides 0
YES! (Because 6·0=0)
No! (Because 4·1<6 and 4·2>6)
1 divides every integer
Yes! (Because 1·6=6)

6. Division Algorithm
Thm: Let a, b∈ℤ with b>0. Then there exist unique integers q, r∈ℤ such that a=b·q+r and 0≤r<b. The integer q is called the quotient and r the remainder upon division of a by b.
If b | a, then r=
If b>a, then q= and r=
a=17 and b=5 ⇒
a=-23 and b=6 ⇒
a=20 and b=5 ⇒ ?? ?? ?? a
????
q=3 and r=2, since 17 =5·3+2
????
q=-4 and r=1, since -23=6·(-4) +1
????
q=4 and r=0, since 20=5·4+0

7. Greatest common divisors (GCDs)
Defn: The greatest common divisor of two non-zero integers a and b is
The greatest common divisor of a and b is denoted gcd(a, b).
the largest positive divisor of a that
is also a divisor of b.
???
gcd(4, 15)=
gcd(4, 10)=
gcd(22 · 32 · 5, 2 · 33 · 72)= 1 ?? 2 ??
2 · 32 ??
Defn: a and b are relatively prime if
gcd(a, b)=1
???
a.k.a co-prime

8. Least common multiples (LCMs)
Defn: The least common multiple of two non-zero integers a and b is
The least common multiple of a and b is denoted by lcm(a, b).
the smallest positive multiple of a
that is also a multiple of b.
???
lcm(4, 15)=
lcm(4, 10)=
lcm(4, 8)=
lcm(22 · 32 · 5, 2 · 33 · 72)=
60 (4=22 and 15=3·5) ??
20 (4=22 and 10=2·5) ??
(4=22 and 8=23) ??
22 · 33 · 5 · 72 ??

9. Properties of GCDs and LCMs
Let a, b∈ℤ with a>0 and b>0, let d=gcd(a, b) and let m=lcm(a, b). Then
a·b=d·m
If c | a and c | b, then c | d
If a | s and b | s, then m | s
Bezout’s identity: There exist integers (unique) s and t such that d=as+bt.
a·b=gcd(a, b)·lcm(a,b)
every common divisor divides the gcd
every common multiple is a multiple of the lcm
“gcd is a linear combination”

10. Extended Euclidean Algorithm
Fact 1: If b | a, then gcd(a, b)=b
Fact 2: If a=b·q+r, then gcd(a, b)=gcd(b, r)
(Assume that a>b; if not, swap a and b)
Idea: Apply Facts 1 and 2 repeatedly, using the Division Algorithm to write a=b·q+r at each step
a = q 0 b +r0
b = q 1 r0+r1
r0 = q 2 r1+r2
r1 = q 3 r2+r3 ⋮
rk-3= q k-1rk-2+rk-1
0 = q k rk-1+rk
Pseudocode:
function gcd(a,b) {
while (a ≠ b)
if (a >b)
a := a − b;
else
b := b − a; }
return a ;
r0 = a-q0b
r1 = b-q 3 r0 ⋮
rk-3= rk-5 – q k-3rk-4
rk-2= rk-4 -q k-2rk-3
gcd(a, b)= rk-3-qk-1rk-2
repeated substitution
yields s,t such that
as+bt=gcd(a,b)
gcd(a, b)

11. Euclid’s Lemma
Euclid’s Lemma: Let a and b be integers. If p is a prime such that p | a·b, then p | a or p | b (or both).
Proof: Assume W.L.O.G. that p is prime and p | a·b but p ∤ a. We prove that p | b.
First note that p ∤ a with p prime implies that gcd(a, p)=1; thus, by Bezout’s Theorem, there exist integer s and t such that as+pt=1.
Multiplying both sides of this expression by b yields bas+bpt=(ab)s+p(bt)=b.
Now, p | (ab)s by assumption and clearly p | p(bt); hence, p | (bas+bpt).
But bas+bpt=b and it therefore follows that p | b, as desired ☐

12. Fundamental Theorem of Arithmetic
Fundamental Theorem of Arithmetic: Let n>1 be an arbitrary integer. Then n is a product of (powers of) primes, and this product is unique (up to the order of the prime powers)
In other words, every n>1 can be “factored” into primes in one—and only one—way.

13. Clock arithmetic
Q:If it is 10:00 am now, what time will it be in 6 hours? A:4:00 pm Q:If it is 2:00 pm now, what time was it 3 hours ago? A:11:00 am Q:It is currently October. What month will it be 25 months from now? A: November Q:On Mercury, each day is 1407 hours long. If it is 1200:00 now, what time will it be in 500 hours? A:293:00 (=1700: :00)

14. Modular arithmetic a=q·n+r, as in the Division Algorithm. ???
Defn: Let a, r, n∈ℤ with n>1 and 0≤r<n. Then a is congruent to r modulo n if We write a≡r mod n to indicate that a is congruent to r modulo n.
a=q·n+r, as in the Division Algorithm.
???
3 ≡ mod 2
6 ≡ mod 2
11≡ mod 3
62 ≡ mod 85 ?? 1
Thm: Let a, b, and n be integers with n≥1, then a≡b mod n if and only if n | a-b. ?? ?? 2 62 ??

15. Notational conventions
p and q typically denote prime numbers
n is always a positive integer, which may or may not (though usually not) be prime
ℤn≔{0, 1, 2, …, n-1} with arithmetic modulo n
℥n ≔{ a∈ℤn | gcd(a, n)=1 } with arithmetic modulo n

16. Modular inversion
Defn: Let a∈ℤn. The (multiplicative) inverse of a modulo n is an element b∈ℤn such that The inverse of a modulo n is denoted by a-1.
a·b≡1 mod n ??
2-1 ≡ mod 3
3-1 ≡ mod 5
2-1 ≡ mod 6
Let n be any odd integer. Then 2-1 ≡ mod n ??
2 (since 2·2=4=3·1+1≡1 mod 3)
2 (since 3·2=6=5·1+1≡1 mod 5) ??
⊥ (i.e., the inverse does not exist!) ??
(n+1)/2 ??

17. Modular inverses Q: Which elements of ℤ have inverses modulo n?
Thus, a∈ℤn has an inverse modulo n if and only if a∈℥n
Q: Given a∈℥n, how do you compute a-1 mod n?
A: By definition, a·a-1≡1 mod n ⇒ a·a-1=1+t·n for some t
Hence, a·a-1+(-t)·n=gcd(a, n) ⇒ a-1≡s from Bezout’s Theorem
Thus, we can use the Extended Euclidean Algorithm
Thm: An integer a∈ℤ has an inverse modulo n if and only if gcd(a, n)=1.
Corollary: If p is a prime, then a has an inverse modulo p if and only if a is not a multiple of p; i.e., ℥p={1, 2, …, p-1}.

18. Solving linear equations modulo n
Thm: For any integer n>1, there exists an integer x such that ax≡b mod n if and only if gcd(a, n) | b.
Q: Does 6x≡18 mod 36 have a solution?
A: YES! Because gcd(6, 18)=6 and 6 | In fact, it has 6 solutions in ℤn:3, 9, 15, 21, 27, 33
Q: Does 2x≡5 mod 10 have a solution?
A: NO! Because gcd(2, 10)=2 and 2 ∤ 5

19. Fermat’s Little Theorem
Fermat’s Little Theorem: Let p be a prime. Then for every a∈℥p, we have ap-1≡1 mod p.
Corollary: For all a∈ℤp, ap≡a mod p.
34≡ mod 5
84112≡ mod 113
79561≡ mod 113
Trick: To compute ax mod p, reduce the base (i.e., a) modulo p and the exponent (i.e., x) modulo p-1 ?? 1 ?? 1
79 (since 561=5·(113-1)+1) ??

20. Proof of Fermat’s Little Theorem
To prove Fermat’s Little Theorem, we first prove the following lemma.
Lemma: If n is an integer and a∈℥n, then a·k1≡a·k2 mod n if and only if k1≡k2 mod n.
Proof: Suppose k1a≡k2a. Since a∈℥n, there exists an inverse a-1∈℥n such that a·a-1≡1 mod n.
Therefore, (k1a) a-1≡ (k2a) a-1 mod n. But
(k1a) a-1=k1 (a·a-1)≡k1 mod p
and (k2 a) a-1= k2 (a·a-1)≡k2 mod p;
hence, it follows that k1≡k2 mod p, as desired ☐

21. Proof of Fermat’s Little Theorem
Fermat’s Little Theorem: Let p be a prime. Then for every a∈℥p, we have ap-1≡1 mod p.
Proof: Consider the product a·(2a)·(3a)⋯((p-1)a) mod p. By the lemma on the previous slide, it follows that the above product is congruent to (p-1)!=1·2·3⋯(p-1) mod p. (Indeed, it is clearly a product of p-1 numbers from ℥p; if these numbers aren’t distinct, then we obtain an immediate contradiction.) In other words, we have that a·(2a)·(3a)⋯((p-1)a)≡1·2·3⋯(p-1) mod p. Rearranging the left-hand side, we get ap-1(p-1)!≡(p-1)! mod p. Thus, a second application of the lemma shows that ap-1≡1 mod p. ☐

22. Generating random (probable) primes
Goal: Generate a random n-bit prime
Step 1: Choose n∈R [2n+1, 2n+1-1]
Step 2: Check if 2n-1 ≡ 1 mod n If so, output n; otherwise, go to Step 1
If n is prime, then Pr[2n-1≡ 1]=1
If n is not prime, then Pr[2n-1≡1] is “small”
- Unless you get unlucky and n is a Carmichael number…

23. Chinese Remainder Theorem
Chinese Remainder Theorem: Let n1, n2, …, nk be positive integers with gcd(ni, nj)=1 whenever i≠j, and let N=n1·n2⋯nk.
Then the system of congruence relations
x≡ c1 mod n1 ⋮
x≡ ck mod nk
has a unique solution in ℤN.
zi≡1 mod ni
zi≡0 mod nj when i≠j
For each i=1,…,k, set yi ≡ (N/ni)-1 mod ni and zi≡(N/ni)·yi mod N
The unique solution is x≡∑zi·ci mod N

24. Euler’s phi function
Defn: Let n be a positive integer. Then Euler’s phi function is φ(n)≔|℥n|. In particular, Euler’s phi function of n indicates the number of positive integer less than and relatively prime to n.
If p is prime, then φ(p)=
If p is prime, then φ(p2)=
If p is prime, then φ(pk)=
If p and q are distinct prime, then φ(p·q)=
If n=p1e1p2e2⋯pkek, then φ(n)=
p-1 ??
p(p-1) ??
pk-1(p-1) ??
(p-1)(q-1) ??
p1e1-1(p1-1) p2e2-1(p2-1) ⋯pkek-1(pk-1) ??
Thm (Fermat restatement): If p is prime, then ∀a∈℥p, aφ(p)≡ 1 mod p.

25. Euler’s Theorem
Generalization of Fermat’s Little Theorem
Euler’s Theorem: For any integer n>1 and a∈℥n, aφ(n)≡1 mod n.
34≡ mod 10
8440≡ mod 100
79441≡ mod 100
The last two digits of are
Trick: To compute ax mod n, reduce the base (i.e., a) modulo n and the exponent (i.e., x) modulo φ(n) 1 ?? ??
1 (since φ(100)=5·(5-1)·2·(2-1)=40)
79 (since 441=11·φ(100)+1) ??
01 (since 10381≡3 mod 100) ??

26. That’s all for today, folks!