COMP 2711H: Lecture 24

Date: 2024-10-28 09:07:24

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Fermat's Little Theorem

Fermat's Little Theorem
- theorem: for every prime $p$ , $p \neq ∣ a ⟹ a^{p - 1} = 1 mod p$
  - $f + a : x \mapsto x \cdot a$
- one to one: as we can cancel $a$ by multiplying $a^{- 1}$ both side
- onto: create a program to check
  - code
```
vector<int> seen;
cin >> p >> a;
// e.g. p = 11, a = 2
int p, a;
for (int i = 0; i < p; ++i)
    if (!seen[i]) {

    }
for (int j = 0; )
```
  - if succeed (primitive root), we will result in: (multiplicative) cycle
    - cycle involving only $0$
    - cycle involving others
  - not always the case: e.g. $a = 3, p = 11$
    - results in cycle size of: $0, 5, 5$
    - cycle starting w/ 1: size $5$
      - as $p = 11$ , $11%5 = 1$ : first element star
- cycles will have the same size, except $0$
  - as each element: can be seen as $b \cdot x$
  - and thus: $p - 1$ elements are divided evenly
    - s.t. it's divisor of $p - 1$
- different proof:
  - let there be additive cycle of $a$
    - ${0, a, 2 a, 3 a, \dots, (p - 1) a}$
    - which is an equivalent set of $[1, p - 1]$
  - and multiplying both sides, we get: $a \times 2 a \times 3 a \dots (p - 1) a (p - 1)! a = a p \times 2 a \times 3 a \dots (p - 1) a$
  - cancelling $(p - 1)!$ from both dies, we get
    - thus $a^{p - 1} = 1 mod p$
- proof by induction on $a$
  - rewrite FLT s.t.
    - theorem: for every prime $p$ , integer $a$ , $a^{p} = a mod p$
    - 👨‍🏫 now works for $a = 0$ too
  - $(a + 1)^{p} = a^{p} + (1 p) a^{p - 1} + \dots + (p - 1 p) a + 1$
  - claim: $1 < k < p ⟹ (k p) = 0 mod p$
    - then: all terms except $a^{p}, 1$ are congruent to $0 mod p$
    - $(a + 1)^{p} = a^{p} + 1 mod p$
  - lemma: $1 < k < p ⟹ (k p) = 0 mod p$ $(k p) = \frac{p !}{k ! ( p - k )!}$
    - claim: it's multiple of $p$ when $1 < k < p$
      - as it involves $p$ in denominator
Computing modular exponentiation
- for $p \neq ∣ a$ , compute $a^{b} mod p$
  - $a^{p - 1} = 1 mod p$
  - $a^{2 (p - 1)} = 1 mod p$
  - $a^{k (p - 1)} = 1 mod p$
  - (depends on the size of cycle)
- theorem: if $b = c mod p - 1$ then $a^{b} = a^{c} mod p$
- simply: take modulus of $p - 1$ on exponent $b, c$
  - then check it using the fast modulus algorithm
Non-prime number FLT
- assumption breaks: all cycle no longer necessarily w/ same size
- let $n \geq 1$ , define $φ (n)$ as
  - $∣ {a ∣1 \leq a \leq n, a ⊥ n} ∣$
  - $φ$ : Euler's totient function
    - 👨‍🏫 not a meaningful name, "how many" or something in latin
- for prime $p$ : $φ (p) = p - 1$
- for $n = \prod_{i} p_{i}^{α_{i}}$
  - claim: $φ (n) = n \cdot \prod_{i} (1 - 1/ p_{i})$
  - intuition: $1 - 1/ p_{1}$ of all numbers: not having $p_{1}$ in prime factorization
    - by multiplying all: we find proportion without common factors
- proof 1: suppose $n = \prod_{i} p_{i}$
  - for CRT: we obtain "unique solution" for linear system of congruent equation
    - $x = a_{1} mod p_{1}$
    - $x = a_{2} mod p_{2}$
    - $x = a_{k} mod p_{k}$
    - 👨‍🏫 no matter how you choose $a_{i}$
  - $n ⊥ x ⟺ x \neq = p_{i}$ for $i = [1, k]$
    - and $p_{i} - 1$ such numbers within $[1, p_{i}]$
  - $φ (n) = (p_{1} - 1) (p_{2} - 1) \dots (p_{k} - 1)$
    - and manipulate it s.t. $(p_{1} - 1) = p_{1} (1 - 1/ p_{1})$
  - now, for $n = \prod_{i} p_{i}^{α_{i}}$
    - $n^{'} = \prod_{i} p_{i}$
    - let $k = n / n^{'}$
    - then we have: $φ (n^{'})$ rel. prime number within $[(a - 1) n^{'} + 1, a n^{'}]$
      - for $a \in [1, k]$
    - thus multiply it by $k$ , we obtain
      - $φ (n) = k φ (n^{'}) = n \cdot \prod_{i} (1 - 1/ p_{i})$
- proof 2:
  - lemma: if $n ⊥ m$ , then $φ (nm) = φ (n) \cdot φ (m)$
    - same as: obtaining solution of system of equation
      - $x = a mod n$
        
        choose $a$ s.t. $a ⊥ n$
      - $x = b mod m$
        
        choose $b$ s.t. $b ⊥ n$
    - s.t. $g cd (n, nm) = 1 ⟺ g cd (n, nm) = 1 \land g cd (n, m) = 1$
    - according to CRF, within $[1, nm]$ , we have unique solution for above system
      - finally: $φ (nm) = φ (n) φ (m)$
  - consider $n = p^{α}$
    - $φ (n) = n - n / p = n (1 - 1/ p)$
      - $n / p$ as $n / p$ numbers are rel. prime to $n$
  - general case: $n = \prod_{i} p_{i}^{α_{i}}$
    - then, for each $n_{i} = p_{i}^{α_{i}}$
    - $\forall i \neq = j, n_{i} ⊥ n_{j}$
    - thus: $φ (n) = n \prod_{i} (1 - 1/ p_{i})$
Euler's theorem
- extension of FLT
- let $a ⊥ n$ , then $a^{φ (n)} \equiv 1 mod n$
- $A = {x ∣1 \leq x \leq n, x ⊥ n}$
  - $f_{a} : A \to A$
  - $x \mapsto x \cdot a$
    - one-to-one as $a$ can be cancelled from both sides
  - $f_{a}$ : provides cycles (excluding $0$ )
    - $a, 2 a^{2}, \dots, a^{k - 1}$
    - $b, ba, 2 b a^{2}, \dots, b a^{k - 1}$
      - $b a^{i} \neq = b a^{0}$
    - $k ∣ φ (n)$
  - sum of sizes on each cycle
    - total size: $φ (n)$
- 👨‍🏫 try to extend other proof of FLT for Euler's theorem
Wilson's theorem
- Wilson "guessed" it, not proved it
  - but British always have to take fame :p
- for every prime $p$ , $(p - 1)! = - 1 mod p$
- e.g. $p = 5$
  - $4! = 2 \times 3 \times 4$
  - $(4!)^{2} = 1$
    - as: $a \cdot a^{- 1} = 1$
      - $2^{- 1} = 3$
      - $3^{- 1} = 2$
      - $4^{- 1} = 4$
- $((p - 1)!)^{2} \equiv 1 mod p$
- second attempt:
  - use similar idea: $(p - 2)! \equiv 1 mod p$
  - as $(p - 1)^{2} \equiv 1 mod p$
    - 👨‍🏫 and is the only case
  - thus $(p - 2)!$ : must include number and its (distinct) MMI as pairs
    - $(p - 2)! = \prod a_{i} \cdot a_{i}^{- 1} = 1$
    - naturally: $(p - 1)! = p - 1 = - 1 mod p$
- lemma: the only $a$ s.t. $a^{2} = 1 mod p$ are $1, - 1$
  - $a^{2} = 1 mod p ⟺ p ∣ a^{2} - 1 ⟺ p ∣ (a + 1) (a = 1)$
  - $⟺ p ∣ a + 1 \lor p ∣ a - 1$ (Euclid's lemma)

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science