Bonus Lecture 2: Eulers Totient Theorem Primes Are Overrated - PowerPoint PPT Presentation

Bonus Lecture 2: Euler’s Totient Theorem Primes Are Overrated Anyways 1 / 10

i i i ai mod p i i i i i 1 exists for all i i i 1 i i Factor out a : a p 1 p Recall From the Future... 1 2 1 Multiply by 1 p What happens if p not prime? p Means “Recall” Fermat’s Little Theorem: p Proof : f x ax p is biject. on 1 2 p 1 So 1 p 1 a p 1 a 2 / 10 Theorem : Let p be prime and a ̸≡ 0 (mod p ) . Then a p − 1 ≡ 1 (mod p ) .

i i i ai mod p i i i i i 1 exists for all i i i 1 i i What happens if p not prime? p 1 Multiply by 1 p 1 2 Recall From the Future... p “Recall” Fermat’s Little Theorem: a p 1 Factor out a : Means Proof : 2 / 10 Theorem : Let p be prime and a ̸≡ 0 (mod p ) . Then a p − 1 ≡ 1 (mod p ) . ▶ f ( x ) = ax (mod p ) is biject. on { 1 , 2 , ..., p − 1 } ▶ So { 1 , ..., p − 1 } = { a , ..., ( p − 1 ) a } (mod p )

i 1 exists for all i i i 1 i i Recall From the Future... What happens if p not prime? p 1 Multiply by 1 p 1 2 2 / 10 Proof : “Recall” Fermat’s Little Theorem: Theorem : Let p be prime and a ̸≡ 0 (mod p ) . Then a p − 1 ≡ 1 (mod p ) . ▶ f ( x ) = ax (mod p ) is biject. on { 1 , 2 , ..., p − 1 } ▶ So { 1 , ..., p − 1 } = { a , ..., ( p − 1 ) a } (mod p ) ▶ Means ∏ i i = ∏ i ( ai mod p ) i i ≡ a p − 1 ∏ ▶ Factor out a : ∏ i i (mod p )

Recall From the Future... “Recall” Fermat’s Little Theorem: What happens if p not prime? 2 / 10 Proof : Theorem : Let p be prime and a ̸≡ 0 (mod p ) . Then a p − 1 ≡ 1 (mod p ) . ▶ f ( x ) = ax (mod p ) is biject. on { 1 , 2 , ..., p − 1 } ▶ So { 1 , ..., p − 1 } = { a , ..., ( p − 1 ) a } (mod p ) ▶ Means ∏ i i = ∏ i ( ai mod p ) i i ≡ a p − 1 ∏ ▶ Factor out a : ∏ i i (mod p ) ▶ i − 1 exists for all i ∈ { 1 , 2 , ..., p − 1 } i i ) − 1 ≡ ∏ i ( i − 1 ) (mod p ) ▶ Multiply by ( ∏

Recall From the Future... “Recall” Fermat’s Little Theorem: What happens if p not prime? 2 / 10 Proof : Theorem : Let p be prime and a ̸≡ 0 (mod p ) . Then a p − 1 ≡ 1 (mod p ) . ▶ f ( x ) = ax (mod p ) is biject. on { 1 , 2 , ..., p − 1 } ▶ So { 1 , ..., p − 1 } = { a , ..., ( p − 1 ) a } (mod p ) ▶ Means ∏ i i = ∏ i ( ai mod p ) i i ≡ a p − 1 ∏ ▶ Factor out a : ∏ i i (mod p ) ▶ i − 1 exists for all i ∈ { 1 , 2 , ..., p − 1 } i i ) − 1 ≡ ∏ i ( i − 1 ) (mod p ) ▶ Multiply by ( ∏

m , a m 2 is a 1 ! Euler Attempt 1 1 2 3 1 Not recoverable: if a m 1 1 a m Generally have issues if 2 0 2 ! to 4 maps 2 x Not necessarily! 1 ? m 1 m a biject. on Is ax “Proof” : 3 / 10 Claim : If a ̸≡ 0 (mod m ) , a m − 1 ≡ 1 (mod m ) .

m , a m 2 is a 1 ! Euler Attempt 1 2 0 2 ! 1 Not recoverable: if a m 1 1 a m Generally have issues if to 1 2 3 4 maps 2 x Not necessarily! “Proof” : 3 / 10 Claim : If a ̸≡ 0 (mod m ) , a m − 1 ≡ 1 (mod m ) . ▶ Is ax (mod m ) a biject. on { 1 , ..., m − 1 } ?

m , a m 2 is a 1 ! Euler Attempt 1 “Proof” : Generally have issues if a m 1 Not recoverable: if a m 1 1 3 / 10 Claim : If a ̸≡ 0 (mod m ) , a m − 1 ≡ 1 (mod m ) . ▶ Is ax (mod m ) a biject. on { 1 , ..., m − 1 } ? ▶ Not necessarily! ▶ 2 x (mod 4 ) maps { 1 , 2 , 3 } to { 2 , 0 , 2 } !

m , a m 2 is a 1 ! Euler Attempt 1 “Proof” : Not recoverable: if a m 1 1 3 / 10 Claim : If a ̸≡ 0 (mod m ) , a m − 1 ≡ 1 (mod m ) . ▶ Is ax (mod m ) a biject. on { 1 , ..., m − 1 } ? ▶ Not necessarily! ▶ 2 x (mod 4 ) maps { 1 , 2 , 3 } to { 2 , 0 , 2 } ! Generally have issues if gcd( a , m ) ̸ = 1

Euler Attempt 1 “Proof” : 3 / 10 Claim : If a ̸≡ 0 (mod m ) , a m − 1 ≡ 1 (mod m ) . ▶ Is ax (mod m ) a biject. on { 1 , ..., m − 1 } ? ▶ Not necessarily! ▶ 2 x (mod 4 ) maps { 1 , 2 , 3 } to { 2 , 0 , 2 } ! Generally have issues if gcd( a , m ) ̸ = 1 Not recoverable: if a m − 1 ≡ 1 (mod m ) , a m − 2 is a − 1 !

i i i ai mod m i i i i 1 DNE! i i Euler Attempt 2 m So Issue: not all i s have inverses m a m 1 Factor out a : Means 1 a m a 1 m 1 So 1 m 1 m is biject. on ax f x Proof : 4 / 10 Claim : If gcd( a , m ) = 1, a m − 1 ≡ 1 (mod m ) .

i i i ai mod m i i i i 1 DNE! i i So Issue: not all i s have inverses m Euler Attempt 2 a m 1 Factor out a : Means Proof : 4 / 10 Claim : If gcd( a , m ) = 1, a m − 1 ≡ 1 (mod m ) . ▶ f ( x ) = ax (mod m ) is biject. on { 1 , ..., m − 1 } ▶ So { 1 , ..., m − 1 } = { a , ..., ( m − 1 ) a } (mod m )

1 DNE! i i Euler Attempt 2 Proof : Issue: not all i s have inverses So 4 / 10 Claim : If gcd( a , m ) = 1, a m − 1 ≡ 1 (mod m ) . ▶ f ( x ) = ax (mod m ) is biject. on { 1 , ..., m − 1 } ▶ So { 1 , ..., m − 1 } = { a , ..., ( m − 1 ) a } (mod m ) ▶ Means ∏ i i = ∏ i ( ai mod m ) i i ≡ a m − 1 ∏ ▶ Factor out a : ∏ i i (mod m )

Euler Attempt 2 Proof : 4 / 10 Claim : If gcd( a , m ) = 1, a m − 1 ≡ 1 (mod m ) . ▶ f ( x ) = ax (mod m ) is biject. on { 1 , ..., m − 1 } ▶ So { 1 , ..., m − 1 } = { a , ..., ( m − 1 ) a } (mod m ) ▶ Means ∏ i i = ∏ i ( ai mod m ) i i ≡ a m − 1 ∏ ▶ Factor out a : ∏ i i (mod m ) ▶ Issue: not all i s have inverses i i ) − 1 DNE! ▶ So ( ∏

i S i i S ai mod m a S i S i i S i i S i 1 i S i m 1 m Multiply to get a m , so exists! 1 m Factor out a : Euler Attempt 3 5 / 10 Hence Proof : Let S x m x m 1 f x ax m is bijection on S So S ax mod m x S Theorem : Let φ ( m ) be |{ x ∈ Z m | gcd( x , m ) = 1 }| . 1 Then for a coprime to m , a φ ( m ) ≡ 1 (mod m ) . 1 φ ( · ) is known as Euler’s Totient Function.

i S i i S ai mod m a S i S i i S i i S i 1 i S i m 1 m Multiply to get a m , so exists! 1 m Euler Attempt 3 Factor out a : Hence S ax mod m x So S Proof : 5 / 10 Theorem : Let φ ( m ) be |{ x ∈ Z m | gcd( x , m ) = 1 }| . 1 Then for a coprime to m , a φ ( m ) ≡ 1 (mod m ) . ▶ Let S = { x ∈ Z m | gcd( x , m ) = 1 } ▶ f ( x ) = ax (mod m ) is bijection on S 1 φ ( · ) is known as Euler’s Totient Function.

a S i S i i S i i S i 1 i S i m 1 m Multiply to get a m , so exists! 1 m Euler Attempt 3 Proof : Factor out a : 5 / 10 Theorem : Let φ ( m ) be |{ x ∈ Z m | gcd( x , m ) = 1 }| . 1 Then for a coprime to m , a φ ( m ) ≡ 1 (mod m ) . ▶ Let S = { x ∈ Z m | gcd( x , m ) = 1 } ▶ f ( x ) = ax (mod m ) is bijection on S ▶ So S = { ax mod m | x ∈ S } ▶ Hence ∏ i ∈ S i = ∏ i ∈ S ( ai mod m ) 1 φ ( · ) is known as Euler’s Totient Function.

Euler Attempt 3 Proof : 5 / 10 Theorem : Let φ ( m ) be |{ x ∈ Z m | gcd( x , m ) = 1 }| . 1 Then for a coprime to m , a φ ( m ) ≡ 1 (mod m ) . ▶ Let S = { x ∈ Z m | gcd( x , m ) = 1 } ▶ f ( x ) = ax (mod m ) is bijection on S ▶ So S = { ax mod m | x ∈ S } ▶ Hence ∏ i ∈ S i = ∏ i ∈ S ( ai mod m ) i ∈ S i ≡ a | S | ∏ ▶ Factor out a : ∏ i ∈ S i (mod m ) ) − 1 ≡ ∏ i ∈ S ( i − 1 ) (mod m ) , so exists! ▶ (∏ i ∈ S i ▶ Multiply to get a φ ( m ) ≡ 1 (mod m ) 1 φ ( · ) is known as Euler’s Totient Function.

2 2 3 2 3 2 5 6 / 10 90 1 11 0 10 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 m 90 2 11 1 2 0 3 1 3 1 5 1 5 0 24 1,7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77,79,83,89 11 m 11 12 1 k . Examples : Claim : Suppose m can be factored as p n 1 m 12 2 1 2 1 3 1 3 0 4 1, 5, 7, 11 Understanding φ 1 · ... · p n k k . Then φ ( m ) = ( p 1 − 1 ) p n 1 − 1 · ... · ( p k − 1 ) p n k − 1

2 3 2 5 6 / 10 10 47, 49, 53, 59, 61, 67, 71, 73, 77,79,83,89 1,7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 24 1 5 0 5 1 3 1 3 1 2 0 2 90 90 m 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 1 11 0 Claim : Suppose m can be factored as p n 1 11 1 k . Examples : m 11 11 Understanding φ 1 · ... · p n k k . Then φ ( m ) = ( p 1 − 1 ) p n 1 − 1 · ... · ( p k − 1 ) p n k − 1 ▶ m = 12 = 2 2 · 3 ▶ φ ( 12 ) = ( 2 − 1 ) 2 1 · ( 3 − 1 ) 3 0 = 4 ▶ 1, 5, 7, 11

2 3 2 5 5 90 m Claim : Suppose m can be factored as p n 1 2 1 2 0 3 1 3 1 6 / 10 90 . k 1 5 0 1 24 1,7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77,79,83,89 Examples : Understanding φ 1 · ... · p n k k . Then φ ( m ) = ( p 1 − 1 ) p n 1 − 1 · ... · ( p k − 1 ) p n k − 1 ▶ m = 12 = 2 2 · 3 ▶ φ ( 12 ) = ( 2 − 1 ) 2 1 · ( 3 − 1 ) 3 0 = 4 ▶ 1, 5, 7, 11 ▶ m = 11 ▶ φ ( 11 ) = ( 11 − 1 ) 11 0 = 10 ▶ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

6 / 10 k 47, 49, 53, 59, 61, 67, 71, 73, 77,79,83,89 Claim : Suppose m can be factored as p n 1 Examples : 1 . Understanding φ 1 · ... · p n k k . Then φ ( m ) = ( p 1 − 1 ) p n 1 − 1 · ... · ( p k − 1 ) p n k − 1 ▶ m = 12 = 2 2 · 3 ▶ φ ( 12 ) = ( 2 − 1 ) 2 1 · ( 3 − 1 ) 3 0 = 4 ▶ 1, 5, 7, 11 ▶ m = 11 ▶ φ ( 11 ) = ( 11 − 1 ) 11 0 = 10 ▶ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ▶ m = 90 = 2 · 3 2 · 5 ▶ φ ( 90 ) = ( 2 − 1 ) 2 0 · ( 3 − 1 ) 3 1 · ( 5 − 1 ) 5 0 = 24 ▶ 1,7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,

n such that CRT gives b 1 xx 1 m , xx 1 m inv. choices for b x 1 , n for b x 2 ax m and ax 1 n , mn 1 If ax Thus, m n inv. choices for b x 1 7 / 10 n 1 1 Claim: x invertible ifg b x is mn n m x mod m x mod n b x m mn Consider b Proof : φ Is Multiplicative Lemma : If gcd( m , n ) = 1, φ ( mn ) = φ ( m ) φ ( n ) .