multiplicative functions - lecture 5 · 2018. 12. 5. · justin stevens multiplicative functions...

Multiplicative FunctionsLecture 5

Justin Stevens

Justin Stevens Multiplicative Functions (Lecture 5) 1 / 40

Outline

1 Fibonacci Numbers

2 Linear Number Theory

3 Multiplicative Functions

Justin Stevens Multiplicative Functions (Lecture 5) 2 / 40

Fibonacci Numbers

The Fibonacci numbers show up in many unexpected situations and havemany beautiful properties. We present a motivating example:

Example 1. Consider a board of length n. How many ways are there to tilethis board with squares (length 1) and dominos (length 2)?

Let the number of tilings of an n-board be f (n). We calculate f (4):

Table 1: The five tilings of a 4-board

Hence f (4) = 5. For larger values of n, we need a different strategy.

Justin Stevens Multiplicative Functions (Lecture 5) 3 / 40

Fibonacci Numbers 2

Observe that if an n-board begins with a square, then we have to tile ann − 1 board. However, if the n-board begins with a domino, then we haveto tile an n − 2 board. Therefore,

f (n) = f (n − 1) + f (n − 2).

We see that f (1) = 1 and f (2) = 2, therefore we can compute:

f (3) = f (2) + f (1) = 2 + 1 = 3f (4) = f (3) + f (2) = 3 + 2 = 5f (5) = f (4) + f (3) = 5 + 3 = 8.

The only difference between this and the Fibonacci numbers are that thelatter begins with an extra 1. Therefore, f (n) = Fn+1 . Using this n-tiling,we can prove several identities regarding Fibonacci numbers.

Justin Stevens Multiplicative Functions (Lecture 5) 4 / 40

Property

Example 2. Show that f (m + n) = f (m)f (n) + f (m − 1)f (n − 1).

The left-hand side is simply the number of tilings of an (m + n)-board.

For the right-hand side, we have two cases:

If there is no domino at square m, we have f (m)f (n) tilings:

· · ·1 2 m − 1 m

· · ·m + 1 m + 2 m + n

Justin Stevens Multiplicative Functions (Lecture 5) 5 / 40

Sum Property

If there is a domino at square m, we have f (m − 1)f (n − 1) tilings:

· · ·1 2 m − 1 m m + 1

· · ·m + 1 m + 2 m + n

Hence, f (m + n) = f (m)f (n) + f (m − 1)f (n − 1).

Substituting m = a and n = b − 1 gives the Fibonacci identity

Fa+b = Fa+1Fb + FaFb−1.

Justin Stevens Multiplicative Functions (Lecture 5) 6 / 40

Other Properties

The below properties can all be proved using the tiling method:

F1 + F2 + F3 + · · ·+ Fn = Fn+2 − 1.F1 + F3 + F5 + · · ·+ F2n−1 = F2n.F 2

1 + F 22 + F 2

3 + · · ·+ F 2n = FnFn+1.

Fn+1 =(n

0)

+(n−1

1)

+(n−2

2)

+ · · · .Try them out yourself!

Justin Stevens Multiplicative Functions (Lecture 5) 7 / 40

Fibonacci Divisibility

Example. Prove that Fm | Fmq for all natural q.

We use induction. For q = 1, Fm | Fm. For q = 2,

F2m = Fm+1Fm + FmFm−1.

Suppose the statement is true for q = k, implying that Fm | Fmk . We showit holds for q = k + 1. From the identity with a = mk and b = m,

Fmk+m = Fmk+1Fm + FmkFm−1.

Since Fm | Fmk (hypothesis), Fm | Fmk+m by the linear combinationtheorem. Hence, the statement is proven for q = k + 1.

Justin Stevens Multiplicative Functions (Lecture 5) 8 / 40

Fibonacci Divisibility

Example. Prove that Fm | Fmq for all natural q.

We use induction. For q = 1, Fm | Fm. For q = 2,

F2m = Fm+1Fm + FmFm−1.

Suppose the statement is true for q = k, implying that Fm | Fmk .

We showit holds for q = k + 1. From the identity with a = mk and b = m,

Fmk+m = Fmk+1Fm + FmkFm−1.

Since Fm | Fmk (hypothesis), Fm | Fmk+m by the linear combinationtheorem. Hence, the statement is proven for q = k + 1.

Justin Stevens Multiplicative Functions (Lecture 5) 8 / 40

Fibonacci Divisibility

Example. Prove that Fm | Fmq for all natural q.

We use induction. For q = 1, Fm | Fm. For q = 2,

F2m = Fm+1Fm + FmFm−1.

Suppose the statement is true for q = k, implying that Fm | Fmk . We showit holds for q = k + 1. From the identity with a = mk and b = m,

Fmk+m = Fmk+1Fm + FmkFm−1.

Since Fm | Fmk (hypothesis), Fm | Fmk+m by the linear combinationtheorem. Hence, the statement is proven for q = k + 1.

Justin Stevens Multiplicative Functions (Lecture 5) 8 / 40

Fibonacci Divisibility

Example. Prove that Fm | Fmq for all natural q.

We use induction. For q = 1, Fm | Fm. For q = 2,

F2m = Fm+1Fm + FmFm−1.

Suppose the statement is true for q = k, implying that Fm | Fmk . We showit holds for q = k + 1. From the identity with a = mk and b = m,

Fmk+m = Fmk+1Fm + FmkFm−1.

Since Fm | Fmk (hypothesis), Fm | Fmk+m by the linear combinationtheorem. Hence, the statement is proven for q = k + 1.

Justin Stevens Multiplicative Functions (Lecture 5) 8 / 40

Fibonacci GCD 1

Example. Show that consecutive Fibonacci numbers are relatively prime.

Justin Stevens Multiplicative Functions (Lecture 5) 9 / 40

Fibonacci GCD

Example. Show that gcd(Fn,Fm) = Fgcd(n,m).Write n = mq + r using the division algorithm. Using the Fibonacci identity,

Fn = Fmq+r = Fmq+1Fr + FmqFr1 .

Since Fm | Fmq, we can subtract multiples of Fm using Euclidean algorithm:

gcd(Fn,Fm) = gcd(Fmq+1Fr + FmqFr−1,Fm) = gcd(Fmq+1Fr ,Fm).

Finally, gcd(Fmq+1,Fm) = 1 since consecutive Fibonacci numbers arerelatively prime:

gcd(Fn,Fm) = gcd(Fr ,Fm).

For example, if n = 182 and m = 65, gcd(182, 65) = 13 and

gcd(F182,F65) = gcd(F65,F52) = gcd(F52,F13) = F13.

The conclusion is equivalent to gcd(Fn,Fm) = Fgcd(n,m).

Justin Stevens Multiplicative Functions (Lecture 5) 10 / 40

Fibonacci GCD

Example. Show that gcd(Fn,Fm) = Fgcd(n,m).Write n = mq + r using the division algorithm. Using the Fibonacci identity,

Fn = Fmq+r = Fmq+1Fr + FmqFr1 .

Since Fm | Fmq, we can subtract multiples of Fm using Euclidean algorithm:

gcd(Fn,Fm) = gcd(Fmq+1Fr + FmqFr−1,Fm) = gcd(Fmq+1Fr ,Fm).

Finally, gcd(Fmq+1,Fm) = 1 since consecutive Fibonacci numbers arerelatively prime:

gcd(Fn,Fm) = gcd(Fr ,Fm).

For example, if n = 182 and m = 65, gcd(182, 65) = 13 and

gcd(F182,F65) = gcd(F65,F52) = gcd(F52,F13) = F13.

The conclusion is equivalent to gcd(Fn,Fm) = Fgcd(n,m).

Justin Stevens Multiplicative Functions (Lecture 5) 10 / 40

Fibonacci GCD

Example. Show that gcd(Fn,Fm) = Fgcd(n,m).Write n = mq + r using the division algorithm. Using the Fibonacci identity,

Fn = Fmq+r = Fmq+1Fr + FmqFr1 .

Since Fm | Fmq, we can subtract multiples of Fm using Euclidean algorithm:

gcd(Fn,Fm) = gcd(Fmq+1Fr + FmqFr−1,Fm) = gcd(Fmq+1Fr ,Fm).

Finally, gcd(Fmq+1,Fm) = 1 since consecutive Fibonacci numbers arerelatively prime:

gcd(Fn,Fm) = gcd(Fr ,Fm).

For example, if n = 182 and m = 65, gcd(182, 65) = 13 and

gcd(F182,F65) = gcd(F65,F52) = gcd(F52,F13) = F13.

The conclusion is equivalent to gcd(Fn,Fm) = Fgcd(n,m).

Justin Stevens Multiplicative Functions (Lecture 5) 10 / 40

Fibonacci GCD

Example. Show that gcd(Fn,Fm) = Fgcd(n,m).Write n = mq + r using the division algorithm. Using the Fibonacci identity,

Fn = Fmq+r = Fmq+1Fr + FmqFr1 .

Since Fm | Fmq, we can subtract multiples of Fm using Euclidean algorithm:

gcd(Fn,Fm) = gcd(Fmq+1Fr + FmqFr−1,Fm) = gcd(Fmq+1Fr ,Fm).

Finally, gcd(Fmq+1,Fm) = 1 since consecutive Fibonacci numbers arerelatively prime:

gcd(Fn,Fm) = gcd(Fr ,Fm).

For example, if n = 182 and m = 65, gcd(182, 65) = 13 and

gcd(F182,F65) = gcd(F65,F52) = gcd(F52,F13) = F13.

The conclusion is equivalent to gcd(Fn,Fm) = Fgcd(n,m).

Justin Stevens Multiplicative Functions (Lecture 5) 10 / 40

Fibonacci GCD

Example. Show that gcd(Fn,Fm) = Fgcd(n,m).Write n = mq + r using the division algorithm. Using the Fibonacci identity,

Fn = Fmq+r = Fmq+1Fr + FmqFr1 .

Since Fm | Fmq, we can subtract multiples of Fm using Euclidean algorithm:

gcd(Fn,Fm) = gcd(Fmq+1Fr + FmqFr−1,Fm) = gcd(Fmq+1Fr ,Fm).

Finally, gcd(Fmq+1,Fm) = 1 since consecutive Fibonacci numbers arerelatively prime:

gcd(Fn,Fm) = gcd(Fr ,Fm).

For example, if n = 182 and m = 65, gcd(182, 65) = 13 and

gcd(F182,F65) = gcd(F65,F52) = gcd(F52,F13) = F13.

The conclusion is equivalent to gcd(Fn,Fm) = Fgcd(n,m).

Justin Stevens Multiplicative Functions (Lecture 5) 10 / 40

Fibonacci GCD

Example. Show that gcd(Fn,Fm) = Fgcd(n,m).Write n = mq + r using the division algorithm. Using the Fibonacci identity,

Fn = Fmq+r = Fmq+1Fr + FmqFr1 .

Since Fm | Fmq, we can subtract multiples of Fm using Euclidean algorithm:

gcd(Fn,Fm) = gcd(Fmq+1Fr + FmqFr−1,Fm) = gcd(Fmq+1Fr ,Fm).

Finally, gcd(Fmq+1,Fm) = 1 since consecutive Fibonacci numbers arerelatively prime:

gcd(Fn,Fm) = gcd(Fr ,Fm).

For example, if n = 182 and m = 65, gcd(182, 65) = 13 and

gcd(F182,F65) = gcd(F65,F52) = gcd(F52,F13) = F13.

The conclusion is equivalent to gcd(Fn,Fm) = Fgcd(n,m).

Justin Stevens Multiplicative Functions (Lecture 5) 10 / 40

Outline

1 Fibonacci Numbers

2 Linear Number TheoryCongruencesChicken Mcnugget Theorem

3 Multiplicative Functions

Justin Stevens Multiplicative Functions (Lecture 5) 11 / 40

Modular Inverses

Theorem. Prove that the inverse of a mod m exists iff gcd(a,m) = 1.

Proof.By Bezout’s theorem, there exists integers x and y such that

ax + my = 1

if and only if gcd(a,m) = 1. Taking this equation mod m, we see thatax ≡ 1 (mod m).

Justin Stevens Multiplicative Functions (Lecture 5) 12 / 40

Modular Inverses

Theorem. Prove that the inverse of a mod m exists iff gcd(a,m) = 1.

Proof.By Bezout’s theorem, there exists integers x and y such that

ax + my = 1

if and only if gcd(a,m) = 1.

Taking this equation mod m, we see thatax ≡ 1 (mod m).

Justin Stevens Multiplicative Functions (Lecture 5) 12 / 40

Modular Inverses

Theorem. Prove that the inverse of a mod m exists iff gcd(a,m) = 1.

Proof.By Bezout’s theorem, there exists integers x and y such that

ax + my = 1

if and only if gcd(a,m) = 1. Taking this equation mod m, we see thatax ≡ 1 (mod m).

Justin Stevens Multiplicative Functions (Lecture 5) 12 / 40

Method

Example 3. Solve 39x ≡ 1 mod 100 by finding integers satisfying39x0 + 100y0 = 1.

Solution. We use the Euclidean algorithm:

100 = 39 · 2 + 22 1 = 16 · 100− 41 · 39 ↑39 = 22 · 1 + 17 1 = −9 · 39 + 16 · 22 ↑22 = 17 · 1 + 5 1 = 7 · 22− 9 · 17 ↑17 = 5 · 3 + 2 1 = −2 · 17 + 7 · 5 ↑5 = 2 · 2 + 1 =⇒ 1 = 5− 2 · 2 ↑

Therefore (x0, y0) = (−41, 16), hence x ≡ −41 ≡ 59 (mod 100).Observe the pair (59,−23) also satisfies the equation. This hints at ageneral solution.

Justin Stevens Multiplicative Functions (Lecture 5) 13 / 40

Method

Example 4. Solve 39x ≡ 1 mod 100 by finding integers satisfying39x0 + 100y0 = 1.Solution. We use the Euclidean algorithm:

100 = 39 · 2 + 22 1 = 16 · 100− 41 · 39 ↑39 = 22 · 1 + 17 1 = −9 · 39 + 16 · 22 ↑22 = 17 · 1 + 5 1 = 7 · 22− 9 · 17 ↑17 = 5 · 3 + 2 1 = −2 · 17 + 7 · 5 ↑5 = 2 · 2 + 1 =⇒ 1 = 5− 2 · 2 ↑

Therefore (x0, y0) = (−41, 16), hence x ≡ −41 ≡ 59 (mod 100).

Observe the pair (59,−23) also satisfies the equation. This hints at ageneral solution.

Justin Stevens Multiplicative Functions (Lecture 5) 13 / 40

Method

Example 5. Solve 39x ≡ 1 mod 100 by finding integers satisfying39x0 + 100y0 = 1.Solution. We use the Euclidean algorithm:

100 = 39 · 2 + 22 1 = 16 · 100− 41 · 39 ↑39 = 22 · 1 + 17 1 = −9 · 39 + 16 · 22 ↑22 = 17 · 1 + 5 1 = 7 · 22− 9 · 17 ↑17 = 5 · 3 + 2 1 = −2 · 17 + 7 · 5 ↑5 = 2 · 2 + 1 =⇒ 1 = 5− 2 · 2 ↑

Therefore (x0, y0) = (−41, 16), hence x ≡ −41 ≡ 59 (mod 100).Observe the pair (59,−23) also satisfies the equation. This hints at ageneral solution.

Justin Stevens Multiplicative Functions (Lecture 5) 13 / 40

Linear Diophantine Equations

Theorem. For coprime a and b, all integer solutions to ax + by = c are

(x , y) = (x0 + bk, y0 − ak), k ∈ Z,

where (x0, y0) is a particular solution.

All integer solutions of 2x + 5y = 1 are (x , y) = (−2 + 5k, 1− 2k).

Justin Stevens Multiplicative Functions (Lecture 5) 14 / 40

Linear Diophantine Equations

Theorem. For coprime a and b, all integer solutions to ax + by = c are

(x , y) = (x0 + bk, y0 − ak), k ∈ Z,

where (x0, y0) is a particular solution.

All integer solutions of 2x + 5y = 1 are (x , y) = (−2 + 5k, 1− 2k).

Justin Stevens Multiplicative Functions (Lecture 5) 14 / 40

Solving Linear Congruences

Example. Solve the linear congruence 7x ≡ 3 (mod 34).

We begin by observing that 7−1 ≡ 5 (mod 34). Muliplying by 3:

x ≡ 3 · 7−1 ≡ 3 · 5 ≡ 15 (mod 34).

Corollary. The congruence ax ≡ c mod m is solved by

x ≡ a−1c mod m

for gcd(a,m) = 1.

Justin Stevens Multiplicative Functions (Lecture 5) 15 / 40

Solving Linear Congruences

Example. Solve the linear congruence 7x ≡ 3 (mod 34).We begin by observing that 7−1 ≡ 5 (mod 34). Muliplying by 3:

x ≡ 3 · 7−1 ≡ 3 · 5 ≡ 15 (mod 34).

Corollary. The congruence ax ≡ c mod m is solved by

x ≡ a−1c mod m

for gcd(a,m) = 1.

Justin Stevens Multiplicative Functions (Lecture 5) 15 / 40

Cancellation Law

Theorem. Prove that if ca ≡ cb (mod m), then a ≡ b (mod m/d),where d = gcd(c,m).We begin by observing that by the definition of modulos,

ca ≡ cb (mod m) ⇐⇒ m | ca − cb ⇐⇒ c (a − b) = mk.

Since d = gcd(c,m), we write c = dr and m = ds where gcd(r , s) = 1:

dr (a − b) = dsk ⇐⇒ r (a − b) = sk.

Therefore, s | r (a − b). By Euclid’s Lemma, since s | a − b.

For instance, 3x ≡ 3 (mod 9) =⇒ x ≡ 1 (mod 3).

Justin Stevens Multiplicative Functions (Lecture 5) 16 / 40

Cancellation Law

Theorem. Prove that if ca ≡ cb (mod m), then a ≡ b (mod m/d),where d = gcd(c,m).We begin by observing that by the definition of modulos,

ca ≡ cb (mod m) ⇐⇒ m | ca − cb ⇐⇒ c (a − b) = mk.

Since d = gcd(c,m), we write c = dr and m = ds where gcd(r , s) = 1:

dr (a − b) = dsk ⇐⇒ r (a − b) = sk.

Therefore, s | r (a − b). By Euclid’s Lemma, since s | a − b.

For instance, 3x ≡ 3 (mod 9) =⇒ x ≡ 1 (mod 3).

Justin Stevens Multiplicative Functions (Lecture 5) 16 / 40

Cancellation Law

Theorem. Prove that if ca ≡ cb (mod m), then a ≡ b (mod m/d),where d = gcd(c,m).We begin by observing that by the definition of modulos,

ca ≡ cb (mod m) ⇐⇒ m | ca − cb ⇐⇒ c (a − b) = mk.

Since d = gcd(c,m), we write c = dr and m = ds where gcd(r , s) = 1:

dr (a − b) = dsk ⇐⇒ r (a − b) = sk.

Therefore, s | r (a − b). By Euclid’s Lemma, since s | a − b.

For instance, 3x ≡ 3 (mod 9) =⇒ x ≡ 1 (mod 3).

Justin Stevens Multiplicative Functions (Lecture 5) 16 / 40

Cancellation Law

Theorem. Prove that if ca ≡ cb (mod m), then a ≡ b (mod m/d),where d = gcd(c,m).We begin by observing that by the definition of modulos,

ca ≡ cb (mod m) ⇐⇒ m | ca − cb ⇐⇒ c (a − b) = mk.

Since d = gcd(c,m), we write c = dr and m = ds where gcd(r , s) = 1:

dr (a − b) = dsk ⇐⇒ r (a − b) = sk.

Therefore, s | r (a − b). By Euclid’s Lemma, since s | a − b.

For instance, 3x ≡ 3 (mod 9) =⇒ x ≡ 1 (mod 3).

Justin Stevens Multiplicative Functions (Lecture 5) 16 / 40

Cancellation Law

Theorem. Prove that if ca ≡ cb (mod m), then a ≡ b (mod m/d),where d = gcd(c,m).We begin by observing that by the definition of modulos,

ca ≡ cb (mod m) ⇐⇒ m | ca − cb ⇐⇒ c (a − b) = mk.

Since d = gcd(c,m), we write c = dr and m = ds where gcd(r , s) = 1:

dr (a − b) = dsk ⇐⇒ r (a − b) = sk.

Therefore, s | r (a − b). By Euclid’s Lemma, since s | a − b.

For instance, 3x ≡ 3 (mod 9) =⇒ x ≡ 1 (mod 3).

Justin Stevens Multiplicative Functions (Lecture 5) 16 / 40

No Solutions

Example. Show that the congruence 5x ≡ 3 (mod 10) has no solutions.

From the definition of modulos, we see that

5x ≡ 3 (mod 10) ⇐⇒ 10 | 5x − 3 ⇐⇒ 5x − 3 = 10y

for some integer y . However, this implies that 5 (x − 2y) = 3,contradiction.

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Justin Stevens Multiplicative Functions (Lecture 5) 17 / 40

No Solutions

Example. Show that the congruence 5x ≡ 3 (mod 10) has no solutions.From the definition of modulos, we see that

5x ≡ 3 (mod 10) ⇐⇒ 10 | 5x − 3 ⇐⇒ 5x − 3 = 10y

for some integer y . However, this implies that 5 (x − 2y) = 3,contradiction.

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Justin Stevens Multiplicative Functions (Lecture 5) 17 / 40

No Solutions

Example. Show that the congruence 5x ≡ 3 (mod 10) has no solutions.From the definition of modulos, we see that

5x ≡ 3 (mod 10) ⇐⇒ 10 | 5x − 3 ⇐⇒ 5x − 3 = 10y

for some integer y . However, this implies that 5 (x − 2y) = 3,contradiction.

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Justin Stevens Multiplicative Functions (Lecture 5) 17 / 40

Proof of Linear Congruence Theorem

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Assume that the congruence has solutions. Then, observe that

ax ≡ c (mod m) ⇐⇒ m | ax − c ⇐⇒ ax − c = my .

Since gcd(a,m) | a and gcd(a,m) | m, we must have gcd(a,m) | b.

We now show that if gcd(a,m) | b, then there are solutions. Letd = gcd(a,m), therefore, there exists relatively prime integers a1 and m1such that a = da1 and m = dm1. Furthermore, since d | b, there exists aninteger b1 such that b = db1. We now rewrite:

da1x ≡ db1 (mod dm1) ⇐⇒ a1x ≡ b1 (mod m1).

This congruence has a solution, namely x ≡ b1a−11 (mod m1).

Justin Stevens Multiplicative Functions (Lecture 5) 18 / 40

Proof of Linear Congruence Theorem

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Assume that the congruence has solutions. Then, observe that

ax ≡ c (mod m) ⇐⇒ m | ax − c ⇐⇒ ax − c = my .

Since gcd(a,m) | a and gcd(a,m) | m, we must have gcd(a,m) | b.

We now show that if gcd(a,m) | b, then there are solutions. Letd = gcd(a,m), therefore, there exists relatively prime integers a1 and m1such that a = da1 and m = dm1. Furthermore, since d | b, there exists aninteger b1 such that b = db1. We now rewrite:

da1x ≡ db1 (mod dm1) ⇐⇒ a1x ≡ b1 (mod m1).

This congruence has a solution, namely x ≡ b1a−11 (mod m1).

Justin Stevens Multiplicative Functions (Lecture 5) 18 / 40

Proof of Linear Congruence Theorem

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Assume that the congruence has solutions. Then, observe that

ax ≡ c (mod m) ⇐⇒ m | ax − c ⇐⇒ ax − c = my .

Since gcd(a,m) | a and gcd(a,m) | m, we must have gcd(a,m) | b.

We now show that if gcd(a,m) | b, then there are solutions. Letd = gcd(a,m), therefore, there exists relatively prime integers a1 and m1such that a = da1 and m = dm1. Furthermore, since d | b, there exists aninteger b1 such that b = db1. We now rewrite:

da1x ≡ db1 (mod dm1) ⇐⇒ a1x ≡ b1 (mod m1).

This congruence has a solution, namely x ≡ b1a−11 (mod m1).

Justin Stevens Multiplicative Functions (Lecture 5) 18 / 40

Proof of Linear Congruence Theorem

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Assume that the congruence has solutions. Then, observe that

ax ≡ c (mod m) ⇐⇒ m | ax − c ⇐⇒ ax − c = my .

Since gcd(a,m) | a and gcd(a,m) | m, we must have gcd(a,m) | b.

We now show that if gcd(a,m) | b, then there are solutions. Letd = gcd(a,m), therefore, there exists relatively prime integers a1 and m1such that a = da1 and m = dm1.

Furthermore, since d | b, there exists aninteger b1 such that b = db1. We now rewrite:

da1x ≡ db1 (mod dm1) ⇐⇒ a1x ≡ b1 (mod m1).

This congruence has a solution, namely x ≡ b1a−11 (mod m1).

Justin Stevens Multiplicative Functions (Lecture 5) 18 / 40

Proof of Linear Congruence Theorem

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Assume that the congruence has solutions. Then, observe that

ax ≡ c (mod m) ⇐⇒ m | ax − c ⇐⇒ ax − c = my .

Since gcd(a,m) | a and gcd(a,m) | m, we must have gcd(a,m) | b.

We now show that if gcd(a,m) | b, then there are solutions. Letd = gcd(a,m), therefore, there exists relatively prime integers a1 and m1such that a = da1 and m = dm1. Furthermore, since d | b, there exists aninteger b1 such that b = db1. We now rewrite:

da1x ≡ db1 (mod dm1) ⇐⇒ a1x ≡ b1 (mod m1).

This congruence has a solution, namely x ≡ b1a−11 (mod m1).

Justin Stevens Multiplicative Functions (Lecture 5) 18 / 40

Proof of Linear Congruence Theorem

Theorem. ax ≡ c mod m has solutions if and only if gcd(a,m) | c.

Assume that the congruence has solutions. Then, observe that

ax ≡ c (mod m) ⇐⇒ m | ax − c ⇐⇒ ax − c = my .

Since gcd(a,m) | a and gcd(a,m) | m, we must have gcd(a,m) | b.

We now show that if gcd(a,m) | b, then there are solutions. Letd = gcd(a,m), therefore, there exists relatively prime integers a1 and m1such that a = da1 and m = dm1. Furthermore, since d | b, there exists aninteger b1 such that b = db1. We now rewrite:

da1x ≡ db1 (mod dm1) ⇐⇒ a1x ≡ b1 (mod m1).

This congruence has a solution, namely x ≡ b1a−11 (mod m1).

Justin Stevens Multiplicative Functions (Lecture 5) 18 / 40

Example

Example. Find all solutions to the congruence 18x ≡ 30 (mod 42).

We can divide the congruence by gcd(18, 42) = 6:

3x ≡ 5 (mod 7).

Listing numbers that are 5 mod 7, namely 5, 12, 19, x ≡ 4 (mod 7):

x ≡ 4, 11, 18, 25, 32, 39 (mod 42).

Justin Stevens Multiplicative Functions (Lecture 5) 19 / 40

Example

Example. Find all solutions to the congruence 18x ≡ 30 (mod 42).We can divide the congruence by gcd(18, 42) = 6:

3x ≡ 5 (mod 7).

Listing numbers that are 5 mod 7, namely 5, 12, 19, x ≡ 4 (mod 7):

x ≡ 4, 11, 18, 25, 32, 39 (mod 42).

Justin Stevens Multiplicative Functions (Lecture 5) 19 / 40

Example

Example. Find all solutions to the congruence 18x ≡ 30 (mod 42).We can divide the congruence by gcd(18, 42) = 6:

3x ≡ 5 (mod 7).

Listing numbers that are 5 mod 7, namely 5, 12, 19, x ≡ 4 (mod 7):

x ≡ 4, 11, 18, 25, 32, 39 (mod 42).

Justin Stevens Multiplicative Functions (Lecture 5) 19 / 40

Example

Example. Find all solutions to the congruence 18x ≡ 30 (mod 42).We can divide the congruence by gcd(18, 42) = 6:

3x ≡ 5 (mod 7).

Listing numbers that are 5 mod 7, namely 5, 12, 19, x ≡ 4 (mod 7):

x ≡ 4, 11, 18, 25, 32, 39 (mod 42).

Justin Stevens Multiplicative Functions (Lecture 5) 19 / 40

Wrapping Up

Theorem. For integers a and b, let d = gcd(a, b). If (x0, y0) is a particularsolution of the Diophantine equation ax + by = c , then the general solutionis given by

(x , y) = (x0 + bd · k, y0 −

ad · k), k ∈ Z.

Corollary. The linear congruence ax ≡ c (mod m) has d = gcd(a,m)mutually incongruent solutions if d | c.

Justin Stevens Multiplicative Functions (Lecture 5) 20 / 40

Chicken Mcnugget Theorem

Definition. Let a1, a2, · · · , an be relatively prime positive integers.Consider the set

S = a1x1 + a2x2 + · · ·+ anxn, xi ≥ 0.

The largest positive integer that is non-representable is called the Frobeniusnumber of the set and is denoted g(a1, a2, · · · , an).

Observe that if the restriction xi ≥ 0 was not in place, then S would simplycontain multiples of gcd(a1, a2, · · · , an) by generalized Bezout’s theorem.Furthermore, if the numbers a1, a2, · · · , an are not relatively prime, then theanswer is infinity.

Theorem. For relatively prime positive integers a and b,

g(a, b) = ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 21 / 40

Chicken Mcnugget Theorem

Definition. Let a1, a2, · · · , an be relatively prime positive integers.Consider the set

S = a1x1 + a2x2 + · · ·+ anxn, xi ≥ 0.

The largest positive integer that is non-representable is called the Frobeniusnumber of the set and is denoted g(a1, a2, · · · , an).

Observe that if the restriction xi ≥ 0 was not in place, then S would simplycontain multiples of gcd(a1, a2, · · · , an) by generalized Bezout’s theorem.Furthermore, if the numbers a1, a2, · · · , an are not relatively prime, then theanswer is infinity.

Theorem. For relatively prime positive integers a and b,

g(a, b) = ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 21 / 40

Chicken Mcnugget Theorem

Definition. Let a1, a2, · · · , an be relatively prime positive integers.Consider the set

S = a1x1 + a2x2 + · · ·+ anxn, xi ≥ 0.

The largest positive integer that is non-representable is called the Frobeniusnumber of the set and is denoted g(a1, a2, · · · , an).

Observe that if the restriction xi ≥ 0 was not in place, then S would simplycontain multiples of gcd(a1, a2, · · · , an) by generalized Bezout’s theorem.Furthermore, if the numbers a1, a2, · · · , an are not relatively prime, then theanswer is infinity.

Theorem. For relatively prime positive integers a and b,

g(a, b) = ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 21 / 40

Proof Part 1

Theorem. For coprime a and b, g(a, b) = ab − a − b.

For the sake of contradiction, assume that for nonnegative integers x and y ,

ab − a − b = ax + by .

Taking mod a gives by ≡ −b (mod a) =⇒ y ≡ −1 (mod a). Similarly,ax ≡ −a (mod b) =⇒ x ≡ −1 (mod b). We see that

ax + by ≥ a (b − 1) + b (a − 1) = 2ab − a − b > ab − a − b.

Therefore, ab − a − b is non-representable, and g(a, b) ≥ ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 22 / 40

Proof Part 1

Theorem. For coprime a and b, g(a, b) = ab − a − b.

For the sake of contradiction, assume that for nonnegative integers x and y ,

ab − a − b = ax + by .

Taking mod a gives by ≡ −b (mod a) =⇒ y ≡ −1 (mod a). Similarly,ax ≡ −a (mod b) =⇒ x ≡ −1 (mod b). We see that

ax + by ≥ a (b − 1) + b (a − 1) = 2ab − a − b > ab − a − b.

Therefore, ab − a − b is non-representable, and g(a, b) ≥ ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 22 / 40

Proof Part 1

Theorem. For coprime a and b, g(a, b) = ab − a − b.

For the sake of contradiction, assume that for nonnegative integers x and y ,

ab − a − b = ax + by .

Taking mod a gives by ≡ −b (mod a) =⇒ y ≡ −1 (mod a).

Similarly,ax ≡ −a (mod b) =⇒ x ≡ −1 (mod b). We see that

ax + by ≥ a (b − 1) + b (a − 1) = 2ab − a − b > ab − a − b.

Therefore, ab − a − b is non-representable, and g(a, b) ≥ ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 22 / 40

Proof Part 1

Theorem. For coprime a and b, g(a, b) = ab − a − b.

For the sake of contradiction, assume that for nonnegative integers x and y ,

ab − a − b = ax + by .

Taking mod a gives by ≡ −b (mod a) =⇒ y ≡ −1 (mod a). Similarly,ax ≡ −a (mod b) =⇒ x ≡ −1 (mod b).

We see that

ax + by ≥ a (b − 1) + b (a − 1) = 2ab − a − b > ab − a − b.

Therefore, ab − a − b is non-representable, and g(a, b) ≥ ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 22 / 40

Proof Part 1

Theorem. For coprime a and b, g(a, b) = ab − a − b.

For the sake of contradiction, assume that for nonnegative integers x and y ,

ab − a − b = ax + by .

Taking mod a gives by ≡ −b (mod a) =⇒ y ≡ −1 (mod a). Similarly,ax ≡ −a (mod b) =⇒ x ≡ −1 (mod b). We see that

ax + by ≥ a (b − 1) + b (a − 1) = 2ab − a − b > ab − a − b.

Therefore, ab − a − b is non-representable, and g(a, b) ≥ ab − a − b.

Justin Stevens Multiplicative Functions (Lecture 5) 22 / 40

Proof Part 2

Theorem. For coprime a and b, g(a, b) = ab − a − b.

Take any number M > ab − a − b and consider ax + by = M. By Bezout’stheorem, we know there exists integers x0 and y0 such that ax0 + by0 = 1.

Multiplying by M gives aMx0 + bMy0 = M. Therefore, for an integer k, thesolutions to the diophantine equation are parametized by

(x , y) = (Mx0 + kb,My0 − ka).

We desire to find a solution such that x and y nonnegative. By the divisionalgorithm, we can choose k such that 0 ≤ x ≤ b − 1. For this specific x ,

ax + by = M > ab − a − b =⇒ b(y + 1) > a (b − 1− x) ≥ 0,

therefore y is also nonnegative. Hence, every integer M > ab − a − b isrepresentable.

Justin Stevens Multiplicative Functions (Lecture 5) 23 / 40

Extension

Example. There are (a − 1)(b − 1)/2 non-representable integers.We try an example with a = 12 and b = 5.

0 1 2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 21 22 2324 25 26 27 28 29 30 31 32 33 34 3536 37 38 39 40 41 42 43 44 45 46 4748 49 50 51 52 53 54 55 56 57 58 59

Observe that all the numbers above the multiples of 5 are non-representable.For every 0 ≤ j ≤ a − 1, let mj be the smallest nonnegative solution to

ax + j ≡ 0 (mod b).

In the above example, m0 = 0,m1 = 2, m2 = 4, m3 = 1, m4 = 3, and soforth. Clearly, there are mj non-representable integers in each column.

Justin Stevens Multiplicative Functions (Lecture 5) 24 / 40

Extension

Example. There are (a − 1)(b − 1)/2 non-representable integers.We try an example with a = 12 and b = 5.

0 1 2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 21 22 2324 25 26 27 28 29 30 31 32 33 34 3536 37 38 39 40 41 42 43 44 45 46 4748 49 50 51 52 53 54 55 56 57 58 59

Observe that all the numbers above the multiples of 5 are non-representable.For every 0 ≤ j ≤ a − 1, let mj be the smallest nonnegative solution to

ax + j ≡ 0 (mod b).

In the above example, m0 = 0,m1 = 2, m2 = 4, m3 = 1, m4 = 3, and soforth. Clearly, there are mj non-representable integers in each column.

Justin Stevens Multiplicative Functions (Lecture 5) 24 / 40

Extension

Example. There are (a − 1)(b − 1)/2 non-representable integers.We try an example with a = 12 and b = 5.

0 1 2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 21 22 2324 25 26 27 28 29 30 31 32 33 34 3536 37 38 39 40 41 42 43 44 45 46 4748 49 50 51 52 53 54 55 56 57 58 59

Observe that all the numbers above the multiples of 5 are non-representable.For every 0 ≤ j ≤ a − 1, let mj be the smallest nonnegative solution to

ax + j ≡ 0 (mod b).

In the above example, m0 = 0,m1 = 2, m2 = 4, m3 = 1, m4 = 3, and soforth. Clearly, there are mj non-representable integers in each column.

Justin Stevens Multiplicative Functions (Lecture 5) 24 / 40

Extension Part 2

Consider the multiples of b, 0, b, 2b, 3b, · · · , (a − 1)b. I claim that everymember of this set lies in a different column. Indeed, if for some integers sand t, bs ≡ bt (mod a), then since gcd(a, b) = 1, s ≡ t (mod a).

For each mj , let amj + j = bnj . We see that nj takes on the values0, 1, 2, 3, · · · , a − 1, each exactly once. Summing over the circled numbers:

a−1∑j=0

(amj + j) =a−1∑j=0

(bnj)

aa−1∑j=0

mj +a−1∑j=0

j = ba−1∑j=0

nj

a−1∑j=0

mj = (a − 1)(b − 1)2 .

Since the left hand side is the number of non-representable integers, ourproof is complete.

Justin Stevens Multiplicative Functions (Lecture 5) 25 / 40

Extension Part 2

Consider the multiples of b, 0, b, 2b, 3b, · · · , (a − 1)b. I claim that everymember of this set lies in a different column. Indeed, if for some integers sand t, bs ≡ bt (mod a), then since gcd(a, b) = 1, s ≡ t (mod a).For each mj , let amj + j = bnj . We see that nj takes on the values0, 1, 2, 3, · · · , a − 1, each exactly once. Summing over the circled numbers:

a−1∑j=0

(amj + j) =a−1∑j=0

(bnj)

aa−1∑j=0

mj +a−1∑j=0

j = ba−1∑j=0

nj

a−1∑j=0

mj = (a − 1)(b − 1)2 .

Since the left hand side is the number of non-representable integers, ourproof is complete.

Justin Stevens Multiplicative Functions (Lecture 5) 25 / 40

Extension Part 2

Consider the multiples of b, 0, b, 2b, 3b, · · · , (a − 1)b. I claim that everymember of this set lies in a different column. Indeed, if for some integers sand t, bs ≡ bt (mod a), then since gcd(a, b) = 1, s ≡ t (mod a).For each mj , let amj + j = bnj . We see that nj takes on the values0, 1, 2, 3, · · · , a − 1, each exactly once. Summing over the circled numbers:

a−1∑j=0

(amj + j) =a−1∑j=0

(bnj)

aa−1∑j=0

mj +a−1∑j=0

j = ba−1∑j=0

nj

a−1∑j=0

mj = (a − 1)(b − 1)2 .

Since the left hand side is the number of non-representable integers, ourproof is complete.

Justin Stevens Multiplicative Functions (Lecture 5) 25 / 40

Outline

1 Fibonacci Numbers

2 Linear Number Theory

3 Multiplicative Functions

Justin Stevens Multiplicative Functions (Lecture 5) 26 / 40

Divisors of a Number

Theorem. If n = pk11 pk2

2 · · · pkrk , then the positive divisors of n are of the

formd = pa1

1 pa22 · · · p

arr , 0 ≤ ai ≤ ki .

Example 6. (AIME) Compute the probability that a randomly chosendivisor of 1099 is an integer multiple of 1088.

Justin Stevens Multiplicative Functions (Lecture 5) 27 / 40

Solution to AIME problem

Example 7. (AIME) Compute the probability that a randomly chosendivisor of 1099 is an integer multiple of 1088.

Solution. The divisors of 1099 = 299 · 599 are of the form d = 2a · 5b where0 ≤ a, b ≤ 99. Hence, 1099 has 100 · 100 divisors. If d is a multiple of 1088,then 88 ≤ a, b ≤ 99. Hence, there are 12 · 12 divisors that are multiples of1088. The desired probability is 12·12

100·100 = 9625 .

Justin Stevens Multiplicative Functions (Lecture 5) 28 / 40

Solution to AIME problem

Example 8. (AIME) Compute the probability that a randomly chosendivisor of 1099 is an integer multiple of 1088.

Solution. The divisors of 1099 = 299 · 599 are of the form d = 2a · 5b where0 ≤ a, b ≤ 99. Hence, 1099 has 100 · 100 divisors. If d is a multiple of 1088,then 88 ≤ a, b ≤ 99. Hence, there are 12 · 12 divisors that are multiples of1088. The desired probability is 12·12

100·100 = 9625 .

Justin Stevens Multiplicative Functions (Lecture 5) 28 / 40

Tau and Sigma

Definition. τ(n) is the number of positive divisors of n and σ(n) is thesum of those divisors.For n = 12, σ(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28 and τ(12) = 6.

Theorem. If n = pk11 pk2

2 · · · pkrk , then the number and sum of positive

divisors of n are

τ(n) = (k1 + 1) (k2 + 1) · · · (kr + 1) .

σ(n) =(

pk1+11 − 1p1 − 1

)(pk2+1

2 − 1p2 − 1

)· · ·(

pkr +1k − 1pk − 1

).

Justin Stevens Multiplicative Functions (Lecture 5) 29 / 40

Proof of Formulas

Proof.Let d be an arbitrary divisor of n, so d = pa1

1 pa22 · · · par

r where 0 ≤ ai ≤ ki .For each prime, we have ki + 1 choices for the exponent of pi . Hence,

τ(n) = (k1 + 1) (k2 + 1) · · · (kr + 1) .

Furthermore, each divisor of n appears exactly once in the product(1 + p1 + p2

1 + · · ·+ pk11

) (1 + p2 + p2

2 + · · ·+ pk22

)· · ·

Therefore, applying the geometric series formula

1 + p1 + p21 + · · ·+ pk1

1 = pk1+11 − 1p1 − 1

the result follows for σ(n).

Justin Stevens Multiplicative Functions (Lecture 5) 30 / 40

Tau and Sigma Problems

Example 9. The cells in a jail are numbered from 1 to 100 and theirdoors are activated from a central button. The activation opens aclosed door and closes an open door. Starting with all the doorsclosed the button is pressed 100 times. When it is pressed the k-thtime the doors that are multiples of k are activated. Which doorsare open at the end?

Example 10. Show that the product of the divisors of a number nis nτ(n)/2.

Justin Stevens Multiplicative Functions (Lecture 5) 31 / 40

Jail Cell Solution

Solution. The number 100 seems arbitrary, so we try the problem for a jailwith 10 cells. We create a table of which jail cells are activated each timewe press the button:

Jail Cells1 2 3 4 5 6 7 8 9 10

1 X X X X X X X X X X2 X X X X X X X X X X3 X X X X X X X X X X4 X X X X X X X X X X5 X X X X X X X X X X

Button

s

6 X X X X X X X X X X7 X X X X X X X X X X8 X X X X X X X X X X9 X X X X X X X X X X

10 X X X X X X X X X X1 2 2 3 2 4 2 4 3 4

Justin Stevens Multiplicative Functions (Lecture 5) 32 / 40

Jail Cell Solution 2

Indeed, door n will be activated when the button is pressed the d-th time ifand only if d | n. Therefore, the number of activations is equal to thenumber of positive divisors of n, or τ(n).

We therefore wish to find when τ(n) is odd. By our formula, we see that

τ(n) =∏

(ki + 1) .

If τ(n) is odd, then every term in the product must be too, therefore, kimust be even. Hence, every exponent in the prime factorization of n is even,and n must be a perfect square.

For the original problem, the open cells are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Justin Stevens Multiplicative Functions (Lecture 5) 33 / 40

Jail Cell Solution 2

Indeed, door n will be activated when the button is pressed the d-th time ifand only if d | n. Therefore, the number of activations is equal to thenumber of positive divisors of n, or τ(n).

We therefore wish to find when τ(n) is odd. By our formula, we see that

τ(n) =∏

(ki + 1) .

If τ(n) is odd, then every term in the product must be too, therefore, kimust be even. Hence, every exponent in the prime factorization of n is even,and n must be a perfect square.

For the original problem, the open cells are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Justin Stevens Multiplicative Functions (Lecture 5) 33 / 40

Jail Cell Solution 2

Indeed, door n will be activated when the button is pressed the d-th time ifand only if d | n. Therefore, the number of activations is equal to thenumber of positive divisors of n, or τ(n).

We therefore wish to find when τ(n) is odd. By our formula, we see that

τ(n) =∏

(ki + 1) .

If τ(n) is odd, then every term in the product must be too, therefore, kimust be even. Hence, every exponent in the prime factorization of n is even,and n must be a perfect square.

For the original problem, the open cells are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Justin Stevens Multiplicative Functions (Lecture 5) 33 / 40

Product of Divisors of n

Example 11. Show that the product of the divisors of a number n isnτ(n)/2.Solution. Let d be an arbitrary divisor of n so that n = dd ′. We thereforesee that ∏

d |nd ·∏d ′|n

d ′ = nτ(n).

As d ranges through the divisors, d ′ does the same in reverse order. Hence,∏d |n d =

∏d ′|n d ′:

∏d |n

d

2

= nτ(n) =⇒∏d |n

d = nτ(n)/2.

Justin Stevens Multiplicative Functions (Lecture 5) 34 / 40

Multiplicative Function Definition

After studying these functions, we introduce multiplicative functions.Definition. A function f is multiplicative if whenever gcd(a, b) = 1,

f (ab) = f (a)f (b).

By induction, if n1, n2, n3, · · · , nr are pairwise relatively prime, then

f (n1n2 · · · nr ) = f (n1)f (n2) · · · f (nr ).

Hence, if n = pk11 pk2

2 · · · pkrr is the prime factorzation of a number,

f (n) = f (pk11 )f (pk2

2 ) · · · f (pkrr ).

Therefore, we only evaluate multiplicative functions up to prime powers.Also, since f (a · 1) = f (a) · f (1), so we must have f (1) = 1 if f 6= 0.

f is completely multiplicative if f (ab) = f (a)f (b) for all natural a, b. Forexample, the functions id(n) = n and 1 are completely multiplicative.

Justin Stevens Multiplicative Functions (Lecture 5) 35 / 40

Multiplicative Function Definition

After studying these functions, we introduce multiplicative functions.Definition. A function f is multiplicative if whenever gcd(a, b) = 1,

f (ab) = f (a)f (b).

By induction, if n1, n2, n3, · · · , nr are pairwise relatively prime, then

f (n1n2 · · · nr ) = f (n1)f (n2) · · · f (nr ).

Hence, if n = pk11 pk2

2 · · · pkrr is the prime factorzation of a number,

f (n) = f (pk11 )f (pk2

2 ) · · · f (pkrr ).

Therefore, we only evaluate multiplicative functions up to prime powers.Also, since f (a · 1) = f (a) · f (1), so we must have f (1) = 1 if f 6= 0.

f is completely multiplicative if f (ab) = f (a)f (b) for all natural a, b. Forexample, the functions id(n) = n and 1 are completely multiplicative.

Justin Stevens Multiplicative Functions (Lecture 5) 35 / 40

Multiplicative Function Definition

After studying these functions, we introduce multiplicative functions.Definition. A function f is multiplicative if whenever gcd(a, b) = 1,

f (ab) = f (a)f (b).

By induction, if n1, n2, n3, · · · , nr are pairwise relatively prime, then

f (n1n2 · · · nr ) = f (n1)f (n2) · · · f (nr ).

Hence, if n = pk11 pk2

2 · · · pkrr is the prime factorzation of a number,

f (n) = f (pk11 )f (pk2

2 ) · · · f (pkrr ).

Therefore, we only evaluate multiplicative functions up to prime powers.Also, since f (a · 1) = f (a) · f (1), so we must have f (1) = 1 if f 6= 0.

f is completely multiplicative if f (ab) = f (a)f (b) for all natural a, b. Forexample, the functions id(n) = n and 1 are completely multiplicative.

Justin Stevens Multiplicative Functions (Lecture 5) 35 / 40

Multiplicative Function Definition

After studying these functions, we introduce multiplicative functions.Definition. A function f is multiplicative if whenever gcd(a, b) = 1,

f (ab) = f (a)f (b).

By induction, if n1, n2, n3, · · · , nr are pairwise relatively prime, then

f (n1n2 · · · nr ) = f (n1)f (n2) · · · f (nr ).

Hence, if n = pk11 pk2

2 · · · pkrr is the prime factorzation of a number,

f (n) = f (pk11 )f (pk2

2 ) · · · f (pkrr ).

Therefore, we only evaluate multiplicative functions up to prime powers.Also, since f (a · 1) = f (a) · f (1), so we must have f (1) = 1 if f 6= 0.

f is completely multiplicative if f (ab) = f (a)f (b) for all natural a, b. Forexample, the functions id(n) = n and 1 are completely multiplicative.

Justin Stevens Multiplicative Functions (Lecture 5) 35 / 40

Sum of Divisors of Multiplicative Function

Theorem. If f is a multiplicative function and F is defined as

F (n) =∑d |n

f (d),

then F is also multiplicative.

Corollary. τ and σ are multiplicative functions since τ(n) =∑

d |1 1 andσ(n) =

∑d |n d .

Justin Stevens Multiplicative Functions (Lecture 5) 36 / 40

Proof

Proof.Let m and n be relatively prime integers, so F (mn) =

∑d |mn f (d). Since

d | mn, we decompose d into d = d1d2 such that d1 | m and d2 | n. Wehence have

F (mn) =∑d1|m

∑d2|n

f (d1d2)

=∑d1|m

∑d2|m

f (d1)f (d2)

=

∑d1|m

f (d1)

∑d2|n

f (d2)

= F (m)F (n).

Justin Stevens Multiplicative Functions (Lecture 5) 37 / 40

ω and Ω

Two other functions worth mentioning are ω(n) and Ω(n). ω(n) counts thenumber of distinct prime divisors of n and Ω(n) counts with multiplicity. Inother words,

ω(n) =∑p|n

1, Ω(n) =∑p|n

vp(n).

For instance, 400 = 24 · 52, so ω(400) = 2 and Ω(540) = 4 + 2 = 6.Definition. A function f is additive if f (ab) = f (a) + f (b) whenevergcd(m, n) = 1.

Example 12. Show that ω(n) is an additive function and 2ω(n) ismultiplicative.

Justin Stevens Multiplicative Functions (Lecture 5) 38 / 40

Proof of Additivity

Example. Show that ω(n) is an additive function and 2ω(n) ismultiplicative.

Proof.Let a =

∏1≤i≤r pei

i and b =∏

1≤j≤s pfjj . Since pi 6= qj , we see that

ab =∏

1≤i≤rpei

i ·∏

1≤j≤spfj

j

has r + s distinct prime factors. Hence, ω(ab) = r + s = ω(a) + ω(b).Furthermore, since ω(n) is additive, f (n) = 2ω(n) is multiplicative:

f (ab) = 2ω(ab) = 2ω(a)+ω(b) = 2ω(a)2ω(b) = f (a)f (b).

Justin Stevens Multiplicative Functions (Lecture 5) 39 / 40

A Weird Identity

Example. Prove the identity τ(n2) =∑

d |n 2ω(d).Solution. From our theorem, we see that the right hand side is amultiplicative function.

Furthermore, for coprime a and b witha =

∏1≤i≤r pei

i , b =∏

1≤j≤s qfjj , we see that

τ((ab)2) =

∏1≤i≤r

(2ei + 1) ·∏

1≤j≤s(2fj + 1) = τ(a2)τ(b2).

Hence, τ(n2) is also multiplicative. Therefore, we only need to show theidentity for n = pk :

τ(n2) = τ(p2k) = 2k + 1∑d |n

2ω(d) = 2ω(1) + 2ω(p) + 2ω(p2) + · · ·+ 2ω(pk) = 2k + 1.

The second equation follows since ω(1) = 1 and ω(pj) = 1.

Justin Stevens Multiplicative Functions (Lecture 5) 40 / 40

A Weird Identity

Example. Prove the identity τ(n2) =∑

d |n 2ω(d).Solution. From our theorem, we see that the right hand side is amultiplicative function. Furthermore, for coprime a and b witha =

∏1≤i≤r pei

i , b =∏

1≤j≤s qfjj , we see that

τ((ab)2) =

∏1≤i≤r

(2ei + 1) ·∏

1≤j≤s(2fj + 1) = τ(a2)τ(b2).

Hence, τ(n2) is also multiplicative. Therefore, we only need to show theidentity for n = pk :

τ(n2) = τ(p2k) = 2k + 1∑d |n

2ω(d) = 2ω(1) + 2ω(p) + 2ω(p2) + · · ·+ 2ω(pk) = 2k + 1.

The second equation follows since ω(1) = 1 and ω(pj) = 1.

Justin Stevens Multiplicative Functions (Lecture 5) 40 / 40

A Weird Identity

Example. Prove the identity τ(n2) =∑

d |n 2ω(d).Solution. From our theorem, we see that the right hand side is amultiplicative function. Furthermore, for coprime a and b witha =

∏1≤i≤r pei

i , b =∏

1≤j≤s qfjj , we see that

τ((ab)2) =

∏1≤i≤r

(2ei + 1) ·∏

1≤j≤s(2fj + 1) = τ(a2)τ(b2).

Hence, τ(n2) is also multiplicative. Therefore, we only need to show theidentity for n = pk :

τ(n2) = τ(p2k) = 2k + 1∑d |n

2ω(d) = 2ω(1) + 2ω(p) + 2ω(p2) + · · ·+ 2ω(pk) = 2k + 1.

The second equation follows since ω(1) = 1 and ω(pj) = 1.

Justin Stevens Multiplicative Functions (Lecture 5) 40 / 40

A Weird Identity

Example. Prove the identity τ(n2) =∑

d |n 2ω(d).Solution. From our theorem, we see that the right hand side is amultiplicative function. Furthermore, for coprime a and b witha =

∏1≤i≤r pei

i , b =∏

1≤j≤s qfjj , we see that

τ((ab)2) =

∏1≤i≤r

(2ei + 1) ·∏

1≤j≤s(2fj + 1) = τ(a2)τ(b2).

Hence, τ(n2) is also multiplicative. Therefore, we only need to show theidentity for n = pk :

τ(n2) = τ(p2k) = 2k + 1∑d |n

2ω(d) = 2ω(1) + 2ω(p) + 2ω(p2) + · · ·+ 2ω(pk) = 2k + 1.

The second equation follows since ω(1) = 1 and ω(pj) = 1.

Justin Stevens Multiplicative Functions (Lecture 5) 40 / 40

A Weird Identity

Example. Prove the identity τ(n2) =∑

d |n 2ω(d).Solution. From our theorem, we see that the right hand side is amultiplicative function. Furthermore, for coprime a and b witha =

∏1≤i≤r pei

i , b =∏

1≤j≤s qfjj , we see that

τ((ab)2) =

∏1≤i≤r

(2ei + 1) ·∏

1≤j≤s(2fj + 1) = τ(a2)τ(b2).

Hence, τ(n2) is also multiplicative. Therefore, we only need to show theidentity for n = pk :

τ(n2) = τ(p2k) = 2k + 1∑d |n

2ω(d) = 2ω(1) + 2ω(p) + 2ω(p2) + · · ·+ 2ω(pk) = 2k + 1.

The second equation follows since ω(1) = 1 and ω(pj) = 1.Justin Stevens Multiplicative Functions (Lecture 5) 40 / 40