My solution

U275

Darij Grinberg May 11, 2018

1. Problem

Let µ be the number-theoretical Möbius function; this is the function { 1, 2, 3, . . . } → Z defined by







µ ( m ) =

( 0, if m is not squarefree;

(− 1 )

, if m is squarefree for every m ∈ { 1, 2, 3, . . . }





 .

Let a

be a real number for every positive integer f . Prove that

d

∑

∑

e|n

∑

g|gcd(d,e)

µ ( g )

g dea

= ∑

f|mn

f a

for any positive integers n and m.

1.1. Remark

The above problem is my Problem U275 published in Mathematical Reflections 2013 (issue 4), slightly rewritten to make it more self-contained.

Of course, the problem can be straightforwardly generalized: Let A be any Q- vector space. Then, we can require the a

to be elements of A instead of requiring them to be real numbers. The problem still remains valid in this generality, and the solution given below still applies.

1Here, as usual, a summation sign of the form ∑

h|N

(where N is a positive integer) denotes a summation in whichhruns over allpositivedivisors of N.

2See https://www.awesomemath.org/wp-content/uploads/mr_4_2013_solutions.pdf for a different solution by Daniel Lasoasa.

2. Solution

Before we start solving the problem, let us collect some useful number-theoretical facts.

For every positive integer N, we let N

denote the set of all positive divisors of N. Thus, for every positive integer N, we have

N

= ( the set of all positive divisors of N )

= { h ∈ { 1, 2, 3, . . . } : h | N } .

Lemma 2.1. Let m and g be two positive integers. Let ( b

)

|m

∈ _R

be a

family of real numbers. Then,

d

∑

g|d

b

= ∑

d|m;

g|m/d

b

.

Proof of Lemma 2.1. The condition g | m/d on a divisor d of m is equivalent to the condition dg | m. Hence,

d

∑

g|m/d

b

= ∑

d|m;

dg|m

b

= ∑

d|m;

g|d

b

( here, we have substituted d/g for d in the sum ) .

Lemma 2.2. Let u and v be two positive integers. Let x be an integer. If x | u and x | _{v, then x} | _gcd ( u, v ) .

Proof of Lemma 2.2. This is a classical result.

Lemma 2.3. Let u and v be two positive integers. Let b

g∈_N|u

∈ _R

be a family of real numbers. Then,

g

∑

g|v

b

= ∑

g|gcd(u,v)

b

.

Proof of Lemma 2.3. Both sums range over the same set of g’s, since the positive

divisors of u which also divide v are precisely the positive divisors of gcd ( u, v )

(because of Lemma 2.2).

Lemma 2.4. Let d, u and v be three positive integers. Then, gcd ( du, dv ) = d gcd ( u, v ) .

Proof of Lemma 2.4. This is easy and left to the reader.

Lemma 2.5. Let u and v be two positive integers. Let b

be a real number for every f ∈ { 1, 2, 3, . . . } . Then,

∑

∑

e|v;

gcd(de,u)=d

b

= ∑

f|uv

b

.

Proof of Lemma 2.5. Let d be a positive divisor of u. Then, we can define a map B from the set

{ e is a positive divisor of v : gcd ( de, u ) = d } to the set

{ f is a positive divisor of uv : gcd ( f , u ) = d } by setting

B ( e ) = de for every e in the former set

3

. This map B is bijective

. Hence, we can substitute B ( e ) for f in the sum

∑

f|uv;

gcd(f,u)=d

b

. We thus obtain

f

∑

gcd(f,u)=d

b

= ∑

e|v;

gcd(de,u)=d

b

| {z }

=b_de (sinceB(e)=de)

= ∑

e|v;

gcd(de,u)=d

b

. (1)

3Proof.To see that this map is well-defined, we merely need to check that ifeis a positive divisor ofvsatisfying gcd(de,u) =d, thendeis a positive divisor ofuvsatisfying gcd(de,u) =d. So letebe a positive divisor ofv satisfying gcd(de,u) =d. Then,de| uv(sinced| uande|v), so thatdeis a positive divisor ofuv; also, clearly, gcd(de,u) =d. Thus, we are done.

4Proof. The injectivity ofBis obvious (sinced6=0). It remains to prove thatBis surjective. In other words, it remains to prove that if f is a positive divisor ofuvsatisfying gcd(f,u) =d, then there exists a positive divisoreofvsatisfying gcd(de,u) =dand f =de.

So let f be a positive divisor ofuvsatisfying gcd(f,u) = d. Clearly,d =gcd(f,u) | f, so that f/d∈Z.

Lemma 2.4 (applied tov, f anduinstead ofd,uandv) yields gcd(v f,vu) =vgcd(f,u)

| {z }

=d

=

vd.

Now, f | uv = vu and f | v f (clearly). Hence, Lemma 2.2 (applied to v f, vu and f instead ofu,vandx) shows that f |gcd(v f,vu) =vd. Now,v/(f/d) = (vd)/f ∈_Z(since f | vd), and thus f/d | v(since f/d ∈ _{Z). Hence,} f/d is a positive divisor ofv. Moreover, gcd





d(f/d)

| {z }

=f

,u





=gcd(f,u) =dand f =d(f/d).

Thus, there exists a positive divisore ofv satisfying gcd(de,u) = d and f = de(namely, e= f/d). This completes the proof thatBis bijective.

Now, let us forget that we fixed d. We thus have shown that (1) holds for every positive divisor d of u.

Now, if f is any positive divisor of uv, then gcd ( f , u ) is a positive divisor of u. Hence, we can split the sum ∑

f|uv

b

into subsums according to the value of gcd ( f , u ) as follows:

f

∑

b

= ∑

d|u

∑

f|uv;

gcd(f,u)=d

b

| {z }

= _∑

e|v;

gcd(de,u)=d b_de

(by (1))

= ∑

d|u

∑

e|v;

gcd(de,u)=d

b

.

This proves Lemma 2.5.

Now, let us introduce the following notation (known as the Iverson bracket nota- tion): If A is any logical statement, then [A] shall denote the integer

( 1, if A is true;

0, if A is false . We shall use the following classical property of the Möbius function:

Proposition 2.6. Let N be a positive integer. Then,

d

∑

µ ( d ) = [ N = 1 ] .

Proposition 2.6 is a classical fact in number theory

. We shall nevertheless prove it here soon; but first, let us prepare by stating some simple facts.

Recall that an integer is said to be squarefree if it is not divisible by any square of a positive integer > 1. We state two simple facts, whose proofs we leave to the reader:

• If u and v are two squarefree integers, and if u and v are coprime, then

uv is squarefree. (2)

• If u and v are two integers such that u | v, and if v is squarefree, then

u is squarefree. (3)

5For example, it appears in [NiZuMo91, Theorem 4.7, second part], in [Hildeb13, Theorem 1.3] and in [GriRei18, version with solutions, Solution to Exercise 2.82, proof of (11.68.3)].

In order to convince yourself that the definitions of the functionµin all of these sources are equivalent to our definition ofµ, you only need to observe that the number of primes dividing a squarefree integer mis precisely the number of prime factors of m(because every prime dividing a squarefree integermappears in the prime factorization ofmwith a multiplicity of exactly 1).

Also, from the definition of the Möbius function µ, we immediately obtain the following two facts:

• If m is a squarefree integer, then

µ ( m ) = (− ₁ )

. (4)

• If m is a integer such that m is not squarefree, then

µ ( m ) = 0. (5)

Applying (4) to m = 1, we obtain µ ( 1 ) = 1 (since the number of primes dividing 1 is 0).

We are now ready for the proof of Proposition 2.6:

Proof of Proposition 2.6. If N = 1, then Proposition 2.6 just states µ ( 1 ) = 1, which is true. Hence, for the rest of the proof of Proposition 2.6, we WLOG assume that we don’t have N = 1.

Thus, we have N 6= 1. Hence, the positive integer N has at least one prime divisor. Pick such a prime divisor, and denote it by p. Clearly, p | N. Also, p is prime, so that p > 1.

Let Sqf denote the set of all squarefree positive integers.

For every g ∈ Sqf ∩ ⁿ h ∈ _N

: p | h o

, we have g/ p ∈ Sqf ∩ ⁿ h ∈ _N

: p - h o

(6)

6

. Hence, we can define a map

A : Sqf ∩ ⁿ h ∈ N

: p | h o

→ Sqf ∩ ⁿ h ∈ N

: p - h o

6Proof of (6):Letg∈Sqf∩ⁿh∈_N|N : p|ho . We have g ∈ Sqf∩ⁿh∈N|N : p|ho

⊆ Sqf. In other words, g is a squarefree positive integer. Hence, for any positive integerm>1, we havem²- g. Applying this tom= p, we obtainp²-^{g(since p}>1). In other words,g/p²∈/_Z.

We have g ∈ Sqf∩ⁿh∈N_|N : p|ho

⊆ ⁿh∈N_|N : p|ho

. In other words, g is an element ofN|Nand satisfies p|g. In particular,g∈N|N, so thatgis a positive divisor ofN.

Thus,g|N.

Now,g/pis a positive integer (sincep|g) and satisfiesg/p| g(sinceg=_p(g/p)). Hence, (3) (applied tou=g/pandv=g) shows that g/pis squarefree. In other words,g/p∈Sqf.

Now,g/pis a positive divisor ofN(sinceg/p|g| N). In other words,g/p∈_N|N. Thus, g/p is an element of N|N and satisfies p - g/p (since (g/p)/p = g/p² ∈/ Z).

In other words, g/p ∈ ⁿh∈_N|N : p-^h

o. Combining this with g/p ∈ Sqf, we obtain g/p∈Sqf∩ⁿh∈_N|N : p-^h

o. This proves (6).

by

A ( g ) = g/p for every g ∈ _Sqf ∩ ⁿ h ∈ _N

_: p | h o . Consider this map A.

For every g ∈ _Sqf ∩ ⁿ h ∈ _N

_: p - h o

, we have gp ∈ Sqf ∩ ⁿ h ∈ _N

: p | h o

(7)

7

. Hence, we can define a map

B : Sqf ∩ ⁿ h ∈ _N

: p - h o

→ Sqf ∩ ⁿ h ∈ _N

: p | h o by

B ( g ) = gp for every g ∈ Sqf ∩ ⁿ h ∈ _N

: p - h o . Consider this map B.

The maps A and B are mutually inverse

. Thus, the map A is invertible, and hence a bijection.

For every g ∈ Sqf ∩ ⁿ h ∈ _N

: p | h o

, we have ( number of primes dividing A ( g ))

= ( number of primes dividing g ) − 1 (8)

7Proof of (7):Letg∈Sqf∩ⁿh∈N|N : p-ho . We have g ∈ Sqf∩ⁿh∈_N|N : p-^h

o ⊆ Sqf. In other words, g is a squarefree positive integer.

We have g ∈ Sqf∩ⁿh∈N|N : p-^h

o ⊆ ⁿh∈N|N : p-^h o

. In other words, g is an element of N|N and satisfies p - g. In particular, g ∈ _N|N; in other words, g is a positive divisor ofN. Thus,g| N.

It is well-known that ifqis a prime and zis an integer satisfyingq - ^{z, then qis coprime} toz. Applying this toq= pand z= g, we conclude that pis coprime to g(since p - g). In other words, the integers g and p are coprime. Also, the integer p is prime, and therefore squarefree (since every prime is squarefree). Hence, (2) (applied tou=g andv = p) shows thatgpis squarefree. In other words,gp∈Sqf (sincegpis clearly a positive integer).

A well-known fact from elementary number theory states the following: Ifx,y andzare three integers such thatx|zandy|z, and if the integersxandyare coprime, then we have xy | z. We can apply this to x = g,y = p and z= N(since we have g | Nand p | N, and since the integersgand pare coprime). As a consequence, we obtaingp| N. Hence,gpis a positive divisor ofN. In other words,gp∈_N|N.

Thus, gp is an element of N|N and satisfies p | gp (obviously). In other words, gp ∈ nh∈N|N : p|ho

. Combining this withgp∈Sqf, we obtaingp∈Sqf∩ⁿh∈N|N : p|ho . This proves (7).

8This follows immediately from their definitions (since the map A divides its input by p, whereas the mapBmultiplies its input byp).

9

and thus

µ ( A ( g )) = − µ ( g ) (10)

10

.

9Proof of (8):Letg∈Sqf∩ⁿh∈_N|N : p|ho . We have g ∈ Sqf∩ⁿh∈N|N : p|ho

⊆ ⁿh∈N|N : p|ho

. In other words, g is an element ofN|N and satisfies p | g. Thus, g/p is a positive integer. Of course, p is a prime dividingg(sincep|g).

We have A(g) = g/p (by the definition of A). Thus, g/p = _A(g) ∈ Sqf∩ⁿh∈_N|N : p-^h

o ⊆ Sqf, so that g/p is a squarefree positive integer. In the proof of (6), we have seen thatp-g/p. In other words,pdoes not divideg/p.

Now, every prime dividing g/p must also divide g (since g/p | g), but must be distinct from p (since pdoes not divideg/p). In other words, every such prime must belong to the set(the set of all primes dividingg)\ {p}. Thus,

(the set of all primes dividingg/p)⊆(the set of all primes dividingg)\ {p}. (9) On the other hand, let q ∈ (the set of all primes dividingg)\ {p}. Then, q is a prime dividingg, but we haveq6=p. The primesqandpare distinct, and therefore coprime. Recall the following general fact: Ifx,yandzare three integers such thatxandyare coprime and such thatx | yz, then x | z. Applying this to x = q,y = p andz = g/p, we obtain q| g/p (sinceq|g= p(g/p)). Thus,qis a prime dividingg/p.

Now, let us forget that we fixed q. We thus have shown that every q ∈ (the set of all primes dividingg)\ {p}is a prime dividingg/p. In other words,

(the set of all primes dividingg)\ {p} ⊆(the set of all primes dividingg/p). Combining this inclusion with (9), we obtain

(the set of all primes dividingg/p) = (the set of all primes dividingg)\ {p}. Now,





number of primes dividing A(g)

| {z }

=g/p







= (number of primes dividingg/p) =

(the set of all primes dividingg/p)

| {z }

=(the set of all primes dividingg)\{p}

=|(the set of all primes dividingg)\ {p}|

=|(the set of all primes dividingg)|

| {z }

=(number of primes dividingg)

−1 (sincepis a prime dividingg)

= (number of primes dividingg)−1, qed.

10Proof of (10):Letg∈Sqf∩ⁿh∈_N|N : p|ho . Thus,g∈Sqf∩ⁿh∈_N|N : p|ho

⊆Sqf, so thatgis a squarefree positive integer. Hence,

Now,

d

∑

µ ( d )

= ∑

g|N

µ ( g ) ( here, we have renamed the summation index d as g )

= ∑

g|N;

gis squarefree

| {z }

= _∑ g|N;

g∈Sqf

(because a positive integerg is squarefree if and only if

it satisfiesg∈Sqf )

µ ( g ) + ∑

g|N;

gis not squarefree

µ ( g )

| {z }

=0

(by (5), applied tom=g)

= ∑

g|N;

g∈Sqf

| {z }

= _∑

g∈_N|N; g∈Sqf

= _∑

g∈Sqf;

g∈_N|N

µ ( g ) + ∑

g∈N|N; gis not squarefree

0

| {z }

=0

= ∑

g∈Sqf;

g∈N_|N

µ ( g )

= ∑

g∈Sqf;

g∈_N|N; p|g

| {z }

= _∑

g∈Sqf;

g∈

{

}

= _∑

g∈Sqf∩

{

}

µ ( g ) + ∑

g∈Sqf;

g∈_N|N; p-g

| {z }

= _∑

g∈Sqf;

g∈

{

}

= _∑

g∈Sqf∩

{

} µ ( g )

(4) (applied tom=g) shows thatµ(g) = (−1)(number of primes dividingg). But the target of the map A is Sqf∩ⁿh∈_N|N : p-^h

o. Thus, A(g) ∈ Sqf∩ⁿh∈N|N : p-^h

o⊆Sqf. In other words,A(g)is a squarefree positive integer. Hence, (4) (applied tom=A(g)) shows that

µ(A(g)) = (−1)(number of primes dividingA(g))= (−1)(number of primes dividingg)−1

(by (8))

=−(−1)(number of primes dividingg)

| {z }

=µ(g)

=−µ(g).

This proves (10).

= ∑

g∈Sqf∩

{

}

µ ( g ) + ∑

g∈Sqf∩

{

} µ ( g )

| {z }

= _∑

g∈Sqf∩

{

}

µ(A(g))

(here, we have substitutedA(g)forgin the sum, since the map

A:Sqf∩

{

}

{

}

is a bijection)

= ∑

g∈Sqf∩

{

}

µ ( g ) + ∑

g∈Sqf∩

{

}

µ ( A ( g ))

= ∑

g∈Sqf∩

{

}

( µ ( g ) + µ ( A ( g )))

| {z }

=0 (by (10))

= ∑

g∈Sqf∩

{

}

0 = 0 = [ N = 1 ]

( since [ N = 1 ] = 0 (since we don’t have N = 1) ) . This proves Proposition 2.6.

Proposition 2.6 leads to a neat corollary

:

Corollary 2.7. Let u, v and d be three positive integers. Then,

g

∑

g|v

µ ( g ) = [ gcd ( du, dv ) = d ] .

Proof of Corollary 2.7. From Lemma 2.4, we have gcd ( du, dv ) = d gcd ( u, v ) . Thus, we have the logical equivalence

( gcd ( du, dv ) = d ) ⇐⇒ ( d gcd ( u, v ) = d ) ⇐⇒ ( gcd ( u, v ) = 1 )

(because the nonzero integer d can be cancelled). Hence, [ gcd ( du, dv ) = d ] = [ gcd ( u, v ) = 1 ] .

On the other hand, Lemma 2.3 (applied to b

= µ ( g ) ) shows that

g

∑

g|v

µ ( g ) = ∑

g|gcd(u,v)

µ ( g ) = ∑

d|gcd(u,v)

µ ( d )

( here, we renamed the summation index g as d )

= [ gcd ( u, v ) = 1 ] ( by Proposition 2.6, applied to N = gcd ( u, v ))

= [ gcd ( du, dv ) = d ] ( since [ gcd ( du, dv ) = d ] = [ gcd ( u, v ) = 1 ]) . This proves Corollary 2.7.

11I have stated this corollary in the form that makes it most useful to the solution of the problem;

hence, the useless-looking variabled.

Now, let us step to the solution of the problem. Let n and m be two positive integers.

Let e be a positive divisor of n. Then, for every positive divisor d of m, we have

∑

g|d

µ ( g )

g dea

= ∑

g|gcd(e,d)

µ ( g )

g dea

by Lemma 2.3, applied to u = e, v = d and b

= ^µ ( g )

g dea

= ∑

g|gcd(d,e)

µ ( g )

g dea

( since gcd ( e, d ) = gcd ( d, e )) . (11) Now,

d

∑

∑

g|gcd(d,e)

µ ( g )

g dea

| {z }

=_∑ g|e;

g|d

µ ( g ) g

(by (11))

= ∑

d|m

∑

g|e;

g|d

| {z }

=_∑

g|e ∑ d|m;

g|d

µ ( g ) g d

| {z }

=µ(g)(d/g)

e a

| {z }

=a_(d/g)e (sincede/g=(d/g)e)

= ∑

g|e

∑

d|m;

g|d

µ ( g ) ( d/g ) ea

= ∑

g|e

µ ( g ) e ∑

d|m;

g|d

( d/g ) a

| {z }

= _∑ d|m;

g|m/d da_de

(by Lemma 2.1, applied tob_d=da_de)

= ∑

g|e

µ ( g ) e ∑

d|m;

g|m/d

da

= ∑

g|e

∑

d|m;

g|m/d

| {z }

=_∑

d|m ∑ g|e;

g|m/d

µ ( g ) eda

= ∑

d|m

∑

g|e;

g|m/d

µ ( g ) eda

= ∑

d|m







 ∑

g|e;

g|m/d

µ ( g )









eda

. (12)

But every positive divisor d of m satisfies

∑

g|m/d

µ ( g ) = [ gcd ( de, m ) = d ] (13)

12

. Thus, (12) becomes

d

∑

∑

g|gcd(d,e)

µ ( g )

g dea

= ∑

d|m







 ∑

g|e;

g|m/d

µ ( g )









| {z }

=[gcd(de,m)=d] (by (13))

ed

|{z}

=de

a

= ∑

d|m

[ gcd ( de, m ) = d ] dea

| {z }

= _∑

d|m;

gcd(de,m)=d

[gcd(de,m)=d]dea_de+ _∑ d|m;

gcd(de,m)6=d

[gcd(de,m)=d]dea_de

(since every positive divisordofmsatisfies either gcd(de,m)=d or gcd(de,m)6=d(but not both))

= ∑

d|m;

gcd(de,m)=d

[ gcd ( de, m ) = d ]

| {z }

=1

(since gcd(de,m)=dholds)

dea

+ ∑

d|m;

gcd(de,m)6=d

[ gcd ( de, m ) = d ]

| {z }

=₀

(since gcd(de,m)=ddoes not hold (since we have gcd(de,m)6=d))

dea

= ∑

d|m;

gcd(de,m)=d

1dea

+ ∑

d|m;

gcd(de,m)6=d

0dea

| {z }

=₀

= ∑

d|m;

gcd(de,m)=d

1de

|{z}

=de

a

= ∑

d|m;

gcd(de,m)=d

dea

. (14)

Now, let us forget that we fixed e. We thus have proven that (14) holds for every positive divisor e of n.

12Proof of (13):Letd be a positive divisor ofm. Thus,d is a positive integer and a divisor ofm.

We haved | m(sinced is a divisor ofm) and thusm/d ∈ _{Z. Also,} m/d is positive (sincem anddare positive) and thus a positive integer (sincem/d∈Z). Hence, Corollary 2.7 (applied tou=eandv=m/d) shows that

g|e;

∑

g|m/d

µ(g) =



gcd



de,d(m/d)

| {z }

=m



=d



= [gcd(de,m) =d].

This proves (13).

Now,

d

∑

∑

e|n

| {z }

=_∑

e|n ∑

d|m

g|gcd

∑

µ ( g )

g dea

= ∑

e|n

∑

d|m

∑

g|gcd(d,e)

µ ( g )

g dea

| {z }

= _∑

d|m;

gcd(de,m)=d dea_de

(by (14))

= ∑

e|n

∑

d|m;

gcd(de,m)=d

| {z }

=_∑

d|m ∑

e|n;

gcd(de,m)=d

dea

= ∑

d|m

∑

e|n;

gcd(de,m)=d

dea

= ∑

f|mn

f a

(by Lemma 2.5, applied to u = m, v = n and b

= f a

). This solves the problem.

References

[GriRei18] Darij Grinberg, Victor Reiner, Hopf algebras in Combinatorics, version of 11 May 2018, arXiv:1409.8356v5.

See also http://www.cip.ifi.lmu.de/~grinberg/algebra/

HopfComb-sols.pdf for a version that gets updated.

[Hildeb13] A. J. Hildebrand, Introduction to Analytic Number Theory: Math 531 Lecture Notes, Fall 2005, version 2013.01.07.

http://www.math.uiuc.edu/~hildebr/ant/main1.pdf

[NiZuMo91] Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery, An In-

troduction to the Theory of Numbers, 5th edition, Wiley 1991.