Power values of the sigma function

Power values of the sigma function

Bachelor’s Project Mathematics

July 2017

Student: Andrea Basso

Primary supervisor: Prof. J. Top

Secondary supervisor: Dr. K. Efstathiou

Abstract

We study when the sum of divisors function attains perfect power values for an

unrestricted argument and when it does so with perfect power arguments. We give

a proof of the existence of infinitely many integers whose sum of divisors is a kth

power, for any integer k, and we present several cubes whose sum of divisors is again

a cube.

Contents

1 Introduction 1

2 Integers whose sum of divisors is a power, or σ(n) = m

4 2.1 Integers whose sum of divisors is a square . . . . 4 2.2 Integers whose sum of divisors is a ninth power . . . . 6 2.3 The general case, or σ(n) = m

. . . . 7 3 Powers of integers whose sum of divisors is a power, or σ(n

) = m

11 3.1 Fermat’s classic problems . . . 11 3.2 Cubes whose sum of divisors is a cube . . . 13 3.3 Higher powers, or l ≥ 4, k ≥ 2 . . . 14

4 A numerical approach 15

4.1 Our algorithm to solve σ(n

) = m

. . . 15 4.2 A generalized algorithm to solve σ(n

) = m

. . . 17

5 Conclusion 19

A An elementary proof on the bounds of π(x) 20

B Source code 24

C List of solutions to σ(n) = m

31

D List of solutions to σ(n

) = m

33

Chapter 1 Introduction

In this article, we investigate when the sum of divisors function attains perfect power values for an unrestricted argument and when it does so with perfect power argu- ments. We give a proof of the existence of infinitely many integers whose sum of divisors is a kth power, for any integer k, and we present several cubes whose sum of divisors is again a cube.

Number theory, the study of integers and prime numbers, is older than west- ern civilization itself. It has been called the “Queen of Mathematics”, due to its importance and its fundamental nature and it has been studied for millennia, with particular interest by Ancient Greeks.

As part of their study, they considered the sum of divisors function, which associates to each positive integer the sum of its positive divisors and is usually denoted by σ(n). They studied its properties and obtained several important results, such as its multiplicativity (σ(nm) = σ(n)σ(m), if gcd(n, m) = 1) and divided numbers according to the ratio between its sum of divisors and the number itself. The num- bers n that had σ(n)/n > 2 were called abundant, whereas those with σ(n)/n < 2 were called deficient. Those particular numbers where σ(n)/n is exactly 2, such as 6 or 28, are called perfect. They also called two numbers n, m friendly whenever σ(n)/n = σ(m)/m.

The study of σ(n) spanned across the whole history of mathematics. Fermat and Wilson took great interest in it in the 17

century, as it will be explained in Section 3.1. In modern times the sigma function is related to the Riemann’s hypothesis through Robin’s theorem, which also establishes an upper bound for σ(n), given a sufficiently large n. There are still several open problems related to the sum of divisors function. To name a few, it is not known whether infinitely many perfect numbers exist, whether there exists any odd perfect number at all or if there is any number which is friendly with 10.

In this thesis we investigate a specific problem related to σ(n). We want to study

when and how often the sum of divisors attains a power value, i.e. when σ(n) = m

,

for some integer m. We also want to study when this happens if we restrict the

argument to power values, i.e. when σ(n

) = m

. What motivates this question is

a theorem by Freiberg. His Theorem 1.1 from [4] applied to the sum of divisors has

a peculiar consequence. Namely, the number of positive integers up to x such that their sum of divisors is a kth power is more than x

, for x sufficiently large. A naive guess could have predicted that this number would be about x

, similar to the number of kth powers below x. This theorem thus tells us that σ attains perfect power values much more frequently than random numbers. The authors of [2] write

“one might say that the values of σ are really eager to be perfect powers”. Hence studying when the sum of divisors attains power values assumes a new meaning beyond the study of power values of a random function.

Our investigation follows two approaches: from one side we studied the prob- lem theoretically, proving statements as much as possible. On the other, we run numerical experiments that lead us to find several solutions. The two approaches are not completely independent. The same reasoning and construction we use to prove Theorems 2.1 and 2.2 is at the basis of our computer algorithm; vice-versa, computer computations help support certain conjectures.

This thesis is organized across three chapters. The first focuses on the study of the power values of σ(n) for any possible input. It culminates with the following important result.

Theorem 1.1. Given a positive integer k, there exists infinitely many positive in- tegers n such that σ(n) is the kth power of an integer.

In the following chapter, we restrict the input to power values as well. In particu- lar we consider two problems proposed by Fermat, namely finding the solutions of σ(n

) = m

and σ(n

) = m

, and the problem of finding cubes such that their sum of divisors is again a cube. We find several new solutions to this problem, including the smallest known one. The last chapter explains in detail how the computer algo- rithm works and which optimizations we have introduced.

Before we go further, it is important to have an overview on the sum of divisors function and its main properties. Let us start with a formal definition.

Definition 1. Let n be a positive integer. The sum of divisors function or sigma function, usually denoted by σ(n), is the sum of the positive divisors of n.

From this definition follow two consequences. The first concerns prime numbers: if p is any prime number, σ(p) = p + 1. This also shows that σ(n) can take values arbitrarily close to 1. The second is slightly more general and covers prime powers.

Since the divisors of p

, with p prime, are only lower powers of p, we have σ(p

) = X

d|p^k

d =

k

X

i=0

p

= p

− 1 p − 1 ,

where the last equality comes from a basic result on geometric series.

Another important property of the sigma function is its multiplicativity, i.e. σ(nm) = σ(n)σ(m) whenever gcd(n, m) = 1. To prove this, write

σ(n)σ(m) = X

d|n

d X

d⁰|m

d

= X

d|n d⁰|m

dd

= X

d^?|nm

d

= σ(nm),

where the equality between P

d|n

d P

d⁰|m

d

and P

d|n,d⁰|m

dd

is motivated by n and m being coprime.

If we combine this result and the precedent together, we can obtain a formula for σ(n) for any n. Let n = p

p

· . . . ·

be the prime factorization of n. It follows that

σ(n) = σ

r

Y

i=1

p

!

=

r

Y

i=1

σ(p

) =

r

Y

i=1

p

− 1 p

− 1 .

Having outlined the most important aspects of the sigma function, we now move

to the chapter dedicated to the integers whose some of divisors is a kth power.

Chapter 2

Integers whose sum of divisors is a power, or σ(n) = m ^k

2.1 Integers whose sum of divisors is a square

We begin with the simplest case, finding those numbers whose sum of divisors is a square. The smallest non-trivial solution (where with trivial we indicate the solution n = m = 1) is clearly 3, as σ(3) = 2

. There exist several other known solutions to the problem. A list of them can be found in the sequence A006532 of the Online Encyclopedia of Integer Sequences [9]. Some of these can be easily computed by hand, due to some observations. For instance, if σ(x) = σ(y) = m and gcd(x, y) = 1, σ(xy) = m

. In particular, there are several combinations of four different primes where σ(pq) = σ(rs). We can find some solutions of this kind. Let us assume, to greatly simplify computations, that p and q are twin primes, q = p + 2. We obtain

σ(p(p + 2)) = σ(rs) ⇒ (p + 1)(p + 3) = (r + 1)(s + 1)

⇒ p

+ 4p + 3 − (r + 1)(s + 1) = 0

⇒ p = −2 + p

1 + (r + 1)(s + 1).

For p to be an integer, the expression under the square root ought to be a square.

Thus,

1 + (r + 1)(s + 1) = n

⇒ (r + 1)(s + 1) = n

− 1

⇒ (r + 1)(s + 1) = (n + 1)(n − 1).

If we go for the obvious r + 1 = n + 1, s + 1 = n − 1, we would obtain r = p, s = q.

However, since r and s are distinct, at least one of the two, say s, must be different from 2. We can thus write

(r + 1)(s + 1) = 2(r + 1) s + 1

2 ⇒ 2(r + 1) = n + 1, s + 1

2 = n − 1

⇒ s = 4r − 1, from which we obtain

p = −2 + p

1 + 4r(r + 1) = −2 + √

4r

+ 4r + 1 = −2 + 2r + 1 = 2r − 1.

Thus if we can find two consecutive primes p and p + 2 such that r = (p + 1)/2 and s = 2p + 1 are also primes, then σ(pqrs) is a square. There are several cases where this condition is met. The first is p = 3, q = 5, from which we obtain r = 2, s = 7 and σ(2 · 3 · 5 · 7) = 24

. Other cases are p = 5, q = 7, r = 3, s = 11, which gives us σ(3 · 5 · 7 · 11) = 48

, and p = 21, q = 23, r = 11, s = 43, from which follows σ(11 · 21 · 23 · 44) = 528

.

We have thus far shown that several solutions are known and that some of them are easily discoverable. It still remains to be seen whether infinitely many solutions exist.

For this case, it is possible to prove that there exist infinitely many solutions. A proof of this statement is contained in [2], which we report here. The main reasoning is the following: given a specific integer N , it is always possible to construct a number n such that its sum of divisors is a square and if n = p

p

. . . p

is its prime factorization, p

> N for every i. Given the arbitrariness of the lower bound N , we can always find new solutions, which shows there exist infinitely many solutions.

Theorem 2.1. There exists infinitely many integers n such that the sum of its divisors is square, i.e. σ(n) = m

, for some integer m.

Proof. Let N be an integer greater than 2 and let p

denote the ith prime number, so that p

= 2, p

= 3, . . .. For each p

, define r

as the smallest integer exponent such that p

> N . Choose t such that p

is a prime larger than any prime factor of Q

pi≤N

σ(p

).

Our goal now is to show that for every prime p

≤ p

the factorization of σ(p

) has at most t − 1 prime factors. This would allow us to use linear algebra to construct a solution.

For p

≤ N , p

is greater than any prime factor of σ(p

) by the definition of p

. Instead, for p

> N , it follows that r

= 1 and thus σ(p

) = σ(p

) = p

+ 1 = 2

, because p

is clearly an odd number. This shows that for i ≤ t each prime factor of σ(p

) is strictly less than p

and in turn less than p

, thus σ(p

) has at most t − 1 prime factors.

For i = 1, 2, . . . , t, write

σ(p

) = p

p

. . . p

.

We now consider the matrix (a

)

, where the column j represent the factorization of σ(p

) and its ith entry represent the exponent of p

in the factorization. This matrix has t − 1 rows and t columns. Since this represents an underdetermined homogeneous system, by linear algebra over the field of two elements F

, there exist a non-null vector  in the kernel of the matrix. This means that each element of  is either 0 or 1 and that



(a

, . . . , a

) + . . . + 

(a

, . . . a

) has even entries. Define

n = p

p

. . . p

.

It follows that σ(n) = σ(p

. . . p

) = σ(p

) . . . σ(p

)) is a square.

We have thus far constructed an example of σ(n) = m

where the prime power

factors of n are all larger than N . Given the arbitrariness of N , if we set N to be larger than the largest known solution, we can construct infinitely many solutions.

The constructive nature of this proof and its relative simplicity make this rea- soning particularly appealing. It is only natural to wonder whether it could be generalized to prove other cases. Unfortunately, this appears not to be possible.

If we use a different field other than F

, say F

, to construct solutions of σ(n) = m

, the same reasoning may be used. However, the solutions found would comprise of number different from 0 and 1. This does not suit the problem, as it is only possible to include or exclude a certain p

and it cannot be counted more than once.

On the other hand, if we tried to generalize the proof to study the case of σ(n

) = m

we would incur in a different problem. The original proof uses the fact that σ(p) = 2

and thus factorize in prime factors smaller than p. However, the same method is not applicable to σ(p

) for k > 1.

It is still possible to prove a more general result, but a different approach is needed.

Such proof will be presented in section 2.3.

2.2 Integers whose sum of divisors is a ninth power

The reason why looking for numbers whose sum of divisors is a ninth power is different from the previous ones is that –to the knowledge of the author– it is the first case where current literature contains no solution.

The online encyclopedia of integer sequences [9] contains several sequences about the solutions of σ(n) = m

for k < 9. Namely, A006532 (σ(n) = m

), A020477 (σ(n) = m

), A019422 (σ(n) = m

), A019423 (σ(n) = m

), A019424 (σ(n) = m

), A048257 (σ(n) = m

) and A048258 (σ(n) = m

).

Limiting ourselves to the primes p ≤ 229 such that the factorization of p + 1 did not contain any prime factor larger than 97, we obtained 1259 numbers whose sum of divisors is a ninth power. We report here the first five of these numbers.

σ(3 · 127) = 2

,

σ(2 · 3 · 5 · 17 · 71 · 107) = 2

· 3

, σ(2 · 3 · 5 · 23 · 53 · 107) = 2

· 3

, σ(2 · 3 · 7 · 17 · 53 · 107) = 2

· 3

, σ(2 · 3 · 11 · 17 · 53 · 71) = 2

· 3

.

We found other 16 numbers whose sum of divisors amounts to 2

· 3

. The first solution that has a different sum of divisors is the following.

σ(2 · 3 · 5 · 7 · 11 · 17 · 23 · 31 · 53) = 2

· 3

.

Out of the 1259 solutions we have found, their sum of divisors attains only 13

different values. The following graph shows these values on a logarithmic scale.

The case of σ(n) = m

for large k’s is computationally simpler, as it will be explained in Section 4.2. As further confirmation and as a curiosity, we report here a solution of σ(n) = m

.

We have found 70 solutions to this equation, among which the smallest is n = 2 · 3 · 5 · 7 · 17 · 19 · 29 · 59 · 79 · 89 · 149 · 179 · 199 · 239 · 269 · 359 · 449 · 479·

499 · 599 · 719 · 809 · 1249 · 1279 · 1439 · 1499 · 1619 · 1999 · 2399 · 2699 · 2879 · 2999·

4049 · 4373 · 4799 · 4999 · 5119.

It has 37 prime factors and it is approximately 5.067 · 10

. Furthermore, σ(n) = 2

· 3

· 5

= 60

.

2.3 The general case, or σ(n) = m ^k

In this section, we will prove the following important result.

Theorem 2.2. Given a positive integer k, there exists infinitely many positive in- tegers n such that σ(n) is the kth power of an integer.

The main reasoning of the proof of this theorem is not very different from the approach used for Theorem 2.1. We consider a specific set of primes whose sum of divisors factors into smaller primes and we obtain many more factorizations than primes in the factorization. We then apply a theorem from group theory to show the existence of a solution. The proof concludes the existence of infinitely many solutions from the existence of infinitely many suitable set of primes.

Firstly, we need an estimate for the number of primes π(n) between 2 and n.

While the Prime Number Theorem grants us that π(n) ∼

for n → ∞, we can

use the following theorem, which is weaker but easier to prove.

Proposition 2.3. Let π(x) denote the number of primes p ≤ x. For x > 221, π(x) is bounded above and below by

2 3

x

log x ≤ π(x) ≤ 2 x log x . A proof of this theorem can be found in Appendix A.

Then, we need to count how many primes p ≤ n have large prime factors in the factorization of σ(p) = p + 1. To do so, we employ the following result.

Proposition 2.4. If M > 1 is given, then there exists C > 0 such that the number of primes p ≤ x, where p + 1 is divisible by a prime ≥ p/M , is bounded from above by Cx/ log

(x) as x → ∞.

A proof of this theorem is contained in [2], which uses a sieve result that is presented in Theorem 3.12 of Sieve Methods [6].

Lastly, we need a result from group theory that gives us a sufficient condition for the existence of a solution.

Proposition 2.5. If k ≥ 3 and r are positive integers, then any subset of (Z/kZ)

of cardinality > rk log k contains a non-empty subset whose sum of elements is 0 mod k.

A proof of this theorem in the language of finite abelian groups can be found in an article on Carmichael numbers by W. Alford [1].

After having presented the main components, we can now formulate a proof of Theorem 2.2.

Proof. Let M > 3k log k and x be sufficiently large. The reasons why M is chosen as such and the source of 3 will be clear at the end of this proof. Let π

(x) denote the number of primes p ≤ x such that p + 1 is divisible by a prime ≥ p/M . By Proposition 2.3 and 2.4, the proportion of these primes is

π

(x)

π(x) ≤ Cx

π(x) log

(x) ≤ Cx

2 3

x

log x

log

(x) = C

log x → 0 for x → ∞,

where C

=

C. Thus, the proportion of primes p ≤ x that do not have large factors in the factorization of p + 1 is still ∼ π(x).

Consider the factorizations of σ(p) for all primes p with √

x ≤ p ≤ x that only contains prime factors ≤ p/M . For x sufficiently large, we have established they are

∼ π(x) and thus we have at least

factorizations. Since the highest factor of p + 1 is at most p/M , every factorization we are considering has at most π(x/M ) factors. However,

π( x

M ) ≤ 2 x/M

log(x/M ) ∼ x M

log(x) ,

where M

= M/2. Hence the number of factorizations is about M

times the num-

ber of prime factors.

We now apply Proposition 2.5. We consider the set of exponent vectors modulo k of our factorizations. There are at least (2/3)(x/ log x) of these vectors and each one contains less than x/(M

log x) elements. Thus we are considering a subset of (Z/kZ)

, where r < x/(M

log x) with cardinality >

. Since

2 3

x

log x > x

M

log x k log k = x log x

k log k M/2 ,

because we chose M > 3k log k, the condition of Proposition 2.5 is satisfied. We can then conclude that there exists a non-empty subset of σ(p)’s such that their product is a k-th power.

We have now found a solution. The arbitrariness of x, for x sufficiently large, guarantees the existence of infinitely many solutions.

The core element of this proof and the major block to generalization is Propo- sition 2.4, which guarantees that the number of primes with large prime factors in σ(p) is relatively small. It would be interesting to generalize the proof for the solu- tions of σ(n

) = m

or more generally σ(n

) = m

. However, there exists no proof of an upper bound on the number of primes p ≤ x such that P

i=1

p

factors into small primes. If such bound existed and it were smaller than x/ log x, the same reasoning of this proof could be applied to show the existence of infinitely many solutions.

Notwithstanding, computer calculations could help give credit either to the in-

finiteness of solutions or its contrary. Let us denote with π

(n) the number of

primes p ≤ n such that σ(p

) has factors larger than p/M . We have thus calculated

π

(n) for k = 2, 3, 4 and with M = 2. We present here the graph of the trends of

π

(n) compared to n/ log n.

We do not know whether the ratio keeps decreasing or stabilizes around a value greater than 0. However, since the rate of decrease slows down for n that gets larger, the second option appears more likely. Hence, our simulations give credit to the following conjectures.

Conjecture 2.1. For every positive integer k, there exist infinitely many positive integers n such that σ(n

) is kth power of an integer.

Conjecture 2.2. For every positive integer k, there exist infinitely many positive integers n such that σ(n

) is kth power of an integer.

Conjecture 2.3. For every positive integer k > 1, for every integer l ≥ 4, there exist only finitely many positive integers n such that σ(n

) is kth power of an integer.

It should be noted that there exist no known solution of σ(n

) = m

, for l ≥ 4, k ≥ 2. Thus “finitely many” may well mean none at all.

These conjectures lead us to the next chapter, where we explore some specific

case of σ(n

) = m

, with l > 1.

Chapter 3

Powers of integers whose sum of divisors is a power, or σ(n ^l ) = m ^k

In this chapter we would like to consider a problem strictly related to those treated in the previous part. The argument of the sigma function is now restricted to power values as well.

Let us first consider the simplest possible case, namely σ(n

) = m

. Current literature does not mention this problem, as it does not pose any particular difficulty, nor does it have any interesting aspect. The solutions of this equation are fairly small and computationally easy to find. For instance, there exists 15 solutions with n < 100000.

For a detailed list, one can consult sequence A008847 of the Online Encyclopedia of Integer Sequences [9]; we report here the first few (omitting the trivial n = m = 1).

σ(3

) = 11

, σ(2

· 5

) = 31

, σ(2

· 3

· 5

) = 11

· 31

,

σ(2

· 653

) = 7

· 13

· 19

, σ(2

· 5

· 191

) = 7

· 13

· 31

, σ(2

· 3

· 653

) = 7

· 11

· 13

· 19

.

In this chapter we would like to introduce three other specific problems: those proposed by Fermat in the 17

century, finding the solutions of σ(n

) = m

and σ(n

) = m

, and the more modern problem of finding cubes whose sum of divisors is again a cube.

3.1 Fermat’s classic problems

We now move to two historical problems. Indeed, they mark the first time in history where mathematicians studied the solutions of σ(n

) = m

.

According to Dickson’s History of the Theory of Numbers [3], on January 3, 1657

Fermat posed two problems to his fellow mathematicians. The first involved finding

a cube which when increased by the sum of its aliquot parts becomes a square, for

example while the second asked to find a square which when increased by the sum of its aliquot parts becomes a cube. With the aliquot parts, he referred to the sum of the proper divisors of a number. Thus in modern language, he asked to find the solutions of

σ(n

) = m

, σ(n

) = m

.

In his question, Fermat noted that the smallest solution for the first problem is 7, as 1+7+7

+7

= 20

. In 1658 Wallis tabulated σ(p

) for each prime p < 100 and small powers of 2, 3 and 5. He then proceeded to eliminate those primes that could not form a square (for instance, those which had a prime factor that did not appear in any other factorization) and obtained several solutions, including the following ones.

n = 2

· 5 · 11 · 13 · 41 · 47, m = 2 · 3 · 5 · 13 · 41 · 47;

n = 17 · 31 · 47 · 191, m = 2

· 3

· 5 · 13 · 17 · 29 · 37;

n = 2 · 3 · 5 · 13 · 17 · 31 · 41 · 191, m = 2

· 3

· 5

· 7 · 13 · 17 · 29

· 37;

n = 3

· 5 · 11 · 13 · 17 · 31 · 41 · 191, m = 2

· 3

· 5 · 7 · 11 · 13 · 17 · 29

· 37 · 61;

He also noted that for each pair n, m, if 7 and n are coprime then 7n, 7m is also a solution, given Fermat’s result and the multiplicativity of σ.

The same year Frenicle found one more solution for the first problem.

σ((2

· 5 · 7 · 31 · 73 · 241 · 243 · 467)

) = (2

· 3

· 5

· 11 · 13

· 17 · 37 · 41 · 113 · 193 · 257)

. Moreover, he gave a solution for the second problem as well.

σ((2

· 5 · 7 · 11 · 37 · 67 · 163 · 191 · 263 · 439 · 499)

) = (3

· 7

· 13 · 19 · 31

· 67 · 109)

. For this second problem, Wallis used a similar technique that led him to find solutions of the first problem, namely he tabulated the factorization of σ(p

) for every prime number p < 500. After a long process of exclusion, he only found two solutions, the one already found by Frenicle and

n = 7 · 11 · 29 · 163 · 191 · 439, m = 3 · 7 · 13 · 19 · 31 · 67.

Much later, in 1878, A. S. Bang found the smallest solution (after σ(1

) = 1

) to Fermat’s second question.

σ((2 · 3 · 11 · 653)

) = (7 · 13 · 19)

.

We can hypothesize that this solution took so long to be discovered due to the pres- ence of 653, a relatively large prime.

A list of solutions to both problems can be found in the Online Encyclopedia of Inte-

ger Sequences [9], sequences A008849 and A008850, respectively for the first and

the second of Fermat’s problem. As a curiosity, in the encyclopedia both sequences

have the keyword nice, which is reserved for “[a]n exceptionally nice sequence”.

For the first problem, it would be useful to find primes p such that σ(p

) is a square. For every solution n with p - n, σ(p · n) would also be a square. Similarly, every product of such primes would be a solution to our problem. Unfortunately, E. Gerono proved in 1877 [5] that the only prime number p such that σ(p

) = m

is p = 7. His proof is quite simple and uses only elementary means, however it is also rather long, as it is splits into several cases. The same result can be easily obtained in a more general context within the framework of elliptic curves. An introduction to the problem of finding integer points on an elliptic curve is presented in Chapter 5 of Rational Points on Elliptic Curves [8], while detailed information on the curve 1+p+p

+p

= m

is contained in the The L-functions and Modular Forms Database [7].

3.2 Cubes whose sum of divisors is a cube

The problem of finding solutions to σ(n

) = m

is probably the most interesting.

Except for the trivial n = m = 1, there are no small solutions.

An initial overview shows that no prime p can satisfy σ(p

) = m

. This is due to the fact that σ(p

) = 1 + p + p

+ p

; thus, for that to be equal to m

, it is necessary that m is larger than p. However, if we take m = p + 1, we obtain that m

= 1 + 3p + 3p

+ p

, which is already larger than σ(p

).

The first solution found in the literature is contained in the article Power Values of Divisor Sums [2]. The authors report the following solution, due to Herman Te Riele.

n = 5 · 11 · 13 · 17 · 41 · 47 · 193 · 239 · 443 · 499 · 701 · 1087 · 3583 · 5507, m = 2

· 3

· 5

· 7 · 13

· 17

· 29 · 37 · 61 · 97 · 149 · 157.

The value of n is about 3.59 · 10

. Shortly afterwards, the authors claim that it was unknown, in 2012 at the time of the publication, whether that was the smallest solution after n = m = 1. It turns out it is not.

With the algorithm explained in Chapter 4, we were able to find several other solutions, some of which smaller than Te Riele’s solution. After privately contacting the authors of the article, it appeared that two more people (Gary Mulkey and Michiel Kosters) had found other solutions since the publication. The first computed them by hand and found six more solutions, whereas the second found other sixty solutions. We have independently found those solutions and many more. Overall we have found 182 solutions (which are listed in Appendix D), including the following

n = 3

· 7 · 11 · 13 · 17 · 41 · 43 · 47 · 443 · 499 · 3583, m = 2

· 3

· 5

· 7 · 11 · 13 · 17 · 29 · 37 · 61 · 157.

This solution is several times smaller than the one found by Te Riele, n has 20

digits and in decimal form is 30154214043975990969. To our knowledge, this is the

smallest solution that has ever been found after n = m = 1. It still remains unknown whether a smaller solution exists. This number is still too large to brute-force any number up to n to check for the existence of a smaller solution.

3.3 Higher powers, or l ≥ 4, k ≥ 2

Unfortunately, not much can be said about the cases with l >≥ 4. At the time of writing, it appears that no solution of σ(n

) = m

, with l ≥ 4, k ≥ 2, has ever been recorded in the literature.

We have tried to find solutions of this kind, but to no avail. The largest problem we encountered is that the factorization of σ(n

) contains large primes very often, as predicted at the end of Section 2.3 by numerical evidence. Given the polynomial factorization of P

i=0

p

, the best candidate to factorize into smaller prime factors

is σ(p

). Besides, fro a computational point of view, the most promising case would

be k = 2, as finding a solution is as fast as computing the kernel of a matrix. More

information on this will be provided in the next chapter. It is thus our guess that it

would be easier to find solutions, if any exist, of σ(n

) = m

.

Chapter 4

A numerical approach

4.1 Our algorithm to solve σ(n ³ ) = m ³

From Fermat’s time till today, every mathematician who looked for solutions of σ(n

) = m

has used a variation of the same method. We could not help but do the same. In this section we will present an overview of our method and the specific optimizations we employed. The complete source code of our PARI/GP program can be found in Appendix B.

Let us first consider the method used to find solutions of σ(n

) = m

. We restrict ourselves to the primes p such that σ(p

) has only small prime factors. Formally, we consider the first t prime numbers and then we eliminate all those primes p such that the prime factorization of σ(p

) contains a factor larger than B. Thus, our program takes in two parameters, t and B.

Since σ(p

) = 1 + p + p

+ p

= (p + 1)(p

+ 1) and division by polynomials shows that gcd(p + 1, p

+ 1) = 2 for p ≥ 3, we can compute the factorizations of (p + 1)/2 and (p

+ 1)/2 separately. If both of them do not contain a prime larger than B, we can combine them together to form the factorization of σ(p

). Then, we list the factorizations we have found in a matrix, where the jth column represent the prime factorization of σ(p

) and the ith row represents the exponent of p

. The result resembles the following matrix

M =

σ(2

) σ(3

) σ(5

) σ(7

) . . .









































2ˆ 0 3 2 4 . . .

3ˆ 1 0 1 0 . . .

5ˆ 1 1 0 2 . . .

7ˆ 0 0 0 0 . . .

11ˆ 0 0 0 0 . . .

13ˆ 0 0 1 0 . . .

.. . .. . .. . .. . .. . . ..

,

because σ(2

) = 3 · 5, σ(3

) = 2

· 5 and so forth. This matrix has π(B) rows and

less or equal to t columns.

Then we consider the matrix M = M mod 3. Our goal is to find non-trivial solutions of M x = 0 that contain only 0 and 1 mod 3. We could not find any intel- ligent way to do so, and –to our knowledge– no such method exists. Thus we had to resort to a brute-force kind of computation. We generate the kernel of the matrix M , with –say– m vectors. Since the vectors in ker(M ) usually contain 2 mod 3, we are forced to consider all the 2

possible subsets of vectors and check whether their sum is (2 mod 3)-free.

Two considerations are now in order. The first is that the bulk of the program, where it spends most of the time, is this brute-force computation. The second is that the length of this computation is heavily dependent on the number of vectors in ker(M ). If a single vector were to be added, it would double the computation time.

Hence, it is only reasonable to focus our attention on reducing the kernel size to improve the computing performance. We perform two types of optimization: first, we optimize the matrix M and then we optimize its kernel.

Let us consider the matrix M . Firstly, we remove empty rows. While this does not reduce the kernel size, it slightly improves computation times and gives us a cleaner matrix. Then we focus our attention on each row individually. If the sum of the elements in a row is less than 3, there is no subset of its non-zero elements whose sum can be a multiple of 3. Thus, we can eliminate the columns that have a non-zero element in such row and eliminate the row itself.

Then we bring our attention to those rows whose elements add up to exactly 3.

The only subset of its non-zero elements which sum up to a multiple of 3 is the set composed of all non-zero elements. Thus the columns with non-zero elements in such rows are removed and replaced by a single column containing their sum.

After considering every row once, several columns have been deleted. Thus several rows whose sum of elements was greater than 3 may now match one of the two criteria for optimization. Thus once every row has been considered, we start again and proceed multiple times, till no further optimization is possible.

After having done so, we compute the kernel of M . Unfortunately, not much further optimization is possible at this stage. We limit ourselves to consider the rows that contain only zeros except for a single two. In that case, we can remove the column containing the 2 element.

Finally, we can only sit back and wait for the program to finish. With this opti- mizations, it is considerably fast. Using t = 1500, B = 2000 as input, the program runs in less than 27 seconds on a modern consumer computer and finds 64 solutions.

The effects of optimizations can be appreciated by looking at this data: the origi- nal matrix M has 275 columns and 303 rows, whereas the reduced matrix has 39 columns and 21 rows. The kernel basis contains 18 vectors.

Lastly, it should be noted that the method described here does not find all

possible solutions, regardless of the input (t, B). So far, we have only considered

σ(p

), which implies that every number n we have found with σ(n

) = m

is square-

free. It is thus sensible to consider higher powers of small primes, replacing the factorization of σ(p

) with that of σ(p

), σ(p

) or even higher. We restrict only to small primes, say p < 7, because the factorization of powers of larger primes contains prime factors well above our bound B. At this moment our algorithm does not support concurrent evaluation of multiple powers of the same prime, the specific exponent is chosen according to manual input. We realize however that it would be more performing to evaluate them together: an algorithm that considers σ(p

), σ(p

), σ(p

), σ(p

) and σ(p

) would run in about 3/16 of the time it would take to run the five different configurations.

4.2 A generalized algorithm to solve σ(n ^l ) = m ^k

After having discussed our algorithm to find solutions of σ(n

) = m

, we would like to introduce a more general approach, capable of finding solutions to any equation of the form σ(n

) = m

, if any exist within the boundaries imposed by the input (t, B). It should be noted that in most cases, our method will be less performing than for l = k = 3.

Firstly, we focus on generalizing our method for l 6= 3. This does not present great difficulties. We simply replace the factorization of 1 + p + p

+ p

with that of P

i=0

p

. However it should be noted that certain powers factorize in smaller primes more easily than others. To formalize this idea, we use some results from the theory of cyclotomic polynomials, which gives us

σ(p

) =

l

X

i=0

p

= p

− 1

p − 1 = 1 p − 1

Y

d|l+1 d≥1

Φ

(p),

where Φ

(p) represents the dth cyclotomic polynomial.

Here is a list of the first 6.

Φ

(x) = x − 1, Φ

(x) = x

+ 1,

Φ

(x) = x + 1, Φ

(x) = x

+ x

+ x

+ x + 1, Φ

(x) = x

+ x + 1, Φ

(x) = x

− x + 1,

For further information and a theoretical background on cyclotomic polynomials, one can consult Chapter 4 of [10].

Since Φ

(x) = x − 1 and one divides any number, we can rewrite σ(p

) = 1

p − 1 Y

d|l+1 d≥1

Φ

(p) = Y

d|l+1 d≥2

Φ

(p).

Thus the polynomial P

i=0

p

has as many factors as the prime factors of l + 1,

which shows why certain factorizations of σ(p

) have small prime factors more easily

than others. For instance, the prime factors of σ(p

1) tend to be small because σ(p

) = (p + 1)(p

+ p + 1)(p

+ 1)(p

− p + 1)(p

− p

+ 1).

Let us know consider the generalization of our algorithm to the case k 6= 3. If k = 2 the case is trivial: we skip the brute-force part of the algorithm as every vector in the kernel of M is a solution. For k ≥ 4, we need to distinguish two cases:

k is prime or k is not.

In the former situation, not much changes from the base algorithm. The main difference is that for larger ks our optimization techniques become much more ag- gressive, as we are able to eliminate several columns whenever the sum of elements of a row is less than k. This allows us to use much bigger input (t, B). However, the solutions for large k tend to be larger, thus the two aspects somewhat balance out.

On the other hand, if k is not prime the situation is quite different. The main

discrepancy lies in the fact that Z/kZ is not a field for k not prime. This implies

that we cannot compute the kernel of our matrix M , as such thing does not even

exist. We are thus forced to directly consider all possible subsets of columns of M

and check whether their sum is 0 mod k. The number of columns of M is obviously

much larger than the dimension of the kernel in the cases of k prime, which force us

to greatly reduce the input size of (t, B). Thus, if we restrict ourselves to reason-

able computation times, this method can only find relatively small solutions, if any

exists.

Chapter 5 Conclusion

In this thesis, we have given an overview of the problems and the results concerning power-values of the sum-of-divisors function σ.

We have first investigated when σ takes on power values for any possible input.

We have shown that for any integer k there exist infinitely many numbers n such that σ(n) = m

and new solutions for specific cases, which were not found in current literature, have been presented.

Subsequently, we looked for powers whose sum of divisor attained a power value.

This case proved to be much harder to tackle. However, we presented a historical perspective on some of these questions. Furthermore, we were not able to prove or disprove the existence of infinitely many solutions for this family of problems, but we presented a reasoning and computer calculations that gave credit to some conjectures; namely, the existence of infinitely many squares and cubes such that their sum of divisors take on a kth power, for any integer k, while only finitely many higher powers do so.

Eventually, we studied the case of σ(n

) = m

, where we were able to lower the value of the smallest known solution, even though there is still no definite answer on whether there exists a smaller solution.

Lastly we gave a detailed explanation of how our numerical algorithm worked and what specific optimizations we have introduced.

The most important open problem is to obtain proofs of conjectures 2.1, 2.2 and

2.3. It would also be interesting to find a complete list of the first few solutions of

σ(n

) = m

.

Appendix A

An elementary proof on the bounds of π(x)

This section contains a proof of an upper and lower bound of π(x), the number of prime numbers less than x, using reasonings from elementary number theory.

Theorem A.1. Let π(x) denote the number of primes p ≤ x. For x > 221, π(x) is bounded above and below by

2 3

x

log x ≤ π(x) ≤ 2 x log x .

Proof. This proof is split into six statements. The first three provide a proof for the existence of a lower bound, whereas the remaining three provide a proof for the existence of an upper bound. They are

i If p

|

and p

-

, then p

≤ n, ii

≤ n

,

iii 2

≤ (n + 1)n

and π(n) > c

, iv There exists B > 0 such that

log 2 log 2 + log n

log n + log 2n 2n < 42

100 B, for all n ≥ k.

v π(2n) < π(n) + 2 log(2

), vi π(n) < C

.

Statement i. To prove statement i, we want to find the largest exponent α such

that p

divides n!. To do so, we count how many numbers between 1 and n are

multiple of p, how many are multiple of p

and so forth. The multiples of p

are

b

c. By noting that b

c = 0 if p

> n, one can conclude that the largest exponent

α is P

b

c.

Since

=

, the value of a, the largest exponent of p such that p

|

, is given by

a =

∞

X

j=1

 b n

p

c − b k

p

c − b n − k p

c



. (A.1)

We now employ division with remainder to estimate b

c − b

c − b

c. Let k = q

· p

+ r

and n − k = q

· p

+ r

, where r

, r

< p

. It follows that n = (q

+ q

) · p

+ r

+ r

, where r

+ r

< 2p

. Thus,

b k

p

c = q

; b n − k

p

c = q

; b n

p

c = q

+ q

+ b r

+ r

p

c ≤ q

+ q

+ 1.

Hence, the value of b

c − b

c − b

c is at most 1 if p

≤ n, 0 otherwise. It follows that a ≤ #{j = 1, 2, . . . | p

≤ n}, which shows p

≤ n.

Statement ii. Consider the prime factorization of

= Q

p

. By what has been proven in statement i,

r

Y

i=1

p

≤

r

Y

i=1

n = n

,

where r is the number of prime factors of

. It follows that r is less than the number of prime factors of n!, which is π(n), as every prime less than n divides n!.

Hence,

≤ n

≤ n

.

Statement iii. The first part of statement iii is easily shown by expressing 2

as the power of a binomial and by using the results of statement ii.

2

= (1 + 1)

=

n

X

k=0

n k



≤

n

X

k=0

n

= (n + 1)n

.

To prove the second part of statement iii, we take the logarithm of both sides.

n log 2 ≤ log(n + 1) + π(n) log(n), which becomes, by rearranging,

π(n) ≥ n log 2

log n − log(n + 1) log n = n

log n ·



log 2 − log(n + 1) n

 .

Statement iv. The expression in statement iv is monotonically decreasing for n > 1. Since we want to find an upper bound for every n > k, taking

B ≥ 100 42



log 2 log 2 + log k

log k + log 2k 2k



will satisfy the condition. In particular, for k = 110 we can take B = 2.

Statement v. Let us consider the primes between n and 2n. Since there are π(2n) − π(n) of them and they all are strictly larger than n, it follows that

n

≤ Y

n<p≤2n

p.

Furthermore every prime between n and 2n divides (2n)!. Since

=

and no prime between n and 2n appears in the denominator, they all divide

as well. Since they are both positive quantities, it is immediate that Q

n<p≤2n

p ≤

.

Lastly,

2n n



≤

2n

X

k=0

2n k



= (1 + 1)

= 2

. By combining these inequalities together, we obtain

n

≤ 2

,

from which follows, by taking the logarithm of both sides, π(2n) − π(n) < 2n log 2

log n .

Statement vi. The last and final statement of this proof is proved by induction.

Suppose that

π(k) ≤ B k

log k for all k ≤ n, our claim is that the same holds for n < l ≤ 2n.

Let k = bl/2c. Thus π(l) ≤ π(2k + 1) ≤ π(2k) + 1. Hence, using the result of statement v we obtain

π(l) ≤ π(2k) + 1 < π(k) + 2k log 2 log k + 1.

If we apply the inductive assumption,

π(k) + 2k log 2

log k + 1 < B k

log k + 2k log 2 log k + 1

= 2k log 2k



B log 2k

2 log k + log 2 log 2k

log k + log 2k 2k



= 2k log 2k









 B( 1

2 + log 2 2 log k

| {z }

≤8/100 for k≥110

) + log 2 log 2k

log k + log 2k 2k

| {z }

≤42B/100, by stat. iii











≤ B 2k

log 2k .

Hence, by induction we obtain that

π(n) ≤ B n log n , for all positive integers n.

Thus far, we have shown that



log 2 − log(n + 1) n

 n

log n ≤ π(n) ≤ B n log n ,

where B depends on k, which is bn/2c. This implies that picking a certain n, we can compute a numerical value for the upper and lower bound of π(x) for all x ≤ n.

For instance, let n = 221. Then, for the lower bound we have log 2 − log(n + 1)

n = log 2 − log(222)

221 ' 0.66870 > 2 3 .

On the other hand, for the upper bound we obtain k = bn/2c = 110. By statement

iv, we know that B = 2. This completes the proof.

Appendix B Source code

We report here the source code of the program, written in PARI\GP, that imple- ments the ideas expressed in Chapter 4.

numberOfPrimes = 1500 pBound = 2000

nMod = 3 m = [ ; ] ;

p r i m e L i s t = L i s t ( ) ; removeEmptyRows (m) = {

m = m~ ; \\ T r a n s p o s e o f m, i t i s e a s i e r t o m a n i p u l a t e columns

m = l i f t (m∗Mod( 1 , nMod) ) ; i = 1 ;

w h i l e ( i <= m a t s i z e (m) [ 2 ] ,

i f ( vecsum (m[ , i ] ) == 0 , \\ The row i s empty m = v e c e x t r a c t (m, S t r ( " ^ " , i ) ) ,

\\ELSE i ++;

) ; ) ;

r e t u r n (m~) ; }

removeUnmatchCols (m) = { i = 1 ;

w h i l e ( i <= m a t s i z e (m) [ 1 ] , i f ( vecsum (m[ i , ] ) < nMod ,

f o r ( j = 1 , m a t s i z e (m) [ 2 ] , i f (m[ i , j ] != 0 ,

m = v e c e x t r a c t (m, S t r ( " ^ " , j ) ) ; \\ Range i s ^2 , i .

e . t h e complement o f t h e 2nd column

l i s t p o p ( p r i m e L i s t , j ) ; b r e a k ;

) ; ) ,

vecsum (m[ i , ] ) == nMod , i n d e x = −1;

j = 1 ;

w h i l e ( j <= m a t s i z e (m) [ 2 ] , i f (m[ i , j ] != 0 ,

i f ( i n d e x == −1, i n d e x = j ,

\\ELSE

m[ , i n d e x ] += m[ , j ] ;

m = v e c e x t r a c t (m, S t r ( " ^ " , j ) ) ; \\ Range i s

^2 , i . e . t h e complement o f t h e 2nd column p r i m e L i s t [ i n d e x ] ∗= p r i m e L i s t [ j ] ;

l i s t p o p ( p r i m e L i s t , j ) ; ) ;

) ; j ++;

) ; ) ; i ++;

) ;

r e t u r n (m) ; }

removeCols ( mat ) = { c o u n t = 0 ;

o l d C o l s = m a t s i z e ( mat ) [ 2 ] ; mat = removeEmptyRows ( mat ) ; mat = removeUnmatchCols ( mat ) ; mat = removeEmptyRows ( mat ) ; newCols = m a t s i z e ( mat ) [ 2 ] ; w h i l e ( newCols != o l d C o l s ,

o l d C o l s = newCols ;

mat = removeUnmatchCols ( mat ) ; mat = removeEmptyRows ( mat ) ; newCols = m a t s i z e ( mat ) [ 2 ] ; c o u n t++;

i f ( m a t s i z e ( mat ) [ 2 ] <= 5 , b r e a k ( ) ; ) ; ) ;

mat = removeEmptyRows ( mat ) ;

r e t u r n ( mat ) ; }

r e m o v e K e r n e l C o l s (m) = { f o r ( i = 1 , m a t s i z e (m) [ 1 ] ,

c o u n t = 0 ; i n d e x = −1;

f o r ( j = 1 , m a t s i z e (m) [ 2 ] , i f (m[ i , j ] != 0 , c o u n t++) ; i f (m[ i , j ] == 2 , i n d e x = j ) ; ) ;

i f ( c o u n t == 1 && i n d e x != −1, m = v e c e x t r a c t (m, S t r ( " ^ " , i n d e x ) ) ) ;

) ;

r e t u r n (m) ; }

subFactorWithBound ( n ) = { l e f t = n ;

l i s t = v e c t o r ( p r i m e p i ( pBound ) ) ; f o r ( i = 1 , l e n g t h ( l i s t ) ,

p = prime ( i ) ;

w h i l e (Mod( l e f t , p ) == 0 , l e f t /= p ;

l i s t [ i ]++;) ; ) ;

i f ( l e f t == 1 , r e t u r n ( l i s t ) ) ; }

factorWithBound ( p ) = {

i f ( 1 == −1, \\ Case \ sigma ( p ) = m^k \\ These a r e f l a g s t o be manually s e t

l 1 = subFactorWithBound ( p + 1 ) ; i f ( l e n g t h ( l 1 ) != 0 ,

l i s t p u t ( p r i m e L i s t , p ) ; r e t u r n ( l 1 ) ;

) ; ) ;

i f ( p == −1, \\ p ^ 2 ; p \ i n { 2 , 3} \\ These a r e f l a g s t o be

manually s e t

l = subFactorWithBound ( p^6 + p^5 + p^4 + p^3 + p^2 + p + 1 ) ;

i f ( l e n g t h ( l ) != 0 ,

l i s t p u t ( p r i m e L i s t , p ) ; r e t u r n ( l ) ,

r e t u r n ( ) ; ) ,

p == 3 | | p == −1 | | p == −1, \\ p^3 p \ i n { 2 , 3 , 5} \\

These a r e f l a g s t o be manually s e t l 1 = subFactorWithBound ( p + 1 ) ;

l 2 = subFactorWithBound ( p^4 − p^3 + p^2 − p + 1 ) ; l 3 = subFactorWithBound ( p^4 + p^3 + p^2 + p + 1 ) ; i f ( l e n g t h ( l 1 ) != 0 && l e n g t h ( l 2 ) != 0 && l e n g t h ( l 3 )

!= 0 ,

l i s t p u t ( p r i m e L i s t , p ) ; r e t u r n ( l 1 + l 2 + l 3 ) , r e t u r n ( ) ;

) ,

p == −1, \\ p ^ 5 ; p \ i n { 2 , 3} \\ These a r e f l a g s t o be manually s e t

i f ( p == 2 , l i s t p u t ( p r i m e L i s t , 2 ) ; r e t u r n (

subFactorWithBound ( 3 ) + subFactorWithBound ( 5 ) \ + subFactorWithBound ( 1 7 ) +

subFactorWithBound ( 2 5 7 ) ) ) ; l 1 = subFactorWithBound ( ( p + 1 ) / 2 ) ;

l 2 = subFactorWithBound ( ( p^2 + 1 ) / 2 ) ; l 3 = subFactorWithBound ( ( p^4 + 1 ) / 2 ) ; l 4 = subFactorWithBound ( ( p^8 + 1 ) / 2 ) ;

i f ( l e n g t h ( l 1 ) != 0 && l e n g t h ( l 2 ) != 0 && l e n g t h ( l 3 )

!= 0 && l e n g t h ( l 4 ) != 0 , l i s t p u t ( p r i m e L i s t , p ) ; l 1 [ 1 ] += 4 ;

r e t u r n ( l 1 + l 2 + l 3 + l 4 ) , r e t u r n ( ) ;

) ,

p == −1, \\ p ^ 9 ; p \ i n {2} \\ These a r e f l a g s t o be manually s e t

l 1 = subFactorWithBound ( 3 ) ; l 2 = subFactorWithBound ( 1 2 7 ) ; l 3 = subFactorWithBound ( 1 6 3 8 5 ) ; l 4 = subFactorWithBound ( 4 3 ) ; l i s t p u t ( p r i m e L i s t , 2 ) ;

r e t u r n ( l 1 + l 2 + l 3 + l 4 ) ,

\\ELSE

i f ( p == 2 , l i s t p u t ( p r i m e L i s t , 2 ) ; r e t u r n (

subFactorWithBound ( 3 ) + subFactorWithBound ( 5 ) ) ) ; l 1 = subFactorWithBound ( ( p + 1 ) / 2 ) ;

l 2 = subFactorWithBound ( ( p^2 + 1 ) / 2 ) ; i f ( l e n g t h ( l 1 ) != 0 && l e n g t h ( l 2 ) != 0 ,

l i s t p u t ( p r i m e L i s t , p ) ; l 1 [ 1 ] += 2 ;

r e t u r n ( l 1 + l 2 ) ; )

) ; }

c o u n t = 0 ; compute ( p ) = {

r e t = factorWithBound ( p ) ;

i f ( l e n g t h ( r e t ) != 0 , c o u n t++; m = matconcat ( [ m, Mat ( r e t )

~ ] ) ; ) ; }

f o r ( i = 1 , numberOfPrimes , compute ( prime ( i ) ) ) ; p r i n t ( " The i n i t i a l m a t r i x s i z e i s : " ) ;

p r i n t ( m a t s i z e (m) ) ;

p r i n t f ( " The rank i s %d . \ n " , matrank (m) ) ; m = removeCols (m) ;

p r i n t ( " The m a t r i x s i z e a f t e r r e d u c t i o n i s : " ) ;

p r i n t ( m a t s i z e (m) ) ;

m = m ∗ Mod( 1 , nMod) ; s o l u t i o n s = L i s t ( ) ; i f ( i s p r i m e (nMod) , {

k e r = matker (m) ;

k e r = r e m o v e K e r n e l C o l s ( k e r ) ; p r i n t ( " The k e r n e l s i z e i s : " ) ; p r i n t ( m a t s i z e ( k e r ) ) ;

c o u n t = 0 ;

f o r ( i = 1 , 2^( m a t s i z e ( k e r ) [ 2 ] ) − 1 ,

i f (Mod( i , 1 0 0 0 0 0 ) == 0 , p r i n t f ( "We a r e a t %02.2 f %%.\n " , i /2^( m a t s i z e ( k e r ) [ 2 ] ) ∗ 1 0 0 ) ; ) ;

submat = v e c e x t r a c t ( ker , i ) ;

o n e s = v e c t o r v ( m a t s i z e ( submat ) [ 2 ] , x , 1 ) ; \\ C r e a t e s a v e c t o r o f o n e s [ 1 , 1 , . . . , 1 ]

columnSum = submat ∗ o n e s ; found = 1 ;

f o r ( j = 1 , l e n g t h ( columnSum ) ,

i f ( columnSum [ j ] != 0 && columnSum [ j ] != 1 , found = 0 ;

b r e a k ( ) ; )

) ;

i f ( found == 1 , c o u n t++;

s o l = 1 ;

f o r ( k = 1 , l e n g t h ( columnSum ) , i f ( columnSum [ k ] != 0 ,

s o l ∗= p r i m e L i s t [ k ] ; )

) ;

l i s t p u t ( s o l u t i o n s , s o l ) ; ) ;

) ;

l i s t s o r t ( s o l u t i o n s ) ;

p r i n t f ("%d s o l u t i o n s have been found . \ n " , c o u n t ) ;

p r i n t f ( " The s m a l l e s t s o l u t i o n i s %d w i t h %d d i g i t s . \ n " , s o l u t i o n s [ 1 ] , ( l o g ( s o l u t i o n s [ 1 ] ) \ l o g ( 1 0 ) ) + 1 ) ;

p r i n t f ( " The l a r g e s t s o l u t i o n i s %d w i t h %d d i g i t s . \ n " , s o l u t i o n s [# s o l u t i o n s ] , ( l o g ( s o l u t i o n s [# s o l u t i o n s ] ) \ l o g ( 1 0 ) ) + 1 ) ;

} , {

c o u n t = 0 ;

f o r ( i = 1 , 2^( m a t s i z e (m) [ 2 ] ) − 1 ,

i f (Mod( i , 1 0 0 0 0 0 ) == 0 , p r i n t f ( "We a r e a t %02.2 f %%.\n " , i /2^( m a t s i z e (m) [ 2 ] ) ∗ 1 0 0 ) ; ) ;

submat = v e c e x t r a c t (m, i ) ;

o n e s = v e c t o r v ( m a t s i z e ( submat ) [ 2 ] , x , 1 ) ; \\ C r e a t e s a v e c t o r o f o n e s [ 1 , 1 , . . . , 1 ]

columnSum = submat ∗ o n e s ; found = 1 ;

f o r ( j = 1 , l e n g t h ( columnSum ) , i f ( columnSum [ j ] != 0 ,

found = 0 ; b r e a k ( ) ; )

) ;

i f ( found == 1 , c o u n t++;

s o l = 1 ;

p r i m e s I n S o l = v e c e x t r a c t ( Vec ( p r i m e L i s t ) , i ) ; f o r ( k = 1 , #p r i m e s I n S o l ,

s o l ∗= p r i m e s I n S o l [ k ] ; ) ;

l i s t p u t ( s o l u t i o n s , s o l ) ; ) ;

) ;

l i s t s o r t ( s o l u t i o n s ) ;

p r i n t f ("%d s o l u t i o n s have been found . \ n " , c o u n t ) ;

p r i n t f ( " The s m a l l e s t s o l u t i o n i s %d w i t h %d d i g i t s . \ n " , s o l u t i o n s [ 1 ] , ( l o g ( s o l u t i o n s [ 1 ] ) \ l o g ( 1 0 ) ) + 1 ) ;

p r i n t f ( " The l a r g e s t s o l u t i o n i s %d w i t h %d d i g i t s . \ n " , s o l u t i o n s [# s o l u t i o n s ] , ( l o g ( s o l u t i o n s [# s o l u t i o n s ] ) \ l o g ( 1 0 ) ) + 1 ) ;

} ) ;

Appendix C

List of solutions to σ(n) = m ⁹

We report here a list with the first 300 solutions of σ(n) = m

we have found. We transcribe only n for space reasons.

n = 381, n = 3874470, n = 3912990, n = 4049094, n = 4222086, n = 4366670, n = 4477970, n = 4602070, n = 4658170, n = 5636974, n = 5863814, n = 5940854, n = 5977234, n = 6013274, n = 6039615, n = 6652083, n = 7103195, n = 7173815, n = 7356665, n = 7423339, n = 9260743, n = 1483974030, n = 1660273230, n = 1663701270, n = 1678082070, n = 1698538170, n = 1706106570, n = 1724594970, n = 1735155870, n = 1749775830, n = 1753388670, n = 1760490930, n = 1774762770, n = 1787498130, n = 1809288030, n = 1828638966, n = 1828695270, n = 1849671978, n = 1868061426, n = 1879324986, n = 1915675566, n = 1992765126, n = 2026502170, n = 2040717910, n = 2045948070, n = 2065924630, n = 2113538770,

n = 2138159814, n = 2147687094, n = 2202658986, n = 2216147406, n = 2229509766, n = 2234113134, n = 2253424494, n = 2263465374, n = 2271121545, n = 2272633746, n = 2277326154, n = 2290481655, n = 2291057394, n = 2336153470, n = 2350459370, n = 2364631570, n = 2369513930, n = 2404111270, n = 2430163670, n = 2440992070, n = 2450879530, n = 2455939970, n = 2470748170, n = 2477168770, n = 2492104970, n = 2497250530, n = 2507365870, n = 2514691102, n = 2518538270, n = 2525896142, n = 2545752286, n = 2545830670, n = 2551008614, n = 2590083195, n = 2604425194, n = 2615833815, n = 2616030074, n = 2634381302, n = 2660572334, n = 2694311895, n = 2699419426, n = 2706317295, n = 2721302914, n = 2727591735, n = 2733223515, n = 2774241646, n = 2790455565,

n = 2802889365, n = 2822464491, n = 2822551395, n = 2828292159, n = 2850613215, n = 2883312201, n = 2931811449, n = 2949764979, n = 3026351834, n = 3038746821, n = 3075789651, n = 3076483795, n = 3103489094, n = 3113986645, n = 3127862045, n = 3137120374, n = 3156331154, n = 3161757445, n = 3170395234, n = 3181119095, n = 3222913295, n = 3262201145, n = 3352607995, n = 3478111719, n = 3512071871, n = 3512691123, n = 3528343083, n = 3750904795, n = 3823456145, n = 3854451905, n = 3861468965, n = 3884751865, n = 3892772885, n = 3919959659, n = 3937426339, n = 3971460899, n = 3995780929, n = 4019873669, n = 4028173681, n = 4106535559, n = 4131278239, n = 4149686519, n = 4166495201, n = 4175097949, n = 4200271889, n = 4327912139, n = 775911464790,