Abstract Diophantine Equations Involving

ABSTRACT

DIOPHANTINE EQUATIONS INVOLVING ARITHMETIC FUNCTIONS OF FACTORIALS

Daniel M. Baczkowski

We examine and classify the solutions to certain Diophantine equations involving factorials and some well known arithmetic functions. F. Luca has showed that there are ﬁnitely many solutions to the equation:

f(n!) = a · m!

where f is one of the arithmetic functions φ or σ (sum of the divisors function) and a is a rational number. We study the solutions for this equation when a is a prime power or a reciprocal of a prime power. Furthermore, we prove that if % is prime and k > 0, then φ(n!) = %k · m! and %k · f(n!) = m! have ﬁnitely many solutions (%, k, m, n), too. Diophantine equations involving arithmetic functions of factorials

A Thesis

submitted in partial fulﬁllment of the requirements for the degree of Master’s of Arts Department of Mathematics and Statistics

Daniel M. Baczkowski Miami University Oxford, OH 2004

Advisor: Dr. Reza Akhtar

Reader: Professor Patrick Dowling

Reader: Dr. Thomas Farmer Contents

1 Introduction 1 1.1 Our Discoveries ...... 1 1.2 Other Intriguing Problems ...... 1 1.3 The Brocard-Ramanujan Diophantine Equation ...... 2

2 Notation and Preliminaries 3 2.1 Customary Notations and Conventions ...... 3 2.2 Elementary Properties of φ and σ ...... 3 2.3 φ and σ Involving Factorials ...... 4

3 Factorials 4 3.1 Exponents of Factorials ...... 4

3.2 The Stopping Point of αp(n)...... 5

3.3 Properties of αp(n)...... 6 3.4 Factorization of n! ...... 7

4 F. Luca’s Results 10 4.1 Arithmetic Functions ...... 10 4.2 Two Diophantine Equations ...... 11

5 Arithmetic Functions and a Prime Power 12 5.1 Lower Bounds for Primitive Prime Factors ...... 12 5.2 Multiplying by %k ...... 13

6 Concluding Remarks 19

ii ACKNOWLEDGEMENTS

First of all, I must thank the Lord for assembling my life and providing me numerous people along the way that continue to encourage and edify my erudition and well being. I wish to thank my advisor, Reza Akhtar, for providing his continuous support and assistance. I am also obligated to thank the other skilled professors at Miami University who contributed towards my education and passion for mathematics, especially Paddy Dowling, Tom Farmer, and Bruce Magurn. Much appreciation is required for the reference and help that Florian Luca pro- vided and speedy responses to my messages. Also, I am grateful to Michael Filaseta who suggested a particular reference. The profound friends I have made at Miami, especially Brian Lehman, Nate Miller, Dave Aukerman, Jesse Thornton, Jared Owens, Megan Wawro, Catherine and Alyssa Shaﬀer, and all the mathematics graduate students. Also, the many other inﬂuential, extraordinary friends not listed deserve recognition for their numerous contributions. My assiduous students at Miami are also entitled to credit for inspiring my journey by listening to my digressions. Finally, I thank my parents for their perpetual support and faith in my abilities. My brother, Bry, for being the coolest brother alive, deserves recognition, too.

iii 1 Introduction

1.1 Our Discoveries

There are numerous peculiar and fascinating facts involving factorials and arithmetic functions. Particularly inﬂuenced by the work of Florian Luca, we shall elaborate some of his discoveries. The paper contains four main results. Let p and q be arbitrary primes and denote by Q pαp the prime factorization of n!. p≤n

Theorem 3.9 If p < q are primes and n >> 0, then pαp > qαq .

Corollary 3.11 If n ≥ 4, then 2α2 > qαq for any prime q > 2.

Corollary 3.12 If n ≥ 6, then 3α3 > qαq for any prime q > 3.

Following this we prove three theorems in §5.2 which extend the work of F. Luca’s paper, Equations Involving Arithmetic Functions of Factorials [Luca]. Luca shows that if a is any rational number and f is one of the arithmetic functions φ or σ, then f(n!) the equation = a has only ﬁnite number of solutions (m, n). He also proves m! that the only solutions of the equations φ(n!) = m! are obtained for n = 0, 1, 2, 3 and of the equation σ(n!) = m! are obtained for n = 0, 1. Our contributing results corresponding to the case that a is a prime power or a reciprocal of a prime power are listed below. (Notice that since k is variable, Luca’s result does not apply directly.)

Let % be a prime and k > 0, then Theorem 5.6 The equation %k · φ(n!) = m! has solutions if and only if 0 ≤ n ≤ 4 and k = 1.

Theorem 5.7 The equation φ(n!) = %k · m! has solutions if and only if 3 ≤ n ≤ 8 and 1 ≤ k ≤ 6.

Theorem 5.9 The equation %k · σ(n!) = m! has solutions if and only if 0 ≤ n ≤ 5 and k = 1.

1.2 Other Intriguing Problems

An assortment of problems involving factorials have already been examined in the literature. An equation similar to Theorem 5.9 has been studied by Pomerance [Pom]

1 who has shown that the only positive integers n such that n! is multiply perfect that is a divisor of σ(n!) are n = 1, 3, 5. Erdösand Graham have studied the equation 2 p p y = a1!a2! ··· ar! [EG]. Erdös and Obláth have investigated the equations n! = x ±y and n! ± m! = xp [EO].

1.3 The Brocard-Ramanujan Diophantine Equation

In the marvellous subject of number theory, many problems which initially appear accessible, in fact turn out to be impenetrable. For some insight on such problems, consider the following conjecture:

Conjecture 1.1. It is believed that the Diophantine equation n! + 1 = m2 has only ﬁnitely many integral solutions (m, n).

This equation is often referred to as the Brocard-Ramanujan Diophantine Equa- tion. First in 1876 [Bro1] and again in 1885 [Bro2], H. Brocard proposed the question. Unaware of Brocard’s work, Ramanujan [Ram] proposed the same problem. The three known solutions are (m, n) = (5, 4), (11, 6), (71, 7). Many mathematicians including A. Geradin, H. Gupta, B. Berndt, and W. Galway have contributed to the search for another solution. Using crafty computer programming, Berndt and Galway claim that there are no other solutions for n ≤ 109 [BG]. √ That is one billion factorial! Stirling’s Approximation says n! ≈ ( n )n 2nπ. So, √ √ e 9 109 109 9 8 109 9 8∗109+4 9 10 ! ≈ ( e ) 2 · 10 π > (10 ) 10 > 10 . This implies that 10 ! has at least eight billion and four digits. Assuming the number of atoms in our universe stays constant, it is suspected the universe contains 1080 atoms. So, take the number of suspected atoms in the universe and raise it to the 100,000,000 power, and we have an approximation that is smaller than 109!. The number is unfathomable. M. Overholt has shown if a weak form of Szpiro’s Conjecture, which is related to the ABC Conjecture, holds, then the Brocard-Ramanujan Diophantine equation has only ﬁnitely many solutions [Ov]. As many mathematicians unravelling the realms of number theory know, proving the ABC Conjecture seems inaccessible using today’s methods.

2 2 Notation and Preliminaries

2.1 Customary Notations and Conventions

Throughout the paper m and n will denote nonnegative integers and k a positive integer. Also, p and q will represent positive prime integers. In this section and

Q αp throughout the rest of the paper, we will let n! = p and µp(n) be the sum of p≤n the digits of n written in base p. As is customary, we will use log x for the natural logarithm function and [·] for the greatest integer function. We will state some basic deﬁnitions and formulate necessary results which will be

used throughout the paper. In all deﬁnitions we will assume that the pj are distinct primes.

i Deﬁnition 2.1. Given n ∈ N, ordp(n) = max{i ≥ 0 : p |n}; equivalently ordp(n) is the exponent of the prime p in the prime factorization of n.

In elementary number theory one encounters the following arithmetic functions, φ, the ”Euler phi function”, and σ, the sum of the divisors function.

Definition 2.2. φ(m) = #{i : 1 ≤ i ≤ m, gcd(i, m) = 1} X Definition 2.3. σ(m) = d d|m 1≤d≤m Notice that the above definitions imply φ(1) = σ(1) = 1. Now we shall derive some simple, yet necessary formulas which will be used throughout the paper.

2.2 Elementary Properties of φ and σ s Y 1 Theorem 2.4. If n = pβ1 pβ2 ...pβs , then φ(n) = n 1 − . 1 2 s p i=1 i Proof. First we will show that φ(pβ1 ) = pβ1 1 − 1 . It follows from the deﬁnition that 1 1 p1 β1 β1 β1 β1 β1 φ(p1 ) = p1 − #{i : 1 ≤ i ≤ p1 , gcd(i, p1 ) > 1}. Clearly gcd(i, p1 ) > 1 if and only

if p1 | i. In this range there are:

β1−1 p1, 2p1, 3p1, . . . , p1 (p1). β1 β1 β1−1 β1 1 Thus, φ(p1 ) = p1 − p1 = p1 1 − . p1

3 Now, a proof that φ is multiplicative; that is if gcd(m, n) = 1, then φ(mn) = φ(m)φ(n), can be found in [GA]. Therefore, the result follows from φ being a multiplicative function.

s β +1 Y p i − 1 Theorem 2.5. If n = pβ1 pβ2 ...pβs , then σ(n) = i . 1 2 s p − 1 i=1 i Proof. β1 2 3 β1 Notice that the divisors of p1 are 1, p1, p1, ··· , p1 . So,

β1+1 β1 2 3 β1 p1 − 1 σ(p1 ) = 1 + p1 + p1 + ··· + p1 = . p1 − 1 Again, a proof that σ is multiplicative can be found in [GA]. Therefore, the result follows from σ being a multiplicative function.

2.3 φ and σ Involving Factorials

It is worthwhile to record the following remarks derived from the previous two theorems. If n ∈ N, then for n! the functions φ and σ may be computed as follows: Y Y pαp+1 − 1 φ(n!) = pαp−1 · (p − 1) and σ(n!) = . (1) p − 1 p≤n p≤n where the products are taken over all primes less than or equal to n and αp = ordp(n!).

In the literature, it is common to denote these exponents as αp(n). We will write just

αp if there is no danger of ambiguity.

3 Factorials

3.1 Exponents of Factorials

Lemma 3.1. Given n ∈ N and ordp(n!) = αp, ∞ X n α = p pi i=1 where · denotes the greatest integer function.

A quintessential argument proving Lemma 3.1 can be found in George Andrews’ classic book entitled Number Theory [GA].

4 n Lemma 3.2. p ∈ ( 2 , n] if and only if αp(n) = 1.

Proof. It is clear that the theorem holds for n = 2, 3, 4. Now, let n ≥ 5. Consider a prime n i n2 n p ∈ , n . Since for i ≥ 2, pi > > n implies that = 0, Lemma 3.1 gives 2 4 pi ∞ X n α = = 1. p pi i=1 n Conversely, suppose that α = 1. Then, Lemma 3.1 implies that 1 = α = , p p p n n i.e. 1 ≤ < 2. It follows that < p ≤ n. p 2

n Corollary 3.3. If α (n) ≥ 2, then p ≤ . p 2 Proof. The statement follows immediately from the previous proposition.

3.2 The Stopping Point of αp(n)

Deﬁnition 3.4. Given n ∈ N, the stopping point of αp(n) denoted by Sp(n), is deﬁned i by Sp(n) = max{i ≥ 0 : p ≤ n}.

∞ X n log n Proposition 3.5. If α (n) = , then S (n) = ; thus, p pi p log p i=1

∞ Sp(n) X n X n α (n) = = . p pi pi i=1 i=1 Proof. k To avoid triviality suppose that k = Sp(n) > 0. Then, n = rp where r is rational log n log r log r and r < p. So, k = − . Since r < p, it follows that < 1. Hence log p log p log p log n k = . log p

Again for simplicity, we will denote Sp(n) by just Sp whenever there is no danger of ambiguity.

5 3.3 Properties of αp(n)

Proposition 3.6. If p and q are primes and p < q, then αp ≥ αq.

Proof. log n log n Let n be as above, then S = ≥ = S since p < q. So, p log p log q q

Sq Sp Sp X n X n X n α = = ≤ = α . q qi qi pi p i=1 i=1 i=1

Recall that µp(n) is the sum of the digits of n written in base p. n − µ (n) Lemma 3.7. For n ∈ , α = p . N p p − 1 Proof. k k−1 1 Write n in base p; that is, let n = ckp + ck−1p + ... + c1p + c0 with ck 6= 0 and n n n 0 ≤ ci ≤ p − 1 for all 1 ≤ i ≤ k − 1. By Lemma 3.1, αp = [ p ] + [ p2 ] + [ p3 ] + ··· Substitute n as written above into the expression for αp to produce the following:

 k−1 (k−1)−1 0 −1 [ckp + ck−1p + ··· + c1p + c0p ]  k−2 (k−1)−2 −1 −2  + [ckp + ck−1p + ··· + c1p + c0p ] αp =  + ···  k−k (k−1)−k 1−k 0−k  + [ckp + ck−1p + ··· + c1p + c0p ] which when evaluating the greatest integer of each summand, we have

 k−1 k−2 1 0  ckp +ck−1p + ··· + c2p + c1p  + c pk−2 +c pk−3 + ··· + c p0 α = k k−1 2 p + ···  0  + ckp Now, by adding columns we obtain that

pk − 1 pk−1 − 1 p1 − 1 p0 − 1 α = c + c + ... + c + c p k p − 1 k−1 p − 1 1 p − 1 0 p − 1

n − Pk c n − µ (n) = i=0 i = p . p − 1 p − 1

6 Lemma 3.8. For n ∈ N, n n − 1 − S ≤ α ≤ . p − 1 p p p − 1 Proof. k k−1 1 0 If we write n = ckp + ck−1p + ... + c1p + c0p as above, then by construction Pk µp(n) = i=1 ci and k = Sp. So, µp(n) ≤ (p − 1)Sp. Thus by Lemma 3.7, we have n n − µ (n) n − 1 − S ≤ α = p ≤ . p − 1 p p p − 1 p − 1

3.4 Factorization of n!

Theorem 3.9. Fix primes p and q such that p < q. For n >> 0, pαp > qαq .

Proof. First assume that p = 2 and q = 3. By Lemmas 3.7 and 3.8, we have that n − µ (n) n − µ (n) n α = 2 ≥ n − S and α = 3 ≤ . 2 2 − 1 2 3 3 − 1 2 One can check that this special case of theorem holds for 4 ≤ n ≤ 22. First note that log 3 1 n log n 1 − > and if n ≥ 23, then − > 0. Hence, for n ≥ 23 2 log 2 5 5 log 2 log 3 n log n n(1 − ) − S > − > 0. 2 log 2 2 5 log 2 In particular by rearranging the terms on the left hand side of the inequality, we obtain the following equivalent statements: log 3 n n − S > · 2 log 2 2 n (n − S ) log 2 > log 3 2 2 n−S n 2 2 > 3 2 . Hence we have that α n−S n α 2 2 ≥ 2 2 > 3 2 ≥ 3 3 . Now, suppose that p and q are odd primes such that 3 ≤ p < q. From Lemmas 3.7 and 3.8, n n α ≥ − S and α ≤ . p p − 1 p q q − 1

7 log x Using the fact that is a decreasing function for x ≥ 3, we have that x − 1 log p log q 1 log q 1 − > 0 ⇐⇒ − · > 0. p − 1 q − 1 p − 1 log p q − 1 By ﬁxing p and q, it follows that for n >> 0, this implies the equivalent statements:

1 log q 1 log n n − · > = S (2) p − 1 log p q − 1 log p p n log q n − S > · (3) p − 1 p log p q − 1 n n − S log p > log q (4) p − 1 p q − 1 n − S n p p−1 p > q q−1 . (5)

n − S n Hence pαp ≥ p p−1 p > q q−1 ≥ qαq for n >> 0. Next, we have the following lemma, derived from the previous theorem, which explicitly gives a bound for n. pp+1(p − 1)(p + 1) Lemma 3.10. Let 3 ≤ p < q. If n ≥ , then pαp > qαq . (p + 2)p−1 Proof. log y log yz First notice that for y, z ≥ 1, > if and only if yy > yz. This statement z yz follows quite easily by the following reasoning:

yy 1 log y log yz log yy yz > ⇔ yz > 0 ⇔ > 1 ⇔ yy > yz. z yz yz yz

pp+1 Let y = and z = (p − 1)(p + 1). Note that y, z > 1. Observe that yy (p + 2)p−1 grows exponentially with p and yy > yz for all p ≥ 3. Therefore let

pp+1(p − 1)(p + 1) n ≥ yz = . (p + 2)p−1

log x Again, since is a decreasing function, we have that x log n log yz log y log p log(p + 2) log p log q ≤ < = − ≤ − . n yz z p − 1 p + 1 p − 1 q − 1

8 log x Note that the last inequality follows because p < p + 2 ≤ q and is a decreasing x − 1 function; that is, the smallest the right hand diﬀerence could be occurs when p and q are twin primes. All together, we have that log n log p log q < − (6) n p − 1 q − 1 and we conclude with the following equivalent statement log n n log q n < − · . log p p − 1 log p q − 1 This implies that

log n 1 log q 1 S = < n − · p log p p − 1 log p q − 1 in which the equivalent statements (2) through (5) from the proof of Theorem 3.9 yield the desired result that pαp > qαq . Combining Theorem 3.9 and the lower bound in Lemma 3.10, we have the following two corollaries. Now, some discussion is required in order to motivate the validity of the corollaries. By Lemma 3.10, Corollary 3.12 follows. We will prove Corollary 3.11 below.

Corollary 3.11. If n ≥ 4, then 2α2 > qαq for any prime q > 2.

Proof. It suﬃces by the reasoning of Lemma 3.10 to show (6); that is, show that log n log p log q < − holds for p = 2 and q > 2. The proof proceeds quite easily by n p − 1 q − 1 noting that log q log 3 log 22 log n log 2 − ≥ log 2 − > ≥ q − 1 2 22 n if n ≥ 22. The curious reader may check the values for 4 ≤ n ≤ 21. A similar analysis gives the following.

Corollary 3.12. If n ≥ 6, then 3α3 > qαq for any prime q > 3.

9 4 F. Luca’s Results

4.1 Arithmetic Functions

In this section, we will display some work of F. Luca’s related to Diophantine equations involving arithmetic functions of factorials.

c n Lemma 4.1. For n ≥ ee , φ(n) > where c represents the Euler- 2 log log n + 1 Mascheroni constant (c ≈ .5772).

Proof. n 2.50637 By [RS] (Theorem 15 on page 72), for n ≥ 3, < ec log log n+ . It follows φ(n) log log n that n φ(n) > c 2.50637 e log log n + log log n n > 2 log log n + 1 for n ≥ eec . Luca proves the following “ﬁniteness” theorem.

Theorem 4.2. (Luca, [Luca]) Let a be any rational number and let f be one of the arithmetic functions φ or σ. Then, the equation

f(n!) = a (7) m! has only ﬁnitely many solutions (m, n).

Proof. The theorem and proof are from [Luca]. We remark that there is a mistake in his proof which we have corrected. First, let f = φ. We will consider the three cases n < m, n > m, and n = m. If n < m, then n ≤ m − 1, so we get

am! = φ(n!) < n! ≤ (m − 1)!,

1 which implies that m ≤ a . Now let n > max(m, eec ). We will use the previous lemma and the fact that x m+1 2 log log x is an increasing function of x. So n ≥ m + 1 and since (m + 1)! < (m + 1) , we have the following implications n! (m + 1)! (m + 1)! am! = φ(n!) > ≥ > 2 log log(n!) 2 log log(m + 1)! 2 log log((m + 1)m+1)

10 m + 1 < 2a log((m + 1) log(m + 1)) m + 1 < 2a log((m + 1) log(m + 1)) Thus, the last inequality says that m is bounded by some constant depending on a. Lastly, we consider the case where m = n. Using Theorem 2.4, we obtain 1 n! Y 1 = = 1 + . a φ(n!) p − 1 p≤n The product on the left tends to infinity as n tends to infinity, consequently the equation has only finitely many solutions for n = m. Therefore, equation (7) above has only finitely many solutions (m, n). The proof for f = σ is completely analogous.

4.2 Two Diophantine Equations

Furthermore, Luca clariﬁes the solutions to the above equations in the case where a = 1.

Corollary 4.3. (Luca, [Luca])

1. The equation φ(n!) = m! has solutions if and only if 0 ≤ n ≤ 3.

2. The equation σ(n!) = m! has solutions if and only if n = 0, 1.

Proof. Again, the corollary and proof are taken from [Luca]. First, we will prove part (1). s Suppose now that n ≥ 5 and suppose that ord2(n!) = s where s > 0. Then, n! = 2 t s−1 Q where t is odd and thus φ(n!) = 2 φ(t). Now, φ(t) is divisible by p≤n(p − 1) and in particular, φ(t) is divisible by (3 − 1)(5 − 1) = 8. It follows that ord2(φ(n!)) ≥ s − 1 + 3 > s. On the other hand, since m! = φ(n!) < n!, it follows that m < n. Thus,

ord2(m!) ≤ ord2(n!) = s. Therefore we have the desired contradiction. One can check the other values of n to conclude the proof. Now for the proof of (2), one can check that the asserted solutions are in fact the only solutions for n ≤ 8. Assume now that n ≥ 9. We know from Theorem 2.4 and Q 1 Q p−1 Q pαp+1−1 Theorem 2.5 that φ(n!) = n! (1 − p ) = n! ( p ) and that σ(n!) = p−1 . p≤n p≤n p≤n So, we have that Y pαp+1 − 1 Y φ(n!) · σ(n!) = n! < n! pαp = n! · n!. p p≤n p≤n

11 It follows that σ(n!) n! Y p < = . n! φ(n!) p − 1 p≤n Thus we obtain that σ(n!) Y p Y k 3 5 7 8 n < ≤ = n · · · · < . n! p − 1 k − 1 4 6 8 9 2 p≤n 2≤k≤n k6=4,6,8,9 n Hence, n! < m! = σ(n!) ≤ · n! < (n + 1)!, which is a contradiction since no factorial 2 can lie strictly between n! and (n + 1)!. Hence, we list the solutions to (1) and (2) below:

n m n m 0 1 0 1 1 1 1 1 2 1 3 2

We remark that a shorter proof for (2) follows from Corollary 5.3 which we introduce in the next section.

5 Arithmetic Functions and a Prime Power

5.1 Lower Bounds for Primitive Prime Factors

Next, we introduce as a lemma a theorem and result of A. Schinzel involving a lower bound for the greatest prime factor of integers of the form aω − bω. In the following, we let P (n) denote the greatest prime factor of n ∈ N. Deﬁnition 5.1. A prime p such that p | aω − bω is a primitive prime factor if p | aω − bω and p - aη − bη for all η < ω. Lemma 5.2. (A. Schinzel, [AS] Theorem 3) For all integers ω > 12, 2ω − 1 contains two distinct primitive prime factors. Moreover, Zsigmondy’s Theorem [Zs] tells us that P (2ω − 1) ≥ 2ω + 1 and some other primitive prime factor is greater than or equal to ω + 1. log n Corollary 5.3. For n ≥ 18,P (σ(n!)) > 2(n − ) + 1. log 2

12 Proof. By deﬁnition σ(n!) contains the factor 2α2(n)+1 − 1. Use Schinzel’s Lemma 5.2 where log n log n ω = α2(n)+1. By Lemma 3.8, α2(n) ≥ n− log 2 . Thus, if ω = α2(n)+1 > n− log 2 > 13 (i.e. n ≥ 18), then log n P (σ(n!)) ≥ P (2ω − 1) ≥ 2(α (n) + 1) + 1 > 2(n − ) + 1. 2 log 2

5.2 Multiplying by %k

The work of F. Luca forms the foundation of our work. Now, we will elaborate his work by discussing several other types of similar equations; in particular, we will examine solutions of the equations of Theorem 4.2 when a is a prime power or its reciprocal. First, we will need the following deﬁnition and lemma derived from the Prime Number Theorem. Recall that k ∈ N. Let % denote an arbitrary prime.

Deﬁnition 5.4. For x ∈ R, deﬁne the function π(x) = #{p : p ≤ x and p is prime }. Lemma 5.5. (Rosser and Schoenfeld, [RS] p.69) For x ≥ 17, x < π(x). log x Theorem 5.6. The equation %k · φ(n!) = m! has solutions if and only if 0 ≤ n ≤ 4 and k = 1.

Proof. First, we will show that the equation % · φ(n!) = m!, has solutions if and only if 0 ≤ n ≤ 4. Suppose there exists a solution for some n ≥ 5. Write % · n! = 2s · t where t is odd, that is, ord2(% · n!) = s. So, s if % 6= 2 ord (n!) = 2 s − 1 if % = 2. Recognize that by note (1), which is subsequent to Theorems 2.4 and 2.5, we have Y Y φ(n!) = pαp(n)−1(p − 1) which implies (p − 1) | φ(n!). (8) p≤n p≤n It follows that ! Y ord2(% · φ(n!)) ≥ ord2(φ(n!)) ≥ s − 2 + ord2 (p − 1) . p≤n

13 Note t has a factor of 3 and 5. So, (3 − 1)(5 − 1) | Q (p − 1), that is 23 | Q (p − 1). p≤n p≤n All together, ord2(% · φ(n!)) ≥ s + 1. On the other hand, from the given equation, % · φ(n!) = m! implies that % ≤ m. Since φ(n!) < n!, m! < % · n! ≤ m · n!. So, (m − 1)! < n! which is equivalent to m! ≤ n!. Hence,

ord2(m!) ≤ ord2(n!) ≤ s < s + 1 ≤ ord2(% · φ(n!))

which yields the necessary contradiction. Therefore the only possible solutions exist for n ≤ 4. One can manually check that the only solutions occur in the following cases: % n m 2 0 2 2 1 2 2 2 2 3 3 3 3 4 4 Now, we will show that for k > 1 there are no solutions. Let n ≥ 6. Notice that n! has a factor of 3 and 5; hence by (8), ord2(φ(n!)) ≥ α2(n) + 2. Suppose that % 6= 2 and a solution exists. It follows that

k ord2(m!) = ord2(% · φ(n!)) = ord2(φ(n!)) ≥ α2(n) + 2 > ord2(n!).

Consequently, this means m > n. Now, Bertrand’s Postulate asserts that we can pick m i a prime q ∈ , m . By Lemma 3.2, we have that ord (%k · φ(n!)) = ord (m!) = 1. 2 q q Notice that % 6= q, since k > 1. So, ordq(φ(n!)) = 1. We claim that this forces

ordq(n!) = 2. Consider the prime factorization of φ(n!) as displayed in note (1). Since n m Q (p−1) is divisible by primes less than < < q, the claim follows. So, Corollary p≤n 2 2 n m n 3.3 tells us that q ≤ . Therefore, < q ≤ which implies that m < n. But this 2 2 2 contradicts the fact that m > n. Now, this forces % = 2, therefore reducing to the equation 2k · φ(n!) = m! where k > 1. Let n ≥ 17. Using Lemma 3.7,

ord2(m!) = α2(m) = m − µ2(m) < m. (9)

Again by note (1), we have that ! k Y ord2 2 · φ(n!) = k + α2(n) − 1 + ord2 (p − 1) . p≤n

14 Since for every odd prime, 2 | (p − 1); and moreover 22 | (5 − 1) and 22 | (13 − 1), we ! Q have that ord2 (p − 1) ≥ π(n) + 1. So, it follows by Lemma 5.5 that p≤n n ord (m!) = ord 2k · φ(n!) ≥ k + α (n) + π(n) > k + α (n) + . (10) 2 2 2 2 log n

n log n Now, by Lemma 3.8 and the fact that > = S (n), we have the following: log n log 2 2 n n k + α (n) + > k + n − S (n) + > k + n. (11) 2 log n 2 log n

Adjoining inequalities (9), (10), and (11) implies that k + n < m, i.e.

k < m − n. (12)

Since k > 0, this implies that m > n. Note for a solution to exist, k ≤ α2(m) ≤ m − 1 where Lemma 3.8 conﬁrms the second inequality. Recall that φ(n!) < n!. So, m! = 2k · φ(n!) < 2k · n! which implies that m! < 2k · n! < m(m − 1) ··· (m − k + 1) · n!. This implies that (m − k)! < n!. Hence, we have a lower bound for k, namely

m − n < k. (13)

Putting together (12) and (13), we obtain a contradiction! Thus, no solutions exist for n ≥ 17. One can check manually that the only solutions are those listed above.

Theorem 5.7. The equation φ(n!) = %k · m!, where k > 0, has solutions if and only if 3 ≤ n ≤ 8 and 1 ≤ k ≤ 6.

Proof. Suppose there exists a solution for some n ≥ 17. Notice that n! > φ(n!) = %k ·m! > m! implies that n > m. Using Lemma 3.7, we have

ord2(m!) = α2(m) = m − µ2(m) < m. (14)

Again by note (1), we have that ! k Y ord2(% · m!) = ord2(φ(n!)) = α2(n) − 1 + ord2 (p − 1) . p≤n

15 Since for every odd prime, 2 | (p − 1), and furthermore 22 | (5 − 1) and 22 | (13 − 1), we have ! Y ord2 (p − 1) > π(n) + 1. p≤n So, it follows by Lemma 5.5 that if % 6= 2, then n ord (m!) = ord (%k · m!) = ord (φ(n!)) > α (n) + π(n) > α (n) + . (15) 2 2 2 2 2 log n Therefore, combining (14) and (15), then using Lemma 3.8, and lastly the fact that n log n > = S (n), we have the following inequalities: log n log 2 2 n n m > α (n) + > n − S (n) + > n. 2 log n 2 log n This contradicts the fact that n > m. Thus, it is required that % = 2. Observe that ! k Y αp(n)−1 ord3(m!) = ord3(% · m!) = ord3(φ(n!)) = ord3 p (p − 1) . p≤n ! Q The right hand side equals α3(n) − 1 + ord3 (p − 1) > α3(n) + 1 > ord3(n!) p≤n where the ﬁrst inequality holds since p = 7 and 13 in Q (p − 1) yield two factors of 3. p≤n Hence, all together, we have that ord3(m!) > ord3(n!) which is a contradiction since m < n. Direct computations show that the only solutions are those asserted in the statement of the theorem, namely

n 3 4 4 5 5 6 6 7 8 % 2 2 2 2 2 2 2 2 2 k 1 3 2 5 4 5 3 3 6 m 0,1 0,1 2 0,1 2 3 4 6 6

Now, switching to the sum of divisors function σ, we have some general results. We are able to ﬁnd all solutions to the equation %k · σ(n!) = m!. Unfortunately, we are not yet able to do the same for σ(n!) = %k · m!, but we formulate a conjecture in this context.

Lemma 5.8. If the equation %k · σ(n!) = m! has a solution, then

%k ≤ m(m − 1) ··· (m − k + 1).

16 Proof. Suppose that %k · σ(n!) = m! where k > 0 has a solution, then %k | m! and so % ≤ m.

Note that k ≤ S%(m). Let ` be the largest integer 1 ≤ ` ≤ m such that % | `. By deﬁnition ` ≥ %k. Then, it follows that % | `, % | (`−%),% | (`−2%),...,% | (`−(k−1)%). We note that ` − (k − 1)% > 0, since %k−1 > k − 1 for all k > 0. Therefore,

%k | `(` − %)(` − 2%) ··· (` − (k − 1)%)

which implies the desired result since ` ≤ m.

Theorem 5.9. The equation %k · σ(n!) = m! has solutions if and only if 0 ≤ n ≤ 5 and k = 1.

Proof. We will begin by proving some general consequences, assuming that a solution exists. Next, we consider the case % 6= 2, 3 then separately consider the cases % = 3 and % = 2. Throughout the proof, we will assume that a solution exists for n ≥ 18. First, notice that m ≥ n + 1, since n! < %k · σ(n!) = m!. By the preceding lemma, we have that m! m! (m − k)! = ≤ = σ(n!) < (n + 1)!. m(m − 1) ··· (m − k + 1) %k

Thus, m ≤ n + k which yields

n + k k ≤ α (m) ≤ α (n + k) < (16) % % % − 1 where the last inequality follows from Lemma 3.8. Finally, using the Corollary to log n Schinzel’s Lemma which says P (σ(n!)) > 2(n − log 2 ) + 1, we are forced for the given n to have

log n 3 m ≥ 2(n − ) + 1 > n. (17) log 2 2 Now, assume that % 6= 2 or 3. After grouping like terms in (16) this implies that n n k < ≤ . Since m ≤ n + k it follows by (17) that % − 2 3 4 3 n ≥ n + k ≥ m > n 3 2 which produces a contradiction.

17 Now, suppose that % = 3 and n ≥ 25. From the above work in (16), we have that k < n. By (16) and (17), it follows that 3 n ! < m! = 3kσ(n!) < 3k · (n + 1)! < 3n · (n + 1)! 2 which implies that 3 n ··· (n + 2) < 3n = (33)n/3. 2 But, this is a contradiction for the given n, since there are more terms on the left hand side of the inequality each of which is bigger than 33. Lastly, suppose that a solution exists for % = 2 and n ≥ 26. First, notice that

k ≤ α2(m) ≤ m − 1. So, hmi m! m! ! ≤ ≤ = σ(n!) < (n + 1)!. 2 2m−1 2k This implies that m ≤ 2(n + 1). Hence, k ≤ m − 1 ≤ 2n + 1. Again the Corollary to 3 Schinzel’s Lemma tells us that m > 2 n. Therefore, 3 n ! < m! = 2kσ(n!) < 2k · (n + 1)! ≤ 22n+1 · (n + 1)! 2 which implies that 3 n ··· (n + 2) < 22n+1 < 64(n+1)/3. 2 But, this results in a contradiction for n ≥ 62, since there are more terms on the left hand side of the inequality each of which is bigger than 64. Simple calculations show that the above statement is also a contradiction for 26 ≤ n ≤ 61. Now, easy calculations provide that the only solutions are those asserted below.

% n m 2 0 2 2 1 2 2 2 3 2 3 4 2 4 5 2 5 6

Unfortunately, we are unable to compute the solution set for the other equation, namely σ(n!) = %k ·m!. Nevertheless, we display our known results about the equation and conjecture the solution set.

18 Conjecture 5.10. The equation σ(n!) = %k ·m! has solutions if and only if n = 2, 3, 5.

We know for certain that if a solution exists for n ≥ 18, then by the Corollary to log n Schinzel’s Lemma, we have % = P (σ(n!)) > 2(n − log 2 ) + 1. Also, utilizing the bound for the other primitive prime factor yields a bound for m; that is,

log n m > n − . (18) log 2 k 3 k So, we have (n + 1)! > σ(n!) = % · m! > ( 2 n) · m! which implies (n + 1)! (n + 1)! (n − k + 1)! = > > m!, (19) 3 k (n + 1) ··· (n − k + 2) ( 2 n) which produces the result n − k ≥ m. Now, (18) and the result of (19) yield that log n k ≤ log 2 . So, we hope that this will help in determining the solution set.

6 Concluding Remarks

It is worthwhile to note that if the theory involving primitive prime factors of aω − bω is improved, then one might be able to prove Conjecture 5.10. In fact, A. Schinzel [AS2] has shown for ω >> 0 that aω − bω has four primitive roots for “almost all” ω. In these speciﬁc cases, a contradiction follows for the given ω. Moreover, the bound for ω happens to be too large; hence making it impossible to check the other values manually. To conclude our work, we leave the reader with some interesting subsequent ques- tions, such as:

1. What other arithmetic functions f of a factorial could we ﬁnd the solution sets f(n!) for = a where a ∈ ? m! Q 2. Could these Diophantine equations help devise proofs for other unsolved problems in number theory?

19 References

[GA] Andrews, George Number Theory (1st Dover edition) (W.B Saunders Com- pany, Philadelphia, 1971).

[BG] Berndt, Bruce and Galway, William On the Brocard-Ramanujan Diophantine Equation. http://www.math.uiuc.edu/vberndt/articles/galway.pdf, Submit- ted.

[Bro1] Brocard, H. Question 166. Nouv. Corresp. Math. 2 (1876), 287.

[Bro2] Brocard, H. Question 1532. Nouv. Ann. Math. 4 (1885), 391.

[EG] Erd¨os, P. and Graham, R. On Products of Factorials Bull. Inst. Math. Acad. Sinica, Taiwan, 4 (1976), 337-355.

[EO] Erd¨os, P. and Obl´ath, R. Uber¨ diophantische Gleichungen der Form n! = xp ± yp und n! ± m! = xp Acta Szeged. 8 (1937), 241-255.

[Luca] Luca, Florian Equations Involving Arithmetic Functions of Factorials. Divul- gaciones Math. Vol.8. (2000), 15-23.

[Ov] Overholt, Marius The Diophantine Equation n!+1 = m2. London Math. Soc. 25 (1993): 104-113.

[Pom] Pomerance, C. Problem 10331 Amer. Math. Monthly 100 (1993), 796.

[Ram] Ramanujan, S. Question 469. J. Indian Math. Soc. 5 (1913), 59.

[RS] Rosser, J.B. and Schoenfeld, L., Approximate Formulas for Some Functions of Prime Numbers. Illinois Journal of Math. (1962), 64-94.

[AS] Schinzel, A. On the primitive prime factors of an − bn. Proc. of Cambridge Phil. Soc. 58 (1962), 555-562.

[AS2] Schinzel, A. On primitive prime factors of Lehmer numbers. II. Acta Arith. 8 (1962/1963), 251-257.

[Zs] Zsigmondy, K. Zur Theorie der Potenzreste. Monatsh. Math 3 (1892), 265- 284.