Remarks on an interpolation between Wilson's theorem and Giuga's conjecture

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Remarks on an interpolation between Wilson’s theorem and Giuga’s conjecture

Thomas Sauvaget

To cite this version:

Thomas Sauvaget. Remarks on an interpolation between Wilson’s theorem and Giuga’s conjecture.

2019. �hal-02373255v3�

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THEOREM AND GIUGA’S CONJECTURE

THOMAS SAUVAGET

Abstract. A family of congruences interpolating between those of Wilson and Giuga is constructed. Several elementary results are presented, in order to present a possible approach to establishing Giuga’s conjecture.

1. Introduction

Let p be a prime, and j a positive integer such that 1 ≤ j ≤ p−1. Then by Fermat’s Little Theorem we have j^p−1 ≡ 1 (modp). And thusPp−1

j=1j^p−1 ≡ p−1≡ −1 (modp), an observation of Sierpi´nski [7].

In 1950, Giuga [5] asked whether the converse holds, i.e. if any integer n such that Pn−1

j=1 jⁿ⁻¹≡ −1 (modn) is necessarily prime. He provided support for that conjecture by showing that any counterexample, i.e. any composite integer satis- fying the congruence, is greater than 10¹⁰⁰⁰. This lower bound was later improved many times over the ensuing decades, by numerical means : to 10¹⁷⁰⁰ by Bedoc- chi [2], to 10¹³⁸⁸⁷ by Borwein, Borwein, Borwein & Girgensohn [3], and finally to 10¹⁹⁹⁰⁸ by Borwein, Maitland & Skerritt [4]. These works nevertheless have arithmetical limitations, in that the method used intrinsically cannot treat any number having more than 8,135 prime factors.

On the other hand, Tipu [13] proved by analytic means that for any real number x, the number of counterexamples to Giuga’s conjecture G(x) := #{n <

x : nis composite and Pn−1

a=1aⁿ⁻¹ ≡ −1 (modn)}, if any, is at best of moder- ate growth: G(x)√

xlogx. This was later improved by Luca, Pomerance and Shparlinski [9] toG(x) =o(√

x).

Agoh [1] showed in 1995 that conjecturing that n is prime iff nB_n−1 ≡ −1 (mod n) (where the (Bn)_n≥0are Bernoulli numbers, see e.g. Granville [6] and Sun [12]) is equivalent to Giuga’s conjecture (see also Kellner [8]).

The paper is organized as follows. In the next section, we introduce a family of congruences that interpolates between Wilson’s and Giuga’s, and prove an important lemma. Then, in section 3, we show that the family of congruences can be rewriten using unsigned Stirling numbers of the first kind. Properties of those numbers observed numerically, combined with the important lemma, would then imply that odd squarefree composite integers cannot satisfy Giuga’s congruence.

2010Mathematics Subject Classification. 11A41, 11A07.

Key words and phrases. Giuga’s conjecture; primality.

This version of the manuscript takes into account helpful comments and encouragements provided by several referees of previous versions, to whom the author is very grateful. The author also wishes to thank all those who contributed, directly or indirectly, to make Sage and its libraries such a useful free software.

1

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2 THOMAS SAUVAGET

Also, in an appendix, we present a combinatorial relation which might be of independent interest (which arised in an earlier incorrect attempt at proving Giuga’s conjecture).

Notations. Let N2 := N− {0,1}. The set of primes is denoted by P :=

{2,3,5,7,11, . . .}. The double-brackets denote sets of consecutive integers: Ja;bK:=

{i ∈ N|a ≤ i ≤ b}. Define also for each k ∈ N the functions Sk : N → N by Sk(n) :=Pn−1

j=1j^k .

2. An interpolation of Wilson’s and Giuga’s congruences 2.1. Basic definitions.

Definition 2.1. LetfW :N27→Nbe defined byfW(n) := 1 + (n−1)!

Then Wilson’s theorem can be recast as:

Theorem 1. ∀n∈N2 we have: n∈ P ⇔n|f_W(n).

Remark 1. Wilson’s theorem allows to change the definition of primality, which is negative in nature (n is prime iff it is not the product of strictly smaller integers), into a positive one (n is prime iff it divides some specific larger integer). This crucial reversal is at the heart of the philosophy of this paper.

Definition 2.2. Letf_G:N27→Nbe defined byf_G(n) := 1 +S_n−1(n).

Then Giuga’s conjecture is:

Conjecture 1. ∀n∈N2 we have: n∈ P ⇔n|fG(n).

2.2. An interpolation family. We introduce the following integer-valued functions.

Definition 2.3. LetHk :N27→Nbe defined byHk(n) := 1+Pn−1 i=1

(n+i−1−k)!

i! i^k fork∈ J1;n−1K

Remark 2. For any givenn∈N2, we have that for anyk∈J1;n−1Kthe function H_k is integer-valued, since for eachi∈J1;n−1K we havei≤n+i−1−k.

Remark 3. We directly obtainH_n−1=fG. The relationship betweenH1 and fW

is described in the proof of the next lemma.

Lemma 1. ∀n∈N2 we have: n∈ P ⇔n|H1(n).

Proof. Since by definitionH1(n) = 1+(n−1)!+^n!_2!2+^(n+1)!_3! 3+· · ·+^(2n−3)!_(n−1)!(n−1) = 1 + (n−1)! +n[^((n−1)!_2! 2 +^(n+1)!_3!n 3 +· · ·+^(2n−3)!_(n)! (n−1)]≡1 + (n−1)!≡fW(n) (mod n). The term in the bracket is necessarily an integer since for anyj∈J1, n−3K we have that (n+j)! = 1×2× · · · ×(j+ 2)

| {z }

=(j+2)!

×(j+3)×· · ·×n×(n+1)×· · ·×(n+j), so this is simultaneously divisible by (j+ 2)! and byn. The result then follows from

Wilson’s theorem.

ThusH1also characterizes the primes. To motivate the strategy employed in the next sections, consider the residues of the values ofHk(n) modulon, fork∈J1; 34K and n∈J5; 35K, which are provided in the following table:

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Figure 1. Residues of the values ofHk(n) modulon.

On this very limited data set, the following observations can be made:

• for any k∈J1;n−1K,Hk(n)≡0 (modn) for allnprime

• for any k∈J1;n−1K,Hk(n)6≡0 (modn) for any compositen

• in the casencomposite, Hk(n)≡1 (modn) fork≤ ⁿ₂

The last point can be easily established beyondn= 35 for squarefree composites by the following lemma:

Lemma 2. Letn∈N2 be composite and squarefree. ThenH_k(n)≡1 (modn)for k≤ⁿ₂.

Proof. Given ncomposite, let its prime factorization be n=QJ

j=1qj, where q1 <

· · ·< q_J are distinct primes. We necessarily have that 2≤q₁<· · ·< q_J≤ ⁿ₂. Letk∈J1;bⁿ₂cK. So, for eachi∈J1;n−1Kwe haven+i−1−k≥ bⁿ₂c+i−1≥ qJ+i−1≥qJ >· · ·> q1 ().

Recall the definition of H_k(n). Since ^{(n+i−1−k)!}_i! is a product of n+i−1−k consecutive integers starting ati+ 1, each term ^{(n+i−1−k)!}_i! i^k contains a product of n+i−kconsecutive integers starting at i.

So, given the inequalities (), we infer that for eachi∈J1;n−1K, each of theqjis a factor of at least one of those consecutive integers, hence each term ^{(n+i−1−k)!}_i! i^k is a multiple ofn. So only the first term 1 inH_k(n) remains modulon.

Going beyond k=bⁿ₂cwould require a more detailed analysis of the structure of those values of H_k(n). But, given the first and second observations, it appears more straighforward, to make the following conjecture: it captures the primality condition (no composite is a false positive), while not entering into the arithmetical details of the values of the residues forncomposite.

Conjecture 2. For eachk∈N2 we have: [n∈ P, withn≥k+ 1]⇔n|H_k(n).

Remark 4. The casek=n−1is Giuga’s conjecture.

In the next section, we show how lemma 2 could make it impossible for odd squarefree composite integers to satisfy Giuga’s congruence.

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4 THOMAS SAUVAGET

3. Towards a path for the proof of Giuga’s conjecture

3.1. The case of n = p prime of conjecture 2. We start with the following elementary lemma, whose proof is taken almost verbatim from Nicolas [10].

Lemma 3. Letpbe an odd prime. Then for anyk∈N^∗such thatgcd(p−1, k) = 1 we have S_k(p)≡0 (modp).

Proof. Letbbe any number relatively prime top. Recall that the numbersb,2b,3b, 4b, . . . ,(p−1)b are a reduced residue class modulop.

It follows thatb^k+ (2b)^k+ (3b)^k+· · ·+ ((p−1)b)^k≡1^k+ 2^k+ 3^k+· · ·+ (p−1)^k (mod p) (∗). But (ib)^k =b^ki^k, so we can rewrite (∗) as (b^k−1)(1^k + 2^k+ 3^k+

· · ·+ (p−1)^k)≡0 (modp) (∗∗).

Let b have order p−1 modulop. So we are lettingb be a primitive root of p.

Since p−1 does not divide k, we have b^k 6≡1 (modp). Then it follows from (∗) that 1^k+ 2^k+ 3^k+· · ·+ (p−1)^k≡0 (modp) .

Recall the following classical definitions.

Definition 3.1. The unsigned Stirling numbers of the first kind, M

j

, are the coefficients of the expansion of the rising factorial into powers: Qj=M−1

j=0 (x+j) = Pj=M

j=0

M j

x^j. The Stirling numbers of the first kind, s(M, j), are the coefficients of the expansion of the falling factorial into powers: Qj=M−1

j=0 (x−j) = Pj=M

j=0 s(M, j)x^j. We have|s(M, j)|= M

j

.

We can use this and lemma 3 to prove the following result, which is one direction of conjecture 2.

Theorem 2. For all odd primep, and allk∈J1;p−2Kwe haveHk(p)≡0 (modp).

Proof. Indeed, for anyn, prime or composite, we can expandH_k(n) as function of H_n−1(n), and{S_k(n), . . . , S_n−2(n)}, and{

n−k 1

, . . . ,

n−k n−k−1

}:

H_k(n) = 1 +Pn−1 i=1

(n+i−1−k)!

i! i^k

= 1 +Pn−1

i=1(n+i−1−k)(n+i−2−k). . .(1 +i)i^k

= 1 +Pn−1 i=1

Pn−k

j=0

n−k j

i^j

i^k−1

= 1 +Pn−1 i=1

n−k 0

i^k−1+· · ·+ n−k

n−k

iⁿ⁻¹

This can the be rearranged as:

Hk(n) = H_n−1(n) +

n−k n−k−1

S_n−2(n) +· · ·+ n−k

1

Sk(n) since

n−k 0

= 0 and n−k

n−k

= 1.

Now, forn=pprime we mentioned at the begining of this paper thatH_p−1(p)≡ 0 (modp). Combining this with lemma 3 finishes the proof.

We shall not attempt to prove conjecture 2 in this work. Let us only mention the following simpler conjecture based on numerical data.

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Conjecture 3. For any odd squarefree compositen=q₁· · ·q_J, one hasH_n−q_J₊₁(n)≡ n+ 1−q1 (mod n).

Remark 5. This would be, for odd squarefree composites, the first “non-trivial value ofk” i.e. for thesenone could show like in lemma 2 thatHk(n)≡1 (modn) for k < n−q_J+ 1, while fork=n−q_J+ 1one would have H_k(n)different from either 0 or 1 modulo n.

3.2. Steps towards Giuga’s conjecture. Recall from [3] that even, and non- squarefree, composite integers cannot satisfy Giuga’s congruence. So in this section we consider only the case of odd squarefree integers, i.e. n = q1. . . qJ with 3 ≤ q1<· · ·< qj≤ ⁿ₃.

The following elementary lemma allows to reduce the problem to the prime factors ofn.

Lemma 4. Let n=q1· · ·qJ be odd composite squarefree. Then for any numberM we have the equivalence:

M ≡0 (modq1· · ·qJ)⇔







M ≡0 (mod q₁) ...

M ≡0 (mod q_J)

Proof. This follows immediately by the Chinese Remainder Theorem.

We can apply this lemma in two special cases. Firstly, the condition H_k(n)≡0 (mod n) is equivalent to having







H_k(n)≡0 (mod q₁) ...

H_k(n)≡0 (mod q_J)

So to show H_k(n)6≡0 (modn) it would suffice to show that for at least oneq_j we have H_k(n)6≡0 (modq_j).

Secondly, the conditionHk(n)≡1 (modn) is equivalent to having







Hk(n)≡1 (mod q1) ...

Hk(n)≡1 (mod qJ)

So to show Hk(n)6≡1 (modn) it would suffice to show that for at least oneqj

we have Hk(n)6≡1 (modqj).

Recall the following result on congruences for unsigned Stirling numbers of the first kind, which we take verbatim from Scott [11].

Lemma 5. pis prime⇔for all k∈J2;p−1Kwe have p

k

≡0 (modp).

Proof. Work in the polynomial ring (Z/pZ)[x]: in that ringQp−1

i=0(x−i) =x^p−x, since both polynomials are monic of degreepand have roots 0, . . . , p−1.

Now just equate the coefficients to get the positive part of the result. For the negative result, suppose that nis not prime, let pbe a prime dividing n, and let m=ⁿ_p. Then

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6 THOMAS SAUVAGET

n−1

Y

i=0

(x−i) =

m−1

Y

k=0 p−1

Y

`=0

(x−kp−`) =

m−1

Y

k=0 p−1

Y

`=0

(x−`) =

p−1

Y

`=0

(x−`)

!^m

= (x^p−x)^m=x^m(x^p−1−1)^m

and it follows thats(n, m)≡(−1)^m6≡0 (modp). Since n

m

=|s(n, m)|the claim

follows.

The general expansion of theorem 2:

H_k(n) = H_n−1(n) +

n−k n−k−1

S_n−2(n) +· · ·+ n−k

1

S_k(n) becomes for k=qj :

Hq_j(n) = Hn−1(n) +

n−q_j n−q_j−1

Sn−2(n) +· · ·+ n−q_j

1

Sq_j(n) So if we assume thatH_n−1(n)≡0 (modn), which impliesH_n−1(n)≡0 (modqj), and combine this with our observation of section 1 that Hk(n) ≡ 1 (modn) for k≤ⁿ₃ (sincenis odd), so in particularHq_j(n)≡1 (modqj), we get:

1≡

n−qj

n−qj−1

S_n−2(n) +· · ·+ n−qj

1

Sq_j(n) (mod qj)

Remark 6. Numerically, we observe that for the largest prime factorqKwe always seem to have

n−q_K n−q_K−1

Sn−2(n) +· · ·+

n−q_K 1

Sq_K(n)≡0 (modqK). If one could prove this, a contradiction would follow, hence one would conclude that one cannot have Hn−1(n)≡0 (modn), i.e. Giuga’s conjecture is true. But the precise way in which that cancellation occurs modulo qK seems non-trivial, in particular most of the time several terms are non-zero. Note also that we cannot use lemma 5, at least not directly, since we are dealing with the composite n−q_K.

4. Appendix

In this appendix we present an earlier incorrect attempt of the author, stressing where the error lies. The author is very grateful for the input from multiple referees to finally help him understand the problem.

4.1. General properties.

Lemma 6. For each (n, k)∈N²2 with n≥k+ 1, we have : Hk(n)< H_k−1(n).

Proof. With thosenandk, since∀i∈J1;n−1Kwe have ^{(n+i−1−k)!}_i! i^k= _n+i−1−kⁱ ×

(n+i−k)!

i! i^k−1, we thus explicitely have Hk(n) = 1 +Pn−1 i=1

(n+i−1−k)!

i! i^k < 1 + Pn−1

i=1

(n+i−k)!

i! i^k−1=H_k−1(n)

Lemma 7. For each (n, k)∈N²2 with n≥k+ 1 we have : H_k(n)≡1 +

k

X

i=1

(n+i−1−k)!

i! i^k (mod n)

Proof. Immediate.

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4.2. Passage from H_k to H_k+1. The following result can be established.

Lemma 8. For each n∈N2 and any1≤k≤n−2 we have : Hk+1(n) =Hk(n)−(n−k−1)

n−1

X

i=1

(n+i−2−k)!

i! i^k Proof. We haveHk+1(n) = 1 +Pn−1

i=1

(n+i−(k+1)−1)!

i! i^k+1

= 1 +Pn−1 i=1

i n+i−k−1

(n+i−k−1)!

i! i^k, so adding and substracting Pn−1

i=1

n−k−1 n+i−k−1

(n+i−k−1)!

i! i^k to that, one obtains the result.

The following functions will be useful.

Definition 4.1. LetU_k:N27→Nbe defined by U_k(n) :=

k+1

X

i=1

(n+i−2−k)!

i! i^k ifk∈J1;n−2K

Remark 7. For any givenn∈N2 we have that for anyk∈J1;n−2Kthe quantity Uk(n)is an integer, since for anyi∈J1;k+ 1Kwe have i≤n+i−2−k.

Remark 8. Since ^{(n+i−2−k)!}_i! i^k ≡ 0 (modn) for i ∈ Jk+ 2;n−1K, we have H_k+1(n)−H_k(n)≡(k+ 1)U_k(n) (modn).

Lemma 9. For anyn ∈N2 we haveU1(n)≡0 (modn). In particular, H2 also characterizes P.

Proof. We have to show that for n≥2 we have ^(n−2)!_1! 1¹+^(n−1)!_2! 2¹ ≡0 (modn).

This is true since it simplifies to showing (n−2)! + (n−1)! ≡ 0 (modn), i.e.

showing that (n−2)![1 + (n−1)] ≡0 (modn), which is true. So for n ≥ 2 we have H₂(n)≡H₁(n) (modn), so lemma 1 implies that H₂ also characterizes the

primes.

4.3. A combinatorial relationship. Let us now introduce the following sums.

Definition 4.2. For eachk∈N^∗ let V_k:=

k

X

i=1

(−1)ⁱ⁻¹

k+ 1 i

i^k+ (−1)^k k

i

kⁱ

Theorem 3. For each k∈N^∗ we have Vk= (−1)^k+1.

This will be proved as a result of the following two lemmas. The author is very grateful to the referee of a previous version of this work for providing this proof.

This result might be of independent interest.

Lemma 10. For eachk∈N^∗ we havePk i=0

k i

kⁱ= (k+ 1)^k.

Proof. This follows immediately from the binomial theorem.

Lemma 11. For eachk∈N^∗ we havePk+1

i=0(−1)^{i k+1}_i i^k= 0.

Proof. Defining for any function f : Z→Z the forward difference operator ∆ by

∆f(j) :=f(j+ 1)−f(j), we obtain by induction onnthat then−fold iteration of

∆ on f is ∆ⁿf(j) =Pn

i=0(−1)^{i n}_i

f(j+n−i).

Forn=k+ 1 and the polynomialf(j) =j^k, sincef is of degreek < n we thus have that ∆ⁿf(j) is identically zero. That is, we havePk+1

i=0(−1)^{i k+1}_i

(j+k+ 1− i)^k = 0.

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8 THOMAS SAUVAGET

In particular, when evaluated atj=−(k+ 1) this givesPk+1

i=0(−1)^{i k+1}_i

(−i)^k = (−1)^kPk+1

i=0(−1)^{i k+1}_i

i^k= 0, hence the result.

We now prove theorem 3. Let A:=

k

X

i=1

k i

kⁱ

and

B :=

k

X

i=1

(−1)ⁱ k+ 1

i

i^k

Then, given the slightly different summation ranges than in the previous lemmas, we get A= (k+ 1)^k−1 andB= (−1)^k(k+ 1)^k. So since (−1)^kA+B= (−1)^k+1 we obtainVk= (−1)^k+1.

4.4. The error in the current attempt. Our aim was then to prove the following:

Conjecture 4. With the previous notations, we have for any n ∈ N2 and any 1≤k≤n−2 that Uk(n)≡0 (modn).

Unfortunately it is not true, as was explained to us by (we now realise) several referees. Our aim was to proceed as follows.

The i = 1 term of U_k(n) is (n−1−k)!. Let us say that (n−1−k)! ≡ a (mod n) for some a ∈ J0;n−1K. Then for any j ∈ J1;kK we have by induction (n−1−k+j)!≡a×(n−k)× · · · ×(n−1−k+j)≡(−1)^ja_(k−j)!^k! (modn). So ais in factor in each term ofU_k(n).

Ifa= 0 thenUk(n)≡0 (modn). Ifa6= 0, thenUk(n) can be factored bya, but that is as far as one can proceed. In particular, onecannotwrite (using the previous paragraph withi=j+1) that we haveUk(n) =a×Pk+1

i=1

(−1)ⁱ⁻¹_(k+1−i)!^k!

i! i^k =a×

Pk i=1

1 k+1

k+1 i

(−1)ⁱ⁻¹i^k+ (−1)^k^(k+1)_k+1^k

, due to the division byi! (non-commutativity of two otherwise true congruences – this is nothing else than the fact thatHk(n)6≡

0 nor 1 (modn) as seen in the numerical data).

Had that been the case, recalling that the term in parenthesis is an integer, and multiplying by the quantityk+ 1 (which is non-zero modulonsince 1≤k≤n−2) we would have obtained (k+1)Uk(n) =a×

Pk i=1

k+1 i

(−1)ⁱ⁻¹i^k+ (−1)^k(k+ 1)^k . Expanding (k+ 1)^k with the binomial theorem and simplifying, we notice that the resulting expression is 0 by theorem 3. So we would have necessarily hadUk(n)≡0 (mod n)...

References

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[2] E. Bedocchi,Nota ad una congettura sui numeri primi, Riv. Mat. Univ. Parma (4)11(1985), 229–236.

[3] D. Borwein, J.M. Borwein, P.B. Borwein, and R. Girgensohn,Giuga’s conjecture on primality, The American Mathematical Monthly103(1996), 40–50.

[4] D. Borwein, C. Maitland, and M. Skerritt, Computation of an improved lower bound to Giuga’s primality conjecture, Integers23(2013), A67.

[5] G. Giuga,Su una presumibile propriet`a caratteristica dei numeri primi, Ist. Lombardo Sci.

Lett. Rend. Cl. Sci. Mat. Nat. (3).14(1950), 511–528.

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[6] A. Granville, Arithmetic properties of binomial coefficients i: Binomial coefficients modulo prime powers, Organic mathematics : proceedings of the organic mathematics work- shop, december 12-14, 1995. simon fraser university, burnaby, british columbia, 2006.

http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/binomial.html.

[7] R. K. Guy,Unsolved problems in number theory (2nd edition), Springer, Berlin, 2004.

[8] B.C. Kellner,The equivalence of Giuga’s and Agoh’s conjectures, 2004. math/0409259.

[9] F. Luca, C. Pomerance, and I. Shparlinski,On Giuga numbers, Int. J. Mod. Math.4(2009), 13–18.

[10] A. Nicolas,Sums of powers below a prime, 2018. https://math.stackexchange.com/questions/433678/sums- of-powers-below-a-prime.

[11] Brian M. Scott, Congruence for stirling number of first kind s(n, k) when n is prime, 2018. https://math.stackexchange.com/questions/1241649/congruence-for-stirling- number-of-first-kind-sn-k-when-n-is-prime.

[12] Z.-W. Sun,Introduction to Bernoulli and Euler polynomials, 2002. Lecture given in Taiwan http://maths.nju.edu.cn/ zwsun/BerE.pdf.

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E-mail address: [email protected]