Download Elementary proof of the Giuga-Agoh conjecture

Elementary proof of the Giuga-Agoh conjecture
Thomas Sauvaget
To cite this version:
Thomas Sauvaget. Elementary proof of the Giuga-Agoh conjecture. 4 pages. 2011.
HAL Id: hal-00599178
https://hal.archives-ouvertes.fr/hal-00599178v1
Submitted on 8 Jun 2011 (v1), last revised 9 Jan 2011 (v2)
HAL is a multi-disciplinary open access
archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d’enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.
ELEMENTARY PROOF OF THE GIUGA-AGOH CONJECTURE
THOMAS SAUVAGET
Abstract. We prove the Giuga-Agoh conjecture byelementary means. Thus,
n is prime iff 1 + 1n−1 + 2n−1 + · · · + (n − 1)n−1 /n is a positive integer.
Equivalently, n is prime iff nBn−1 ≡ −1 (mod n), where the Bk are the
Bernoulli numbers. A few other consequences are discussed.
1. Introduction
The aim of this paper is to prove the two following characterisations of prime
numbers:
Theorem 1.1. n is prime iff
1+1n−1 +2n−1 +···+(n−1)n−1
n
is a positive integer.
and
Theorem 1.2. n is prime iff nBn−1 ≡ −1 (mod n), where the Bk are the Bernoulli
numbers (which can be defined as the coefficients appearing in the expansion et t−1 =
P∞
tk
k=0 Bk k! ).
Here is the background. Let p be a prime and a an integer such that 1 ≤ a ≤
p − 1. Then by Fermat’s little theorem we have ap−1 ≡ 1 (mod p). And thus
Pp−1 p−1
≡ p − 1 ≡ −1 (mod p), an observation of Sierpiński [6].
a=1 a
In 1950, Giuga [5] asked whether the converse holds, i.e. if any integer n
Pn−1 n−1
such that a=1
a
≡ −1 (mod n) is necessarily prime. He provided support
for that conjecture by showing that any counterexample, i.e. any composite integer satisfying the congruence, is greater than 101000 . This lower bound was
later improved by Bedocchi [3] to 101700 and more recently by Borwein, Borwein, Borwein & Girgensohn [4] to 1013887 . On the other hand, Tipu [9] proved
that for any real number x, the number P
of counterexamples to Giuga’s conjecture
n−1 n−1
G(x) := #{n < x : n is composite√and
≡ −1 (mod n)}, if any, is at
a=1 a
best of moderate growth: G(x) ≪ x log x.√This was recently improved by Luca,
Pomerance and Shparlinski [8] to G(x) = o( x).
Independently, Agoh [1] had arrived at a conjecture on Bernoulli numbers,
namely that p is prime iff pBp−1 ≡ −1 (mod p), and it has been shown by Kellner
[7] that it is equivalent to Giuga’s, hence the name Giuga-Agoh conjecture found
in the litterature.
Notations. N := {0, 1, 2, . . . } is the set of non-negative integers, while P :=
{2, 3, 5, 7, 11, . . . } is the set of primes and pk the k−th prime. Any integer n admits
Date: 8 June 2011.
2010 Mathematics Subject Classification. Primary 11A41.
1
2
THOMAS SAUVAGET
a unique prime factorisation n = q1 q2 · · · qk , where the prime factors will always,
without loss of generality, be assumed ordered: q1 ≤ q2 ≤ · · · ≤ qk . The integer n
is squarefree if its prime factors are all pairwise different (i.e. all the inequalities
are strict).
2. Proof of the conjecture
Define the sums σk (n) := 1k + 2k + · · · + nk . In particular, σ1 (n) =
. Recall the following result
σ2 = n(n+1)(2n+1)
6
n(n+1)
2
and
Theorem 2.1. (i) For odd k = 3, 5, . . . there exists a polynomial Fk , of degree
1
2 (k + 1) and with a double zero at the origin, such that σk (n) = Fk (σ1 (n)).
(ii) For even k = 2, 4, . . . there exists a polynomial Fk , of degree 12 (k − 2), such
that σk (n) = σ2 (n)Fk (σ1 (n)).
Proof. See Theorem 3.1 in Beardon’s paper [2].
Since in the case (i) the polynomials Fk have no constant term, we deduce the
following
Corollary 2.2. Pick an integer n ∈ N with n ≥ 5, and let i ∈ P\{2, 3} be an (odd)
prime such that i|n. Then i|σ1 (n) and i|σ2 (n), so in fact for all integer k > 1 we
have i|σk (n).
We now prove the following
Theorem 2.3. Let n ∈ N\{0, 1} be an odd integer. Assume there exists a positive
integer b such that 1 + 1n−1 + 2n−1 + · · · + (n − 1)n−1 = bn. Then n is prime.
Proof. We proceed by contradiction.
Case 1: if n is not a multiple of 3. Suppose our odd n is not prime, then there
exists an odd prime i ∈ P\{2, 3} such that i|n, i.e. there exists an integer w such
that n = iw. The assumption in the statement of the theorem can be rewritten
as: there exists b such that 1 + σn−1 (n − 1) = bn. So, assuming i|n implies that
i|(1 + σn−1 (n − 1)), i.e. that 1 + σn−1 (n − 1) ≡ 0 (mod i).
On the other hand, we have that σn−1 (n − 1) = σn−1 (n) − nn−1 . But, since
i|n, we have by corollary 2.2 that in particular σn−1 (n) ≡ 0 (mod i). So since
nn−1 = (iw)n−1 ≡ 0 (mod i), we obtain that 1 + σn−1 (n − 1) ≡ 1 (mod i), a
contradiction. So no such i exists, thus n is prime and this concludes the proof of
this case.
Case 2: if n is a multiple of 3. Then there exists an integer w such that n = 3w,
and by assumption we have that 3|(1 + σn−1 (n − 1), i.e. that 1 + σn−1 (n − 1) ≡ 0
(mod 3).
On the other hand, by Euclidean division we know that for each integer 1 < a ≤
n − 1 there exists two integers 0 < u < n and 0 ≤ v < 3 such that a = 3u + v.
in particular, a ≡ v (mod 3). Now, we can regroup the integers from 1 to n − 1
into (w − 1) groups of 3 consecutive integers, and a last group of two consecutive
integers: 1, . . . , 3, . . . , 3(w − 1), n − 2, n − 1. And by using this decomposition and
the previous remark we thus find that σn−1 (n − 1) ≡ w × (1 + 2) ≡ 0 (mod 3).
Hence 1 + σn−1 (n − 1) ≡ 1 (mod 3), a contradiction. So n cannot be a multiple
ELEMENTARY PROOF OF THE GIUGA-AGOH CONJECTURE
3
of 3 when satisfying the assumption in the statement of the theorem, and this case
never occurs.
The proof is complete.
3. Consequences
We now quickly summarize a few consequences of the validity of Giuga’s conjecture. Recall the following definitions and results.
A Carmichael number isa squarefree
positive integer n such that for any prime
divisor p one has (p − 1)| np − 1 . A Giuga number is a positive integer n such
that for any prime divisor p one has p| np − 1 . In particular, a Giuga number is
necessarily squarefree, or else there would exist a prime divisorp of n which also
divides the integer np , which contradicts the condition p| np − 1 .
Lemma 3.1. Any Carmichael number n is odd.
Proof. (After [4])
Firstly, 2 is not a Carmichael number, otherwise applying the definition one gets
the contradiction 1|0. Assume n > 2 is an even Carmichael number. It must then
have at least two prime factors, let p 6= 2 be one of them. But then by definition
one would have that the even number p − 1 divides the odd number np − 1, a
contradiction.
Theorem 3.2 (Giuga). Let n be a positive integer. Then
n−1
X
a=1
an−1 ≡ −1 (mod n) ⇔ ∀p ∈ P s.t. p|n : (p − 1)|
n
−1
p
and p|
n
−1 .
p
Proof. See the proof of Theorem 1 in [4].
So one can say that the validity of Giuga’s conjecture implies that there does
not exist any integer which is both a Carmichael number and a Giuga number. In
particular, any Giuga number is even.
Finally, recall
Lemma 3.3 (Giuga). An integer n is a Giuga number iff
positive integer.
P
1
p|n p
Proof. This is a particular case of the proof of Theorem 2 in [4].
So, we now know that no odd integer satisfies this condition.
−
1
p|n p
Q
is a
4
THOMAS SAUVAGET
References
1. Takashi Agoh, On Giuga’s conjecture Manuscripta Math. 87 (1995), 501-510.
2. A. F. Beardon, Sums of powers of integers Amer. Math. Montly 103 (1996) 201-213.
3. E. Bedocchi, Nota ad una congettura sui numeri primi Riv. Math. Univ. Parma(4) 11 (1985),
229-236.
4. David Borwein, Jonathan M. Borwein, Peter B. Borwein & Roland Girgensohn, Giuga’s conjecture on primality Amer. math. Monthly 103 (1996) 40-50.
5. Giuseppe Giuga, Su una presumibile proprietà caratteristica dei numeri primi Ist. Lombardo
Sci. Lett. Rend. Cl. Sci. Mat. Nat.(3) 14:83 (1950) 511-528.
6. Richard K. Guy, Unsolved problems in number theory (2nd edition), Springer, 2004.
7. Bernd C. Kellner, The equivalence of Giuga’s and Agoh’s conjectures arXiv:math.NT/0409259
8. Florian Luca, Carl Pomerance & Igor Shparlinki, On Giuga numbers Int. J. Mod. Math. 4
(2009) 13-18.
9. Vicentiu Tipu, A note on Giuga’s conjecture Canad. Math. Bull. 50 (2007) 158-160.
E-mail address: [email protected]