PERFECT POLYNOMIALS OVER F p WITH p + 1 IRREDUCIBLE DIVISORS. 1. Introduction. Let p be a prime number. For a monic polynomial A F p [x] let d

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1 PERFECT POLYNOMIALS OVER F p WITH p + 1 IRREDUCIBLE DIVISORS L. H. GALLARDO and O. RAHAVANDRAINY Abstract. We consider, for a fixed prime number p, monic polynomials in one variable over the finite field F p which are equal to the sum of their monic divisors. We give necessary conditions for the existence of such polynomials, called perfect polynomials, having p + 1 irreducible factors. These conditions allow us to describe the set of all perfect polynomials with p + 1 irreducible divisors in the first unknown case, namely, the case p = Introduction Let p be a prime number. For a monic polynomial A F p [x] let σ(a) = d d A, d monic be the sum of all monic divisors of A (1 and A included). The restriction to monic polynomials is necessary since the sum of all divisors of A that have a given degree is zero. Observe that A and σ(a) have the same degree. Let us call ω(a) the number of distinct monic irreducible polynomials that divide A. The function σ is multiplicative on co-prime polynomials while the function ω is additive (on co-prime polynomials). This fact is used many times without more reference in the rest of the paper. Received March 2, Mathematics Subject Classification. Primary 11T55, 11T06. Key words and phrases. Sum of divisors; polynomials; finite fields; characteristic p.

2 A perfect polynomial is a monic polynomial A such that σ(a) = A. This notion is a good function field analogue of the notion of a multiperfect natural number n that satisfies that n divides σ(n). For example, 120 is a multiperfect number since 120 divides 360 = σ(120). Indeed, since deg(a) = deg(σ(a)), if a monic polynomial A F p [x] divides σ(a), then both are forced to be equal. We say that a polynomial A is odd (resp. even) if it has no root in F p (that is: gcd(a, x p x) = 1) (resp. it is not odd). This definition is natural in the understanding that a polynomial P F p [x], with absolute value P := p deg(p ) is even if and only if it has a divisor d with absolute value d = p. Throughout the paper, we assume that a polynomial means a monic polynomial and that the notion of polynomial irreducibility is defined over F p. Important results about perfect polynomials appear in the work of Canaday [1] and Beard et al. ([2], [3]). Indeed Canaday introduced the subject, working in the case p = 2 in his thesis under Carlitz while Beard et al. extended these results to F p with odd p in the special case where the polynomials considered split completely over F p. Trivially, there is no odd perfect polynomial over F 2 with ω(a) = 1. Canaday [1, Theorem 17] proved the inexistence of odd perfect polynomials over F 2 with two irreducible factors, i.e., with ω(a) = 2. We obtained recently some results about even or splitting perfect polynomials that generalize the work of Canaday and Beard et al., (see [9] and the references therein). Nevertheless, providing complete lists of perfect polynomials satisfying some properties (even polynomials, odd polynomials, splitting polynomials) remains difficult because it is difficult to know precisely the manner in which a given polynomial factorizes over F p, (like the difficulty of factorization of special type of integers prevents to know more about the multiperfect numbers).

3 Observe that for any given positive integer w, there is an infinity of polynomials A F p [x] with ω(a) = w, so potentially an infinity of perfect polynomials with ω(a) = w may exist. The following restriction is important. A perfect polynomial over F p must have a multiple of p number of minimal irreducible divisors (see Lemma 2.2), so trivially there is no perfect polynomial over F p with less than p irreducible factors. We proved in [6], [8] (resp. [7]) the inexistence of odd perfect polynomials over F 2 with ω(a) {3, 4} (resp. over F 3 with ω(a) = 3). In particular, this settles the case p = 2 of the present paper. We should take then p as an odd prime in all the paper. We proved also [10] some general results about odd perfect polynomials over F p with p irreducible factors, leaving unknown the list of such polynomials. However, we got the following explicit result (see [10, Theorem 1.2]): The unique odd perfect polynomial over F p, with p irreducible factors of degree 2 for which all exponents do not exceed two is A(x) := ( (x + a) 2 3/8 ) 2, a F p where either (p 11 mod 24) or (p 17 mod 24). It is natural to consider the following case. What can we say about perfect polynomials with p + 1 irreducible factors? Is it possible to provide the complete list L(p) of such polynomials? In particular, is this list finite? We know only L(2) (see [1, Theorem 9 ] and [6, Theorem 3.1]) that consists of the four even polynomials in F 2 [x] S 1 (x) = x(x + 1) 2 (x 2 + x + 1), S 2 (x) = S 1 (x + 1), S 3 (x) = x 3 (x + 1) 4 (x 4 + x 3 + 1), S 4 (x) = S 3 (x + 1). From some computations reported in [2], the list L(3) contains the following three perfect polynomials of degree 8 in F 3 [x] which are also even: A 1 (x) := x 3 (x + 1) 2 (x + 2)(x 2 + 1), A 2 (x) := A 1 (x + 1), A 3 (x) := A 1 (x + 2).

4 In this paper in Theorem 1.1, we first establish some necessary conditions for the non-vacuity of the list L(p), for a fixed odd prime number. Secondly, we prove Theorem 1.2 by means of Theorem 1.1 that L(3) does not contain anything else. Theorem 1.1. Let p be an odd prime number. Let A = P a P p ap Q b be a perfect polynomial over F p with p + 1 irreducible factors. Then d := deg(p 1 ) = = deg(p p ) and i) (A is even) or (a j is even for at least one j {1,..., p}), ii) for at least one j {1,..., p}, a j is of the form N j p nj 1 with N j, n j N, N j 1, p N j and N j (p 1), iii) either (p b + 1) or (b {p 1, 2p 1} and d deg(q)). Theorem 1.2. The only perfect polynomials over F 3 with four irreducible factors are: x 3 (x + 1) 2 (x + 2)(x 2 + 1), x(x + 1) 3 (x + 2) 2 (x 2 + 2x + 2), and x 2 (x + 1)(x + 2) 3 (x 2 + x + 2). By contrast with the integer perfect numbers, observe that Sylvester [12] already proved in 1888 that every odd perfect number has at least five prime factors. Later, Dickson [4] proved that there is a finite number of odd perfect numbers with given number ω of prime divisors. For polynomials over finite fields, we have not yet an analogue of these important results. 2. Some useful facts We denote, the set of nonnegative integers (resp. of positive integers) as usual by N (resp. by N ). For a set Λ, we denote the cardinal of Λ by #Λ. For polynomials A, B F p [x], we write: A n B if A n B but A n+1 B.

5 Definition 2.1. We say that a polynomial P is a minimal irreducible divisor of A if P is an irreducible divisor of A such that deg(p ) deg(r) for any irreducible divisor R of A. A basic but important result is the following. Lemma 2.2. (see [5, Lemma 2.5]) Let p be a prime number. Let A F p [x] be a perfect polynomial. Then the number of minimal irreducible divisors of A is a multiple of p. We immediately get the corollary. Corollary 2.3. Any perfect polynomial A over F p, with exactly p + 1 irreducible factors may be written as A = P a1 1 P p ap Q b, where a j, b N and deg(p 1 ) = = deg(p p ) < deg(q). Notation 2.4. In the rest of the paper, we fix an odd prime number p. According to Corollary 2.3 for a perfect polynomial A F p [x] with ω(a) = p + 1, we put A = P a1 1 P ap p Q b, where,..., a p, b N and deg(p 1 ) = = deg(p p ) < deg(q), a i = N i p ni 1, b = Mp m 1, N i, n i, M, m N, N i, M 1, p N i, p M. Note that P 1,..., P p may be even whereas Q is always odd. For S {Q, P 1,..., P p } and for s {b,,..., a p }, we would like to understand how σ(s s ) = 1 + S + + S s may be factorized into irreducible divisors of A σ(s s ) = P c P cp p Q c, where c, c l 0 for any l {1,..., p}. We may write s := Np n 1 for some N, n N such that N 1 and p N. In that case, we put d := gcd(n, p 1) and we denote by L N the splitting field of x N 1 over F p which is a strict subset of the algebraic closure F p of F p.

6 Moreover, since p N, the polynomial x N 1 has no multiple root (in L N ). The set U N := {λ L N : λ N = 1} of N-th roots of unity in L N is a cyclic group of order N (see [11, Theorem 2.42]). Consider the Frobenius map φ p (t) = t p for t L N, acting over L N. The action is extended trivially to L N [x] by sending x to x. The Galois group G of the extension L N over F p is generated by φ p. The Galois group G e of the extension ring L N [x] over F p [x] is isomorphic to G and acts as G on the coefficients of any element A L N [x]. We recall that a polynomial P L N [x] lies in F p [x] if and only if φ p (P ) = P. In Sections 3 and 4 we will use the following facts very often. Lemma 2.5. (see [10, Lemma 2.4]) Let S F p [x] be an irreducible polynomial such that S µ is irreducible for any µ F p. Then deg(s) = 1, so that S is even. Lemma 2.6. Let S F p [x] be an irreducible polynomial and N N with p N. If σ(s N 1 ) = Q c1 1 Qct t, where Q l is irreducible, gcd(s, Q l ) = 1 and deg(s) deg(q l ) for any l, then c l {0, 1} for any l. Proof. If c l 2 for some l, then put 1 + S + + S N 1 = R m C, where m = c l and R = Q l. We get (1) S N 1 = (S 1)R m C. Since p N by taking derivatives on both sides of (1) one has 0 NS N 1 S = R m 1 (S RC + (S 1)(mR C + RC )),

7 so with the observation gcd(s, R) = 1, R m 1 S. Thus, we get the contradiction deg(s) (m 1) deg(s) = (m 1) deg(r) deg(s ) < deg(s). 3. The proof of Theorem 1.1 We recall (see Notation 2.4) that we are interested in perfect polynomials of the form A = P a1 1 P ap p Q b, where,..., a p, b N and deg(p 1 ) = = deg(p p ) < deg(q), a i = N i p ni 1, b = Mp m 1, N i, n i, M, m N, N i, M 1, p N i, p M. We give more precisions about Q and its exponent b below Necessary conditions on Q and on b In this section we consider the following subsets of {1,..., p} Λ := {i : n i = 0}, Σ 1 := {i : Q σ(p ai i )}, Σ 2 := {i : Q σ(p ai i )}. We see that Σ 1, Σ 1 Σ 2 = and Σ 1 Σ 2 = {1,..., p}. Lemma 3.1. If i Λ Σ 1, then a i p 2. Proof. One has where α ji {0, 1} by Lemma 2.6. a i = N i 1, p N i, σ(p i a i ) = j i P j α ji,.

8 Thus a i = j i α ji p 1 and a i = N i 1 p 1 since p N i. (2) Lemma 3.2. i) If i Σ 1, then Q pn i σ(p i a i ). ii) One has b = Mp m 1 = Proof. i) One has i Σ 1 p ni = #(Λ Σ 1 ) + i Σ 1 Λ a σ(p i i ) = (P i 1) pn i 1 N (σ(p i 1 i )) pn i, N where σ(p i 1 N i ) is square free by Lemma 2.6. Hence, Q σ(p i 1 i ). ii) The exponent of Q in A is b. The exponent of Q in σ(a) is that of Q in a i Σ 1 σ(p i i ). We get (2) from i). p ni. Corollary 3.3. Let A = P a1 1 P ap p Q b F p [x] be perfect with b = Mp m 1, p M. If m 1, then #(Λ Σ 1 ) = p 1 and deg(p ) deg(q). More precisely, we must have: i) M = m = 1 if (#Λ = p, #Σ 1 = p 1) or (#Λ = #Σ 1 = p 1). ii) M = p n k 1 + 1, m = 1, Q(α) { 1, 1} for any α F p if #Λ = p 1 and #Σ 1 = p, where k is the unique integer not lying on Λ (n k 1). Proof. We see that We apply Relations (2) in Lemma 3.2. #(Λ Σ 1 ) #Λ p.

9 If m 1, then #(Λ Σ 1 ) 1 mod p. So, #(Λ Σ 1 ) = p 1. We get four cases: If #Λ = p = #Σ 1, then #Λ = #Σ 1 = {1,..., p} and p 1 = #(Λ Σ 1 ) = p, which is impossible. If #Λ = p and #Σ 1 = p 1, then b = Mp m 1 = p 1. So, M = m = 1. If #Λ = #Σ 1 = p 1, then b = Mp m 1 = p 1. So, M = m = 1. If #Λ = p 1 and #Σ 1 = p, let k be the unique integer such that k Λ. We get b = Mp m 1 = p 1 + p n k = p(p n k 1 + 1) 1, where p (p n k 1 + 1). It follows that σ(q b ) = Qb+1 1 = (Q 1) p 1 (1 + Q + + Q pn k 1 ) p. Q 1 We see that Q 1 divides σ(a) = A which is odd, so for any α F p, Q(α) 1. Remark also that for any v 1 and for any α F p, one has ( Q p v +1 ) 1 +1 (1 + + Q pv )(α) = (α) = (Q(α))pv 1 Q 1 Q(α) 1 Thus = (Q(α))2 1 Q(α) 1 = Q(α) + 1. (σ(q b ))(α) = 0 whenever Q(α) = 1. It is impossible since σ(q b ) divides σ(a) = A and A is odd. Corollary 3.4. Let A = P a1 1 P ap p Q b F p [x] be perfect with b = Mp m 1, p M. If m = 0, then

10 i) b = M 1 = p ni = #(Λ Σ 1 ) + p ni, i Σ 1 i Σ 1 Λ ii) we have either (Λ = Σ 1 = {1,..., p}) or #(Λ Σ 1 ) p 2. Proof. Relations (2) in Lemma 3.2 give i) and imply that p divides M if #(Λ Σ 1 ) = p 1. It is impossible. Lemma 3.5. One has σ(x p ) = (1 + x) 2 (x 2 u i x + 1), p 1 i=1 where u i = ξ i + 1 ξ with ξ a primitive (p + 1)-root of unity, u i i { 2, 2}, and x 2 u i x + 1 is irreducible over F p. Proof. Since the group of roots (in an algebraic closure of F p ) of x p+1 1 is a cyclic group of order p + 1 (see [11, Theorem 2.42]) with a generator ξ, such roots are So 1, 1, ξ, 1 ξ = ξp, ξ 2, σ(x p ) = (1 + x) 1 ξ 2 = ξp 1,..., ξ p 1 1 2, ξ p p 1 i=1 (x ξ i ) (x 1ξ ) i. = ξ p+3 2. For any i { 1,..., p 1 } 2, x 2 u i x + 1 = (x ξ i ) ( ) x 1 ξ lies in i Fp [x] since u i = ξ i + 1 ξ is invariant i by the Frobenius morphism φ p : x x p. Note that (ξ i ) 2 1 for any i { 1,..., p 1 } ( ) 2 and ui { 2, 2}. Remark also that φ p ξ i 1 ξ = i ξ i + 1 ξ ξ i 1 i ξ so that ξ i 1 i ξ F i p.

11 Each polynomial x 2 u i x+1 is then irreducible over F p, because its discriminant δ i := ( ξ i + 1 ξ i ) 2 4 = ( ξ i 1 ξ i ) 2 is not a square in Fp. Lemma 3.6. Let v 2 and let ξ be a primitive p v + 1-root of unity. Then, for any i, j {1,..., p 1} and for any k, r {0,..., 2v 1} whenever (i, k) (j, r). ξ ipk ξ jpr Proof. If ξ ipk = ξ jpr for some (i, k) (j, r), then ip k jp r 0 mod (p v + 1). We may suppose that k r, so that ip k r j 0 mod (p v + 1). If k = r, then i j 0 mod (p v + 1) and we must have i j = 0 since v 2. So, k > r. Put ip k r j = c (p v + 1). We easily see that p c and we may write c in base p expansion c = e 0 p z + e 1 p z e z 1 p + e z, where e l {0,..., p 1} and z 0. Therefore, z (3) ip k r = p v e l p z l + e 0 p z + e 1 p z e z 1 p + e z + j. Hence l=0 z + v = k r, e 0 = i 0. If z = 0, then v = k r and ip v = i (p v + 1) + j, which is impossible. If z 1, then ip z occurs in the right hand side of Relation (3), but not in the left. It is also impossible.

12 Lemma 3.7. For any v 2, σ(x pv ) is divisible at least by p 1 polynomials (irreducible or not) R 1,..., R p 1 of degree 2v such that gcd(r i (S), R j (S)) = 1 if i j for any S F p [x]. Proof. As above, let ξ be a primitive p v + 1-root of unity. Put N := p v + 1 and R 1 = (x ξ)(x ξ p ) (x ξ p2v 1 ), R 2 = (x ξ 2 )(x ξ 2p ) (x ξ 2p2v 1 ),. R l = (x ξ l )(x ξ lp ) (x ξ lp2v 1 ), For S F p [x] and for l {1,..., p 1}, we get. R l (S) = (S ξ l )(S ξ lp )... (S ξ lp2v 1 ). For any l, let S 1,l,..., S 2v,l be the elementary symmetric polynomials in (ξ l, ξ lp,... ξ lp2v 1 ). For any l and for any 1 k 2v, we get so that φ p (S k,l ) = (S k,l ) p = S k,l, φ p (R l ) = R l and R l F p [x]. We see that deg(r l ) = 2v and if gcd(r i (S), R j (S)) = 1 in L N [x], then gcd(r i (S), R j (S)) = 1 in F p [x]. Let us prove that gcd(r i (S), R j (S)) = 1 in L N [x]. If not, let W L N [x] be an irreducible (over L N ) common divisor of R i (S) and R j (S). We must have modulo W ξ ipk S ξ jpr

13 for some k {0,..., 2i 1} and for some r {0,..., 2j 1}. Thus ξ ipk = ξ jpr, which contradicts Lemma 3.6. Lemma 3.8. If A = P 1 a1 P p ap Q p is perfect (with deg(q) > deg(p j )) and if Q 2 uq + 1 divides σ(q p ) for some u F p { 2, 2}, then (4) Proof. If ω(q 2 uq + 1) = 1, then ω(q 2 uq + 1) 2. Q 2 uq + 1 = P w for some P {P 1,..., P p }, because Q (Q 2 uq + 1) and (Q 2 uq + 1) σ(q p ) σ(a) = A. Since deg(q) > deg(p ), we see that w = 2 deg(q) deg(p ) 3. By taking derivatives in both sides of (4), one has Q (2Q u) = wp w 1 P. If P divides 2Q u, then 2Q u mod P and u = u2 4 u Q2 uq mod P. Thus u 2 = 4 mod P and u { 2, 2}, which is impossible. So, P (2Q u), P w 1 Q, Q = P w 1 R for some R F p [x] and R (2Q u) = wp, which is impossible by considering degrees.

14 Remark 3.9. The conclusion in Lemma 3.8 does not hold in a more general context. For example, take p = 3, Q = x which is odd and irreducible over F 3, ξ a primitive 4-root of 1. One has ξ { 2, 2}, u := ξ + 1 ξ = 0 and Q 2 uq + 1 = Q = x 4 + 2x which is irreducible over F 3 with ω(q 2 uq + 1) = 1. Corollary Let A = P 1... P p ap Q p F p [x] be perfect (with deg(q) > deg(p j )). Then i) The polynomial σ(q p ) has at least p + 1 irreducible factors. ii) More generally, σ(q pv ) has at least p + 1 irreducible factors for any v 1. Proof. i) We get σ(q p ) = (Q + 1) 2 (Q 2 u i Q + 1). Remark that for any i, u i { 2, 2}, gcd(q + 1, Q 2 u i Q + 1) = 1, for any i, j, u i u j and gcd(q 2 u i Q + 1, Q 2 u j Q + 1) = 1, for any i, ω(q 2 u i Q + 1) 2 by Lemma 3.8 and ω(q + 1) 2. It follows that ω(σ(q p )) 2 p = p ii) Each polynomial R l (Q) divides σ(q pv ) and gcd(r i (Q), R j (Q)) = 1 in F p [x] if i j. Moreover, ω(r l (Q)) 2 for any l. So p 1 i=1 ω(σ(q pv )) 2(p 1) p + 1. From Corollaries 3.3, 3.4 and 3.10, we get the following corollary

15 Corollary Let A = P 1... P p ap Q p F p [x] be perfect (with deg(q) > deg(p j )) and b = Mp m 1, p M. i) If m = 0, then p b + 1 and #(Λ Σ 1 ) p 2. ii) If m 1, then m = 1, M {1, 2}, so b {p 1, 2p 1}. Proof. i) If m = 0 and if Λ = Σ 1 = {1,..., p}, then b = p. We get p ω(σ(q b )) p + 1, which is impossible. ii) If m 1, then m = 1. If M = p nk with n k 2, then σ(q M 1 ) = σ(q pn k 1 ) has at least p + 1 irreducible factors. It is impossible as in the proof of Corollary The proof By using Notation 2.4, Propositions 3.12 and 3.13 give the first and second part of our theorem. Corollary 3.11 gives the third part. Proposition There are no odd perfect polynomials over F p of the form P a P p ap Q b with p + 1 irreducible divisors where a i is odd for any i {1,..., p}. Proof. Since is odd, P divides σ(p a1 1 ). P cannot be composite since any of its irreducible factors should have degree < d. So, P is an irreducible divisor of A. By applying the same argument to P 1 + 1, we see that P is also an irreducible divisor of A, and so on. Thus, {P 1,..., P p } = {P 1, P 1 +1, P 1 +2,..., P 1 +(p 1)} and hence P µ is irreducible for any µ F p. This contradicts Lemma 2.5. Proposition There exist no perfect polynomials over F p of the form P a1 1 P p ap Q b with p + 1 irreducible divisors where for any i {1,..., p}, a i = N i p ni 1, p N i, N i p 1.

16 Proof. Since N i p 1, we may write If A is perfect, then Therefore, we may put A = σ(a) = i A = i σ(p ai i ) = µ Ω Ni (P i µ) cµ. σ(p ai i ) σ(q b ) = i A i σ(q b ) = i µ Ω Ni (P i µ) cµ σ(q b ). ξ F p (P i ξ) bξ σ(q b ), where b ξ N (may be equal to 0) and P i P j F p if i j. Since Q σ(q b ), we see that Q does not divide A, which is impossible. 4. The proof of Theorem 1.2 In this section, we take p = 3, so (see Notation 2.4) A = P a1 1 P a2 2 P a3 3 Q b, where, a 2, a 3, b N, deg(p 1 ) = deg(p 2 ) = deg(p 3 ) < deg(q), a i = N i 3 ni 1, b = M 3 m 1, N i, n i, M, m N, N i, M 1, 3 N i, 3 M. As in Section 3.1, we put Λ = {i {1, 2, 3} : n i = 0} and Σ 1 = {i {1, 2, 3} : Q σ(p ai i )}.

17 The following results are useful. Lemma 4.1. Let A = P a1 1 P a2 2 P a3 3 Q b F 3 [x] be perfect. If j Λ, then a j = N j 1 2. If j Λ \ Σ 1, then a j = N j 1 = 1. Proof. We suppose that j = 2 without loss of generality. From Lemma 2.6, we get σ(p a2 2 ) = P β1 1 P β3 3, a 2 = β 1 + β 3 2. Since 3 N 2, we must have a 2 = 1. Lemma 4.2. Let p be an odd prime number such that p 3 mod 4 and let v be a positive integer. Then the polynomial 1 + (x 2 ) (x 2 ) v is irreducible over F p if and only if v = 1. Proof. The sufficiency is obvious since 1 + x 2 is irreducible over F p whenever p 3 mod 4. Now, we prove the necessity. One has If v 2 and if v is odd, then S(x) := 1 + (x 2 ) (x 2 ) v = (x2 ) v+1 1 x 2 1 S(x) = xv + + x + 1 x + 1 = (xv + + x + 1) (x v+1 + 1). x + 1 (x v+1 + 1) = (1 + (x 2 ) (x 2 ) v 1 2 ) (x v+1 + 1),

18 which is reducible. If v 2 and if v is even, then which is also reducible. S(x) = xv (x v + + x + 1) x + 1 = (x v x v 1 + x + 1) (x v + + x + 1), Now, we are ready to prove Theorem 1.2. According to Corollaries 3.3 and 3.4, since Q(α) { 1, 1} for any α F 3, it remains to consider the following cases: ( ) m = 0, #Λ Σ 1 {0, 1}, ( ) M = m = 1, b = 2, Λ = {1, 2, 3}, Σ 1 = {1, 2}, ( ) M = m = 1, b = 2, Λ = {1, 2} = Σ 1. We shall see that only Case ( ) may happen and A must be even. We retrieve (Section 4.2.1, Case n 2 1, n 3 = 0), then the three polynomials described in Theorem Case A odd In this section, we may write A = P a1 1 P a2 2 P a3 3 Q b with a i = N i 3 ni 1, b = M 3 m 1, deg(q) > deg(p i ) 2. Lemma 4.3. If A = P a1 1 P a2 2 P a3 3 Q b F 3 [x] is odd and perfect, then there exists at most one i {1, 2, 3} such that n i = 1 (i.e. #Λ {2, 3}). Proof. Suppose to the contrary that there exist two distinct integers j 1, j 2 {1, 2, 3} such that n j1, n j2 1. Put (without loss of generality) j 1 = 1 and j 2 = 2. We see that P 1 1 and P 2 1 divide σ(a) = A. Thus P 1 1, P 2 1 {P 1, P 2, P 3 }.

19 If P 1 1 = P 2, then P 2 1 = P 3, so P 1, P 1 1 and P 1 2 = P 3 are both irreducible, which contradicts Lemma 2.5. If P 1 1 = P 3, then P 2 1 = P 1, so P 1, P 1 1 and P 1 2 = P 2 are both irreducible, which is also impossible Case ( ) m = 0, #Λ Σ 1 1. If Λ Σ 1 = since Σ 1, from Lemma 4.3, we get #Λ = 2. We may put #Λ = {1, 2}. So 1, 2 Σ 1 and thus Q σ(p a1 a2 1 ) and Q σ(p2 ). Therefore, by Lemma 4.1, = a 2 = 1. Hence, σ(p 1 ) = 1 + P 1 {P 2, P 3 } and σ(p 2 ) = 1 + P 2 {P 1, P 3 }, which contradicts Lemma 2.5. Now, we suppose that #Λ Σ 1 = 1. We may put Λ Σ 1 = {1} so that n 1 = 0 and Q σ(p a1 1 ), (n 2 1 or Q σ(p a2 2 )) and (n 3 1 or Q σ(p a3 3 )). Lemma 4.4. We must have either (n 2 1) or (n 3 1). Proof. First, if n 2 1 and n 3 1, then Λ = {1}, which contradicts Lemma 4.3. If n 2 = 0 = n 3, then Q σ(p a2 2 ) and Q σ(p a3 3 ). So, by Lemma 4.1, we must have a 2 = a 3 = 1. Thus, σ(p 2 ) = 1 + P 2 {P 1, P 3 } and σ(p 3 ) = 1 + P 3 {P 1, P 2 }, which contradicts Lemma 2.5 as above. According to Lemma 4.4, we may suppose that n 3 1, n 2 = 0 and Q σ(p a2 2 ). Thus, a 2 = 1 by Lemma 4.1. Therefore, P 3 1 and σ(p 2 ) divides σ(a) = A, σ(p 2 ) = 1 + P 2 {P 1, P 3 } and P 3 1 {P 1, P 2 }. If 1 + P 2 = P 3, then P 3 1 = P 2 and 1 = a 2 3 n3 1. It is impossible. If 1 + P 2 = P 1, then P 3 1 = P 1. Again, this contradicts Lemma Case ( ) M = m = 1, b = 2, Λ = {1, 2, 3}, Σ 1 = {1, 2}.

20 By Lemma 4.1, a 3 = 1. So, σ(p 3 ) = 1 + P 3 {P 1, P 2 }. We may suppose that 1 + P 3 = P 1 so P 3 (ξ) = 1 = P 1 (ξ) for any ξ F 3. Since Q σ(p ai i ) for i {1, 2} and since σ(a) = A, we get where σ(p 1 ) = P α2 2 P α3 3 Q, σ(p 2 a2 ) = P β1 1 P β3 3 Q, σ(p 3 a 3 ) = σ(p 3 ) = P 1, σ(q b ) = σ(q 2 ) = (Q 1) 2 = (P w1 1 P w2 2 P w3 3 )2, β w 1 =, α 2 + 2w 2 = a 2, α 3 + β 3 + 2w 3 = a 3 = 1, α 2, α 3, β 1, β 3 {0, 1}, (α 2 + α 3 ) = deg(q) deg(p ) = a 2 (β 1 + β 3 ) = w 1 + w 2 + w 3. It follows that w 3 = 0 and (α 3, β 3 ) {(1, 0), (0, 1)}. Moreover, since (σ(p a1 1 ))(ξ) 0 and P 1(ξ) = 1, must be even and (σ(p a1 1 ))(ξ) = 1, since Q 1 = P w1 1 P w2 2, we must have Q(ξ) = 1 for any ξ F 3. Since β w 1 = is even, we get β 1 = 1 and = 2 + 2w 1. If α 3 = 1, β 3 = 0, then a 2 = 1 + deg(q) deg(p ) = 1 + w 1 + w 2. Since 1 = (σ(p a1 1 ))(ξ) = (P 2(ξ)) α2 1 ( 1) = (P 2 (ξ)) α2, we must have α 2 = 1, P 2 (ξ) = 1 and a a 2 = 1 + 2w 2 is odd. In this case, (σ(p 2 2 ))(ξ) = 0 for any ξ. It is impossible, because A is odd. If α 3 = 0, β 3 = 1, then a 2 = deg(q) deg(p ) = 2 + w 1 + w 2. As above, since 1 = (σ(p a1 1 ))(ξ) = (P 2(ξ)) α2 ( 1) = (P 2 (ξ)) α2, we must have α 2 = 1, P 2 (ξ) = 1 a and a 2 = 1 + 2w 2 is odd. In this case, (σ(p 2 2 ))(ξ) = 0 for any ξ. It is impossible, because A is odd.

21 Case ( ) M = m = 1, b = 2, Λ = {1, 2} = Σ 1. In this case, P 3 1 divides σ(a) = A, so P 3 1 {P 1, P 2 }. We may suppose that P 3 1 = P 1. We get where σ(p 1 ) = P 2 α 2 P 3 α 3 Q, σ(p 2 a 2 ) = P 1 β 1 P 3 β 3 Q, a σ(p ) = P n 3 1 γ 1 P 2 3 n 3 2, σ(q b ) = σ(q 2 ) = (Q 1) 2 w = (P 1 w 1 P 2 w 2 P 3 3 ) 2 β n w 1 =, α 2 + γ 2 3 n3 + 2w 2 = a 2, α 3 + β 3 + 2w 3 = a 3, α 2, α 3, β 1, β 3, γ 2 {0, 1}, (α 2 + α 3 ) = deg(q) deg(p ) = a 2 (β 1 + β 3 ) = w 1 + w 2 + w 3, N 3 3 n3 1 = a 3 = 3 n3 1 + γ 2 3 n3, so N 3 = γ {1, 2}. Remark that for any ξ F 3, P 3 (ξ) = 1 = P 1 (ξ) since P 3 1 = P 1. a Since (σ(p 3 3 ))(ξ) 0 and P 3 (ξ) = 1, a 3 must be even and (σ(p a3 3 ))(ξ) = 1. w Since Q 1 = P 1 w 1 P 2 w 2 P 3 3, we must have Q(ξ) = 1 for any ξ F 3. It follows that N 3 = 1, γ 2 = 0 and α 3 + β 3 = a 3 2w 3 = 3 n3 1 2w 3 {0, 2}. Hence, either (α 3 = β 3 = 0) or (α 3 = β 3 = 1). Case α 3 = β 3 = 0 One has the following. Lemma 4.5. For any ξ F 3, P 2 (ξ) = 1. Proof. If P 2 (ξ) = 1 for some ξ F 3, then (σ(p a2 2 ))(ξ) = 1, which contradicts the fact (σ(p a2 2 ))(ξ) = (P 1 β1 Q)(ξ) = 1 β1 ( 1) = 1.

22 Lemma 4.6. Let P F 3 [x] be irreducible and a N, then P + 1 divides σ(p a ) if and only if a is odd. Proof. If a is odd, then we may write a = 2s + 1 and σ(p a ) = 1 + P + + P 2s + P 2s+1 = (1 + P )(1 + (P 2 ) (P 2 ) s ). If a is even, then a 1 is odd and P + 1 divides σ(p a 1 ). Hence σ(p a ) = σ(p a 1 ) + P a is not divisible by P + 1. Corollary 4.7. The integers and a 2 must be even, so that β 1 = α 2 = 0 and = a 2. Proof. a If is odd, then by Lemma 4.6, P divides σ(p 1 1 ), so α 2 = 1 and P 2 = P = P 3, which is impossible. Thus, = β 1 + 2w 3 + 2w 1 is even and β 1 = 0. a If a 2 is odd, then as above, P divides σ(p 2 2 ), so β 1 = 1 and P 1 = P Hence P 1, P 1 1 = P 2, P 1 2 = P 3 are both irreducible. It contradicts Lemma 2.5. So, a 2 = α 2 + 2w 2 is even and α 2 = 0. a From Corollary 4.7, β 1 = α 2, so σ(p 1 a 1 ) = Q = σ(p 2 2 ). Hence w P 1 w 1 P 2 w 2 P 3 a 3 =Q 1 = P 1 (1+P P 1 1 a 1 )=P 2 (1+P P ). Thus, w 1 = w 2 = 1 and 2 = 2w 2 = a 2 = = 3 n w 1 = 3 n3 + 1, which is impossible, because n 3 1.

23 Case α 3 = β 3 = 1 We get where σ(p 1 ) = P 2 α 2 P 3 Q, σ(p 2 a 2 ) = P 1 β 1 P 3 Q, σ(p 3 a 3 ) = P 1 3 n 3 1, σ(q b ) = σ(q 2 ) = (Q 1) 2 = (P 1 w 1 P 2 w 2 P 3 w 3 ) 2 β n w 1 =, α 2 + 2w 2 = a 2, 2 + 2w 3 = a 3 = 3 n3 1, α 2, β 1 {0, 1}, (α 2 + 1) = deg(q) deg(p ) = a 2 (β 1 + 1) = w 1 + w 2 + w 3. Lemma 4.8. The integer is odd and a 2 is even, so that β 1 = 1, α 2 = 0 and a 2 = + 1. a Proof. The integer is odd by Lemma 4.6 since P 3 = P divides σ(p 1 1 ). Again, if a 2 is a odd, then P divides σ(p 2 2 ). So, P = P 1. Thus, P 1, P 1 1 = P 2 and P 1 2 = P 3 are both irreducible. This contradicts Lemma 2.5. Corollary 4.9. i) For any ξ F 3, P 2 (ξ) = 1. ii) w 2 + w 3, w 1 and w 1 + w 2 + w 3 = deg(q) deg(p ) are both even. iii) = 6l + 3, a 2 = 2w 2 = + 1 = 6l + 4 for some l N. Proof. i) If P 2 (ξ) = 1, then modulo 3 we get a = (σ(p 2 a 2 ))(ξ) = (P 1 P 3 Q)(ξ) = (P 1 σ(p 1 ))(ξ) = 1 ( + 1).

24 Hence by Lemma 4.8, we get modulo 3 = a 2 = + 1. It is impossible. ii) We get modulo 3 1 = (Q 1)(ξ) = (P 1 w 1 P 2 w 2 P 3 w 3 )(ξ) = 1 ( 1) w2+w3. We are done. iii) Since P 2 (ξ) = 1 and a 2 is even, we get (σ(p 2 a 2 ))(ξ) = 1. But modulo 3 (σ(p 2 a 2 ))(ξ) = (P 1 σ(p 1 ))(ξ) = 1 ( + 1). We see that 0 mod 3 and 3 mod 6 since is odd. So, Now, in order to end the proof for the odd case, we see that (1 + P 1 )(1 + (P 1 2 ) (P 1 2 ) 3l+1 ) = σ(p 1 ) = P 3 Q = (P 1 + 1) Q. and l must be equal to 0 by Lemma 4.2. Hence Q = 1 + (P 1 2 ) (P 1 2 ) 3l+1 P 1 w 1 P 2 w 2 P 3 w 3 = Q 1 = P 1 2. Thus, w 1 = 2 and = n w 1 = 3 n It is impossible, because 0 mod Case A even In this section, we put A = P a1 1 P a2 2 P a3 3 Q b with P 1 := x + 1, P 2 := x + 2, P 3 := x + 3 = x, a i = N i 3 ni 1, b = M 3 m 1, 3 N i, 3 M. For S F 3 [x], we denote by S (resp. S) the polynomial obtained from S by substituting x by x + 1 (resp., by x + 2).

25 We need the following facts that are more precise than Lemma 4.6. Lemma Let a N and P {P 1, P 2, P 3 }. Then i) P divides σ(p a ) if and only if a is odd, ii) P divides σ(p a ) if and only if 3 divides a + 1. Proof. We may suppose that P = P 1. i) follows from the facts ii) follows from the facts P 2 (1) = 0, P 1 (1) = 2 = 1, (σ(p 1 ))(1) = ( 1)a = ( 1) a P 3 (0) = 0, P 1 (0) = 1, (σ(p 1 ))(0) = + 1. Lemma i) If is odd, n 1 = 0 and σ(p 1 ) = P α2 2 P α3 3 Q, then = 3, α 2 = 1, α 3 = 0 and Q = 1 + P 1 2 = 1 + (x + 1) 2. ii) If is odd, n 1 1 and σ(p 1 ) = P α2 2 P α3 3 Qα, then either (N 1 = 2, α 2 = 3 n1 = α 3 + 1, α = 0) or (N 1 = 4, α 2 = 3 n1 = α 3 + 1, α = 3 n1, Q = 1 + P 1 2 = 1 + (x + 1) 2. Proof. i) if is odd then P 2 = 1 + P 1 divides σ(p 1 ) and α 2 = 1 since α 2 {0, 1}.

26 If α 3 0, then α 3 = 1 and P 3 divides σ(p 1 ). Thus, we get in F 3 N 1 = + 1 = (σ(p 1 ))(0) = (P 2 P 3 Q)(0) = 0, which is impossible since 3 N 1. Therefore, α 3 = 0 and (1 + P 1 )(1 + (P 2 1 ) (P a 1 ) ) = σ(p 1 1 ) = P 2 Q. Hence 1 + (P 2 1 ) (P ) = Q is irreducible. So, we must have = 3 by Lemma 4.2. ii) Since is odd, we may put = (2c 1 ) 3 n1 1 where c 1 N. Hence a σ(p ) = P n 1 1 c 3 (1 + P P ) 3n 1 c (P ) 3n 1. If c 1 = 1, then N 1 = 2 and α = 0. If c 1 = 2, then N 1 = 4, Q = 1 + P 1 2, α 2 = 1 and α = 3 n1. If c 1 3, then P 1 c is reducible over F 3 and thus 3 ω(σ(p 1 )) 4. It is impossible. Lemma i) If is even, n 1 = 0 and σ(p 1 ) = P α2 2 P α3 3 Q, then + 1 is a prime number, α 2 = α 3 = 0 and Q = σ(p 1 ). ii) If is even, n 1 1 and σ(p 1 ) = P α2 2 P α3 3 Qα, then either (N 1 = 1, α 2 = 0, α 3 = 3 n1 1, α = 0) or (N 1 is an odd prime number, α 2 = 0, α 3 = 3 n1 1, α = 3 n1, Q = σ(p 1 N 1 1 )).

27 Proof. i) even implies α 2 = 0. As above, α 3 0 implies = N 1, which is impossible. So, Q = σ(p 1 ) is irreducible and + 1 must be an odd prime number. ii) even implies N 1 odd and α 2 = 0, and n 1 1 implies α 3 = 3 n1 1. If N 1 = 1, then α = 0. If N 1 3, then P 3 3 n 1 1 (1 + P P 1 N 1 1 ) 3n 1 a = σ(p ) = P n Q 3n 1. Thus Q = σ(p 1 N 1 1 ) is irreducible and N 1 must be an odd prime number. Lemma Let p be an odd prime number. If σ(x a ) is irreducible over F p and if σ(x a ) = σ((x + µ) a ) for some µ F p, then µ = 0. Proof. Let ξ be a primitive (a + 1)-root of unity. By hypothesis, a S(x) := σ(x a ) = (x ξ i ) is the minimal polynomial of ξ. If S(x) = S(x+µ) with µ 0, then x ξ = x+µ ξ k for some 2 k a. Thus, the polynomial R(x) := x k x µ F p [x] satisfies R(ξ) = 0. Hence, S divides R and S = R, which is impossible since p 2. Corollary If A = P a1 1 P a2 2 P a3 3 Qb F 3 [x] is even and perfect, then there exists a unique j {1, 2, 3} such that Q σ(p j a j ), so #Σ 1 = 1. i=1

28 Proof. We know that Σ 1. If #Σ 1 2, then we may suppose that 1, 2 Σ 1. According to Lemmata 4.11 and 4.12, we get by Lemma Q {1 + P 1 2, σ(p 1 ), σ(p 1 N 1 1 )} {1 + P 2 2, σ(p 2 a 2 ), σ(p 2 N 2 1 )} = According to Corollary 4.14, it remains to consider only the case ( ) Case m = 0, #Λ Σ 1 = 1. m = 0, #Λ Σ 1 {0, 1}. We may put Λ Σ 1 = {1}, so n 1 = 0, Q σ(p a1 1 Case n 2 = 0 = n 3 One has a 2 = 1 = a 3 by Lemma 4.1. Thus, we get where ), Q σ(p a2 2 σ(p 1 ) = P α2 2 Q, σ(p 2 a2 ) = 1 + P 2 = P 3, σ(p 3 a 3 ) = 1 + P 3 = P 1, σ(q b ) = 1 + Q = P w1 1 P w w 1 = = α 2 + deg(q), α 2 + w 2 = a 2 = 1, α 2 {0, 1}. a3 ), Q σ(p3 ) and thus b = 1. If is odd, then α 2 = 1 and w 2 = 0. So, Q = P 1 w 1 1 is odd and irreducible. It is impossible. If is even, then α 2 = 0 and 1 + P P 1 = σ(p 1 ) = Q. Thus, P 1 does not divide 2 + P P 1 = 1 + Q = σ(q). Hence w 1 = 0 and = 1, which is impossible. Case n 2 1, n 3 1 We get σ(p 1 ) = P α2 2 P α3 3 Q, σ(p 2 a2 ) = P 1 3 n 2 1 P 3 β 3 3 n 2, σ(p 3 a 3 ) = P 2 3 n 3 1 P 1 γ 1 3 n 3, σ(q b ) = 1 + Q = P w1 1 P w2 2 P w3 3

29 where 3 n2 1 + γ 1 3 n3 + w 1 = = α 2 + α 3 + deg(q), α 2, α 3, β 3, γ 1 {0, 1}. If is odd, then by Lemma 4.11, = 3, α 2 = 1, α 3 = 0 and Q = 1 + (x + 1) 2. Hence σ(q b ) = 1 +Q = x 2 + 2x = P 2 P 3, w 1 = 0, w 2 = w 3 = 1 and = 3 n2 1+γ 1 3 n3. It is impossible since 3 ( + 1). If is even, then σ(p a1 1 ) = Q and + 1 is an odd prime number. Therefore, P 1 does not a divide 2 + P P 1 1 = σ(q), w 1 = 0 and = 3 n2 1 + γ 1 3 n3, which is impossible. Case n 2 = 0, n 3 1 In this case, a 2 = 1 by Lemma 4.1. We get and the contradiction 1 = a 2 3 n σ(p 3 a 3 ) = P 2 3 n 3 1 P 1 γ 1 3 n 3 Case n 2 1, n 3 = 0 In this case, a 3 = 1 by Lemma 4.1. If is odd, then by Lemma 4.11, = 3 and σ(p a1 1 ) = P 2 (1 + P 2 1 ) = P 2 Q. Since Q does not divide σ(p a2 2 ), one has σ(q) = 1 + Q = P 2 P 3 and a 2 = 2 = We get the three even perfect polynomials of Theorem 1.2 A = x(x + 1) 3 (x + 2) 2 (1 + (x + 1) 2 ), A and A. If is even, then σ(p a1 1 ) = Q and + 1 is an odd prime number. We get a σ(p 1 a 1 ) = Q, σ(p ) = P n 2 1 β 1 P 3 3 n 2 3, σ(p 3 a 3 ) = 1 + P 3 = P 1, σ(q b ) = 1 + Q = P w1 1 P w2 2 P w3 3,

30 β 3 = 0 since n 2 1 and 1 = a 3 β 3 3 n2. So, a 2 is even. Hence, w 3 = a 3 = 1, a 2 = w 2 = 3 n2 1 and w 1 = 0. Thus, 2 + P P 1 = σ(q) = P 2 a 2 P 3. So, = a It is impossible, because and a 2 are both even Case m = 0, Λ Σ 1 =. We need the following general result. Lemma For any T F 3 [x], one has gcd(1 + T, 1 + (T 2 ) (T 2 ) 3n ) = 1. Proof. If S is a common irreducible divisor of 1 + T and of 1 + (T 2 ) (T 2 ) 3n 1 1 2, then T 1 mod S and T 2 1 mod S. Thus 3 n (T 2 ) (T 2 ) 3n mod S. 2 It is impossible since 3 does not divide 3 n Since Σ 1, one has #Λ 2. By Corollary 4.14, we may consider only two cases: (I) Σ 1 = {1} and Λ = {2, 3}, (II) Σ 1 = {1} and Λ =. Case (I) Since Q σ(p a2 a3 2 ) and Q σ(p3 ), from Lemma 4.1, we get a 2 = a 3 = 1. Moreover, n 1 1, so 3 P n 1 1 a 3 divides σ(p 1 1 ). Hence 1 = a 3 3 n1 1 2, which is impossible.

31 Case (II) n 1 1, n 2 1 and n 3 1. From Corollary 4.14, We get σ(p 1 ) = P 3 3 n 1 1 (P α2 2 Q)3n 1 σ(p 3 a 3 ) = P 2 3 n 3 1 P 1 γ 1 3 n 3 σ(p 2 a 2 ) = P 1 3 n 2 1 P 3 β 3 3 n 2 σ(q b ) = P w1 1 P w2 2 P w3 3. so b = 3 n1. If is odd, then by Lemma 4.11, α 2 = 1, = 4 3 n1 1 and a σ(p ) = P n (P 2 Q) 3n 1, where Q = 1 + P 2 1. So, 1 + Q = P 2 P 3 and P 2 P 3 (1 + (Q 2 ) (Q 2 ) 3n ) = σ(q b ) = P w1 1 P w2 Lemma 4.15, w 2 = w 3 = 1 and 1 + (Q 2 ) (Q 2 ) 3n = P 1 w 1. 2 P w3 3. Thus, by Thus We get Q 1 mod P 1 and w 1 = deg(p 1 w 1 ) = (3 n1 1) deg(q) 4. 3 n It is impossible since 3 does not divide 3 n (Q 2 ) 1 + (Q 2 ) 3n mod P 1. If is even, then by Lemma 4.12, α 2 = 0 and Q = 1 + P P 1 N 1 1, where N 1 is an odd prime number. Hence, (1 + Q) (1 + (Q 2 ) (Q 2 ) 3n ) = σ(q b ) = P w1 1 P w2 2 P w3 3.

32 Since P 1 does not divide 2 + P P 1 N 1 1 = 1 + Q, we have But Q 1 mod P 1, so we get It is impossible as above. 3 n (Q 2 ) (Q 2 ) 3n mod P (Q 2 ) (Q 2 ) 3n mod P Canaday E. F., The sum of the divisors of a polynomial, Duke Math. Journal 8 (1941), Beard Jr. J. T. B., O Connell Jr. J. R. and West K. I., Perfect polynomials over GF (q) Rend. Accad. Lincei, 62 (1977), Beard Jr. J. T. B., Perfect polynomials revisited, Publ. Math. 38 (1 2 (1991), L. E. Dickson, Finiteness of the odd perfect and primitive abundant numbers with n distinct prime factors American J. 35 (1913), Gallardo L. H. and Rahavandrainy O., Perfect polynomials over F 4 with less than five prime factors Portugaliae Mathematica 64(1) (2007), , Odd perfect polynomials over F 2 J. Théor. Nombres Bordeaux 19 (2007), , Perfect polynomials over F 3 Int. J. of Algebra 2(10) (2008), , There is no odd perfect polynomial over F 2 with four prime factors Portugaliae Mathematica 66(2) (2009), Gallardo L. H. and Rahavandrainy O., On even (unitary) perfect polynomials over F 2 Finite Fields Appl. 18(5) (2012), , On perfect polynomials over F p with p irreducible factors Portugaliae Mathematica 69(4) (2012), Lidl R. and Niederreiter H., Finite Fields, Encyclopedia of Mathematics and its applications, Cambridge University Press, 1983 (Reprinted 1987).

33 12. Sylvester J. J., Sur l impossibilité de l existence d un nombre parfait impair qui ne contient pas au moins 5 diviseurs premiers distincts Comptes Rendus Paris 106 (1888), L. H. Gallardo, Mathematics, University of Brest, 6, Avenue Le Gorgeu, C.S , Brest Cedex 3, France, luisgall@univ-brest.fr O. Rahavandrainy, Mathematics, University of Brest, 6, Avenue Le Gorgeu, C.S , Brest Cedex 3, France, rahavand@univ-brest.fr