Odd multiperfect numbers of abundancy four

Odd multiperfect numbers of abundancy four Kevin A. Broughan and Qizhi Zhou University of Waikato, Hamilton, New Zealand Version: th December 006 E-mail: kab@waikato.ac.nz MSC000: A05, A5, N99. Key words: odd multiply perfect number, abundancy, Euler factor.. INTRODUCTION We say a number is multiperfect of abundancy k (or k-perfect) if (N) = kn. No k-perfect odd numbers are known for any k, and it is believed that none exist. For a survey of known results see [6] or [3] and the references given there. For example, if N is odd and 4-perfect then N has at least distinct prime factors. If it is also not divisible by 3 then it has at least 4 prime factors. In this paper we consider properties of classes of odd numbers which must be satised if they are to be 4-perfect. Conversely, properties of classes which can never be 4-perfect. In a number of cases theorems follow, with some changes, in the pattern of corresponding results for -perfect numbers. However, mostly because of the number of primes involved, some of those techniques, from the theory of -perfect numbers, are not so readily available. We show that Euler's structure theorem, that every odd -perfect number has the shape N = q e p pm m, where q e mod 4, has an extension to odd 4-perfect numbers, and then to odd k -perfect numbers. For 4-perfect numbers there are three possible shapes like Euler's form, (A) with q's instead of, (B) with q 3 mod 8 and e mod 4, and (C) with q mod 4 and e 3 mod 8. An immediate corollary is that no square or square free number is 4-perfect. For k -perfect numbers we need to derive a fact, which could be of independent interest. For j, odd primes p and odd e, we have j jj(p e ) if and only if j+ jj(p + )(e + ). We include negative results (i.e. shapes which no odd 4-perfect number can have) for odd 4-perfect cubes, numbers with 9 being the maximum

BROUGHAN AND ZHOU power of 3 dividing N, numbers with each of the p i occurring to the power, and a positive result on the power of 3 dividing any odd k -perfect number.. RESTRICTED FORMS FOR AN ODD PERFECT NUMBER OF ABUNDANCY 4 We begin with two lemmas, summarizing well known results. Lemma.. Let d and n be whole numbers and p a prime number. If d + j n + then (p d ) j (p n ). Lemma.. (Congruences modulo 3) Let p be an odd prime with p 6= 3. () Let j be any natural number and p > 3 an odd prime. Then (p 6j ) mod 3. () Let the whole number j be odd. If p mod 3 then (p 3j ) mod 3. If p mod 3 then (p 3j ) 0 mod 3. (3) Let j be a natural number. If p mod 3 then (p +3j ) mod 3. If p mod 3 then (p +3j ) 0 mod 3 if j is even and (p +3j ) mod 3 if j is odd. (4) Let j be any natural number. If p mod 3 then (p +3j ) 0 mod 3. If p mod 3 then (p +3j ) mod 3 if j is even and (p +3j ) 0 mod 3 if j is odd. This set of results is best summarized in a table, with the rows corresponding to values of p modulo 3, in the rst column, and the columns the values of (p e ) mod 3 for values of e modulo 6 which are in the rst row: p=e 0 3 4 5 0 0 0 0 0 Theorem.. (Euler equivalent) Let N be an odd 4-perfect number. Then N has one of the following forms, where the i are whole numbers and the p i odd primes: (A) N = q e qe p p m m for primes q i and whole numbers e i with q i e i mod 4. In the remaining types N = q e p p m m where:

ODD FOUR-PERFECT NUMBERS 3 (B) q mod 4 and e 3 mod 8 or (C) q 3 mod 8 and e mod 4. Proof. () Let N = p pm m where the p i are odd primes and the i whole numbers. Then (N) = 4N implies jj(p ) (pm m ) so either is the maximum power of two dividing two distinct terms in the product and the remaining terms are odd, or is the maximum power dividing one term and the remaining terms are odd. So type (A) is the former shape and (B) and (C) the latter. Therefore we need only consider primes q and powers such that jj(q ). () Claim: If q mod 4 and 3 mod 8 then 4 j (q ). To see this let = 3 + 8e and q = + 4x then (where f and y are integers) (q ) = ( + 4x)4+8e 4x = ( + 4x)4f 4x = 4f 4x (4x 4f + = 4g (4x) + (4x) 3 y) so 4 j (q ). (3) In the same situation as in (), 8 - (q ): Write (q ) = + q + q + q 3 + + q 3+8e ; group the 4 + 8e terms in + e sets of 4 terms, so that (q ) ( + q + q + q 3 )( + e) mod 8; where we have used q 4 mod 8. Replacing q by + 4x and reducing modulo 8 we get (q ) 4 ( + e) mod 8, which is non-zero, so 8 - (q ). (4) Claim: If q 3 mod 8 and mod 4 then 4 j (q ). Let = +4e and q = 3 + 8x then (where f; y; z and w are integers) (q ) = (3 + 8x)f where f is odd + 8x = ( + y)f where y is odd y = f y (y f + (y) + ) = f + f(f )y + 4z = 4w

4 BROUGHAN AND ZHOU so 4 j (q ). (5) In the same situation as in (4) 8 - (q ): write (q ) = q+4e + 8x q (3 + 8x) 4 mod 8: so, again 8 - (q ). (6) The remainder of the proof consists in showing the above cases constitute the only possibilities by examining in turn the possible additional values of fq; eg modulo 8. In summary, using the notation q e for the values of q and e modulo 8, and using the same techniques as used in parts (), (3) and (4) of the proof, the cases ; 5 ; 5 ; 5 5 give 4 - (q e ). The cases 7 ; 3 3 ; 3 7 ; 5 7 ; 7 ; 7 3 ; 7 5 ; 7 7 give 8 j (p e ), so cannot occur. The remaining cases 5 3 ; 3 5 are covered by (B) and (C). Corollary.. No square or square free number is odd and 4-perfect. Proof. Since the exponents of the leading primes are odd, and one of the three forms is always present, the rst part of the claim is immediate. For the second part we need only consider the special forms N = q q and N = q, where the q i are odd primes to see that m 6= 0, so no odd 4-perfect number is square free. It might be of interest to speculate, on the basis of Euler's theorem and the above, on the general form for division of (p ) by powers of. However for powers 3, and beyond, the situation appears to be well structured but mysterious. For example, in the following each pair corresponds to the classes modulo 4 of an odd prime and odd exponent (p; e) such that 3 jj(p e ). The list appears to be complete for this power of : (; 7); (3; 3); (3; ); (5; 7); (7; ); (7; 5); (7; 9); (7; 3); (9; 7); (; 3); (; ); (3; 7): Note that in each case 4 jj(p + )(e + ). It is a beautiful fact that this is true in general for all powers of. Theorem.. (P + E + ) For all odd primes p, powers j and odd exponents e > 0 we have j jj(p e ) () j+ jj(p + )(e + ):

ODD FOUR-PERFECT NUMBERS 5 Proof. () Let j jj(p e ). First expand p to base : p = + e + e + + j+ ; where f0g [ N and e i f0; g. There exits a minimum i with i j so that since otherwise so p = + + + + i + 0: i + + j+ p = + + + j + j+ mod j+ (p e ) = + p + p + + p e + 0 mod j+ so j+ j(p e ) which is impossible. Hence we can write p = i + i+ ; f0g [ N: Therefore p + = i o where here, and in what follows, \o" represents a generic odd integer, with not necessarily the same value in a given expression. Since e + is even, there exists an l such that e + = l o. Since j jj(p e ) we have p l o p = j o and therefore ( i o ) l o = j o ( i o ). Call this equation (). (a) If i > examine both sides of equation () in base and equate the lowest powers of. This leads to i + l = j + since i o = o. Therefore l = j i +. (b) If i = write p + = :o so p = k o for some k. Hence, because j jj(p e ), p l o ( + k o) l o = j k o = j+k o so, again comparing the lowest powers of on both sides, k + l = j + k so l = j = j +. Hence, for all i, l = j i + and we can write p = i + i+ x e = j i+ + j+ i+ y

6 BROUGHAN AND ZHOU where x; y are integers. Hence (p + )(e + ) = j+ ( + x)( + y) so j+ jj(p + )(e + ). () Conversely, let j+ jj(p + )(e + ) so for some i > 0, i jjp + and j+ i jje +. We now consider two cases, depending on the values if i and j. (a) Let i = and j =. (This is really Euler's theorem). In this case p + = o = (x + ) so p = 4x + and e + = o. Therefore (p e ) = p o p = po p (po + ) = ( + p + + p o )((4x + ) o + ) = o (4y + ) = o so jj(p e ). (b) Let i = and j >. Again p = 4x +. The inductive hypothesis is that for all j 0 < j, j0 jj(p j0 o )=(p j o (p e ) = p p ). Then (p j o + ) = j o((4x + ) j o + ) = j o(4y + ) = j o so in this case also j jj(p e ). (3) First we make some preliminary polynomial constructions where all polynomials are in Z[x]. For n N dene f n ; q n ; s n ; r n by Then f n (x) = ( + x) n = xq n (x) s n (x) = ( + x) n + = (x + )r n (x) for n odd: f o = (( + x) o )(( + x) o + ) = x r o (x) (x + ) q o (x); and for l f l o = f l o(x) s l o(x) = s l o(x) s l o(x) s o (x)x(x + ) r o (x) q o (x) s l o = ((( + x) l ) o ( )) = (( + x) l ( ))((( + x) l ) o + + ) = (( + x) l + )( )

ODD FOUR-PERFECT NUMBERS 7 Since i >, x = i o = o. Hence x + = o + = o and s l o = (o l + )(an even number of odd terms + ) = o o = o; and x + = i o. Note also that r o (x) = (( + y) o )=(y) = o + z = o and q o (x) = (( + x) o )=(x + ) = (o o o o + ) = o. Therefore, with this value of x l o x = l o i o o o = l+i o: Now, at last, we can complete the proof. Let x = p = i o and l = j + i. Then so j jj(p e ). (p e ) = pe+ p = l ( + x) o x = f l o (x) = l+i o = j o Corollary.. Let M q be a Mersenne prime and e odd with e. If j jj(mq e ) then j q. From the theorem we also get the following corollary, which is an extension of Euler's theorem to perfect numbers of abundancy k. Corollary.3. Let N be odd and k -perfect. Then there exists a partition of k, k = k + + k n, with k i, such that N = ny i= p ei i where the e i are odd, the p i ; q j odd primes, and for each i with i n there exist positive integers l i and m i such that li jjp i +, mi jje i + and l i + m i = k i +. my j= q fj j Theorem.3. (Cubes) Let N be an odd cube with 3 - N. (A) If N has shape N = q +4e q +4e p pm m and q 5 mod (i.e. q mod 4 and mod 3) and q mod 4, then N is not a 4-perfect number.

8 BROUGHAN AND ZHOU and q mod 4 (i.e. (B) If N has shape N = q +4e p p m m q 3 mod 8 and mod 3), then N is not a 4-perfect number. (C) If N has shape N = q 3+8e p p m m mod 4 and mod 3), then N is not a 4-perfect number. and q 5 mod (i.e. q Proof. (C) Let N be an odd and 4-perfect cube with q mod 3. Then we can write: (N) = (q 3+4e )(p 6 ) (p 6m m ): By Lemma., the rst factor on the right is congruent to 0 modulo 3, so (N) 0 mod 3. Since (N) = 4N, 0 q 3+4e p 6 p 6m m mod 3; but each factor on the right hand side is non-zero modulo three. Hence N is not 4-perfect. In parts (A) and (B) the result also follows since (q 3+6f ) 0 mod 3. Theorem.4. (Nine is the maximum power of three dividing N) If N is a whole number with 3 jjn and such that if 3, 6 and 97 appear in the prime factorization of N, they do so to powers congruent to modulo 6. Then N is not an odd 4-perfect number. Proof. Let the hypotheses of the theorem hold for N, but let it also be odd and 4-perfect. Then 3 jjn implies (3 ) = 3 j N. So 3 must appear, and by the argument given below, 6 and 97 must also appear. Now, by Lemma., for all primes p and natural numbers e, (p ) j (p +6e ). So 3 6 = (3 ) j (3 ) j N, for some, which implies 3 6 j N. Again 3 3 97 = (6 ) j (6 ) j N, for some, which implies 3 3 97 j N. Finally 3 369 = (97 ) j (97 ) j N, for some, so 3 j N. Now if 3, 6 or 97 appear, even though each is congruent to modulo 4, their powers, being congruent to modulo 6, are even, so must appear amongst the p i in each of the three shapes given in Theorem.. Therefore 3 3 j N, which is a contradiction. Therefore N is not 4-perfect. The following result uses techniques similar to those developed for - perfect numbers by Steuerwald in [8]. Theorem.5. (Small powers) () If N is odd, 3 j N and N has the shape either (a) N = q +4e q +4e 3 p p m or (b) N = q3+8e 3 p p m where, in either case, q i mod 4, or (c) q +4e 3 p p m, where q 3 mod 8, where the primes are distinct, then N is not an odd 4-perfect number.

ODD FOUR-PERFECT NUMBERS 9 () If N is odd, 3 - N and N has the shape either (a) N = q 3+8e p p m with q mod 4 or (b) N = q+4e p p m with q 3 mod 8, or (c) N = q +4e q +4e p p m, with q i mod 4, then N is not a 4-perfect number. Proof. () Let N satisfy (N) = 4N. Then (3 ) = 3 j N. In case (c), q is not in the set f3; 6; 97g. Assume rst that the q i are not in this set in cases (a) and (b). (Below we consider the situation which arises when a q i is in this set.) Under this assumption we obtain the chain: (3 ) = 3 6; (6 ) = 3 3 97; (97 ) = 3 369; so 3 3 j N, which is false. Hence N is not 4-perfect. Since the exponent of each q i is odd, for q = q or q, e = e or e, q + j (q e ). If q = 3, since q + j N we obtain the chain: (7 ) = 3 9 j N; (9 ) = 3 7 j N; (7 ) = 3 549 j N; giving 3 3 j N, which is false. If q = 6 we can assume also (3 ) = 3 6 j N. Again, since q + j N We obtain the chain: (3 ) = 3 33 j N; (33 ) = 3 7 533 j N; (7 ) = 3 549 j N; again giving 3 3 j N, which is false. If q = 97 then (q + )= = 7 j N and the same chain as in the q = 3 case can be derived with the same conclusion. Thus our assumption that no q i is in the set f3; 6; 97g is valid and the proof is complete. (a) and (b): Let N satisfy (N) = 4N and 3 - N, with shape N = q f p p m; where 3 < p < < p m and f is odd. Since, for each i, (p i ) = + p i + p i and 3 - N, we must have p i mod 3. By Theorem., q is congruent to modulo 4 or 3 modulo 8. Because f is odd, q + j (q f ) j N and since also 3 - N we cannot have q mod 3, so must have q mod 3. Since (p ) < (p + ) < p, (p ) is divisible by at most one p i. Therefore either (a) (p ) = q g with g or (b) (p ) = q g p i for some i. Case (b) is impossible, since it is invalid modulo 3. In case (a), [, Lemma ] shows the only possibility is g =.

0 BROUGHAN AND ZHOU Let x = (q + )=. Then x mod 3. Since x is too small to include a power of at least two q's, it must be a product of the p i. We cannot have x = p i since p i mod 3, so it must have at least prime factors, with the smallest factor being less than or equal to p x, and therefore p i p x for some i. But then q = + p + p + p i + p i + p x + x q + 3 r q + + so q = 5 or q = 7. Each of these is impossible since q mod 4 and mod 3 or q 3 mod 8. (c): Now let N = q +4e q +4e p p m, be odd and 4-perfect with 3 - N. Since (N) = 4N we can write: (q +4e )(q +4e )(+p +p )( )(+p m +p m) = 4q +4e q +4e p p m: Considering this equation modulo 3 shows each p i mod 3 and then (q +4e )(q +4e ) q +4e q +4e mod 3. But q i mod 3 implies, by Lemma., 3 j (qi +4ei ), which is impossible. This means q mod 3, q mod 3. (Now we modify the argument of Steuerwald, and nd that the Lemma of Brauer is not needed.) Since (p ) < p, (p ) is divisible by at most one of the p i, so we can write (p ) = q g qg p i or (p ) = q g qg or (p ) = q g or (p ) = q g ; where q < q, g i except in the rst case where g i 0. Consideration of these possibilities modulo 3 shows that the rst case cannot occur. Since e is odd, by Lemma., x = q+ j N and x mod 3. Now x is too small to include a q i in its prime factorization, so must be a product of the p i. We cannot have x = p i (consider modulo 3 again), so there must be two or more of the p i in the factorization of x, so there exists an i with p i p x. But then, in all remaining cases, r q + p + p + p i + p i + p q + x + x = + + q + ; so q + q q+ + q+. But this means q must be ; 3; 5 or 7. Each of these is impossible, since q mod 4 and mod 3. This contradiction veries our conclusion (that no such 4-perfect number exists) in this nal case. If we call the leading prime(s) to odd power(s) with special shape the \Euler part" and the rest the \squared part", then the previous result says

ODD FOUR-PERFECT NUMBERS that \no odd 4-perfect number exists with squared part a square of a square free number". The following result is based on the technique of Starni [7] whose theorem, for -perfect numbers, had uniform powers for the p i. This, in turn depended on a result of McDaniel [4] (incorrectly cited), where the powers are not uniform. Theorem.6. Let N = 3 Q M i= pi i be odd and k -perfect, where the p i are distinct odd primes with p i > 3, > 0, the Euler factor has any of the forms given by Theorem., and, for all i i 6 mod 3. Then 3 j ( ). Proof. Firstly ((3 ); 3 ) =. Since i 6 mod 3, + i ; 5 mod 6. Since p i ; mod 6, (p i i ) mod 6 if p i mod 6 or (p i i ) + i mod 6 if p i mod 6. But then, subject maybe to some reordering, there exists an m 0 with P := MY i= (p i i ) my i= my i= + i mod 6 + i mod 3: By the given assumption, + i 6 0 mod 3, so P 6 0 mod 3, and thus (P; 3 ) =. But for some whole number k, (N) = k N so therefore Therefore 3 j ( ). ( )(3 )P = k 3 MY i= p i i : REFERENCES. A. Brauer, On the non-existence of odd perfect numbers of the form p a q q t q 4 t,, Bull Amer. Math. Soc. 49 (943), p7-78.. P. Erdos, Remarks on the number theory II, some problems on the function, Acta Arith. 5 (959), 7-77. 3. A. Flammenkamp, The Multiply perfect numbers page, http://wwwhomes.unibielefeld.de/achim/mpn.html.

BROUGHAN AND ZHOU 4. W. L. McDaniel, The non-existence of odd perfect numbers of a certain form, Arch. Math. 5 (970), 5-53. 5. W. L. McDaniel and P. Hagis Jr, Some results concerning the non-existence of odd perfect numbers of the form p M, Fibbonacci Quarterly 3 (975), 5-8. 6. R. Sorli, Multiperfect numbers, http://www-sta.maths.uts.edu.au/rons/mpfn/mpfn.htm. 7. P. Starni, On some properties of the Euler's factor of certain odd perfect numbers, J. of Number Theory 6 (006), 483-486. 8. R. Steuerwald, Verschaeung einer notwendigen Bedingung fuer die Existenz einer ungeraden vollkommenen, Zahl. S.-B Math.-Nat.Kl.Bayer. Akad. Wiss., (937), 68-73.