On odd deficient-perfect numbers with four distinct prime divisors
aa r X i v : . [ m a t h . N T ] A ug ON ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIMEDIVISORS
CUI-FANG SUN AND ZHAO-CHENG HEA
BSTRACT . For a positive integer n , let σ ( n ) denote the sum of the positive divisors of n . Let d bea proper divisor of n . We call n a deficient-perfect number if σ ( n ) = 2 n − d . In this paper, we showthat the only odd deficient-perfect number with four distinct prime divisors is · · · . Keywords: deficient-perfect number, deficient divisor, multiplicative arithmetic function.
1. I
NTRODUCTION
For a positive integer n , let arithmetic functions σ ( n ) and ω ( n ) denote the sum of the positivedivisors of n and the number of distinct prime divisors of n , respectively. Let d be a proper divisorof n . We call n a deficient-perfect number with deficient divisor d if σ ( n ) = 2 n − d . If d = 1 , thensuch a deficient-perfect number is called an almost perfect number. In 1978, Kishore [5] proved thatif n is an odd almost perfect number, then ω ( n ) > . In 2013, Tang, Ren and Li [11] determined alldeficient-perfect numbers with at most two distinct prime factors. In a similar vein, Tang and Feng[9] showed that no odd deficient-perfect number exists with three distinct prime factors. For relatedproblems, see [1-3, 5, 6, 8].In this paper, we obtain the following result: theorem 1.1. The only odd deficient-perfect number with four distinct prime divisors is · · · . For convenience, let ( · p ) denote the Legendre symbol. Let m be a positive integer and a be anyinteger relatively prime to m . If h is the least positive integer such that a h ≡ m ) , then h is called the order of a modulo m , denoted by ord m ( a ) . We take n = p α p α p α p α be an odddeficient-perfect number with deficient divisor d = p β p β p β p β , where p < p < p < p areprimes, α i ’s are positive integers and β i ’s are nonnegative integers with β i α i and P i =1 β i < P i =1 α i .By [5] and [9], we have d > and α i ’s are all even. Let D = p α − β p α − β p α − β p α − β . Then(1.1) p α +11 − p − · p α +12 − p − · p α +13 − p − · p α +14 − p − σ ( n ) = 2 n − d = (2 D − d or equally σ ( n ) n + dn = σ ( n ) n + 1 D .
Date : 2019-8-15E-mail: [email protected], [email protected].
2. T
HE CASE OF p = 5 In this section, we consider the case of n = p α p α p α p α with p = 3 and p = 5 . Lemma 2.1.
There is no odd deficient-perfect number of the form n = 3 α α α p α .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If α > , then σ ( n ) n + dn > − · · − · · − · > , which is clearly false. Thus α = 2 . If p , then σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. Thus p > . By (1.1), we have(2.1) · α +1 − · α +1 − · p α +14 − p − · · α α p α − β β β p β and | (2 D − . Thus D = 26 k + 7 for some positive integer k .If k = 1 , then D = 33 . If α > , then σ ( n ) n + dn > − · · − · · − · + 133 > , which is clearly false. Thus α = 2 and p = 31 , which contradicts with p > .If k ∈ { , , , , , , , , , , , } , then p < , this is impossible.If k ∈ { , , , , , , , } , then p ∈ { , , , , , , , } . Notingthat ord (11) = 5 and ord ( p ) are even, we have | ( α + 1) and (11 − | (11 α +1 − .However, | (11 − , a contradiction.If k = 9 , then D = p = 241 . Noting that ord (5) = 5 and ord (241) = 2 , we have | ( α + 1) and (5 − | (5 α +1 − . However, | (5 − , a contradiction.If k > , then D > . If p > , then σ ( n ) n + dn < − · · · · < , which is false. Thus p . If p , then α = 2 and (13 · | (2 D − .Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. Now we divide into the following four cases according to p . Case 1. p = 67 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. Thus α = 2 and (13 · · · | (2 D − . Thus D > and σ ( n ) n + dn < − · · − · · − · · D < , which is absurd. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 3
Case 2. p = 71 . Since ord (11) = ord (71) = 2 , we have β = 0 and | D . Thus D > and σ ( n ) n + dn < − · · − · · · D < , which is clearly false. Case 3. p . By α = 2 , we have σ ( n ) n + dn < − · · − · · · < , which is impossible. Case 4. p . If α = 2 , then σ ( n ) n + dn < − · · − · · · < , which is absurd. Thus α > . Subcase 4.1 p = 127 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is clearly false. Thus α = 4 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 . Since ord (127) = 4 , ord (127) = 10 , we have β = β =0 , β = 1 , D > · · and σ ( n ) n + dn < − · · − · · − · · · · < , which is a contradiction. Subcase 4.2 p ∈ { , } . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 and (7 · · | (2 D − . Since ord (5) = ord (131) =ord (137) = 2 , we have β = 0 and | D . Thus D > and σ ( n ) n + dn < − · · · − · · < , which is a contradiction. Subcase 4.3 p = 139 . If α = 4 , then (71 · | (2 D − . Thus D > and σ ( n ) n + dn < − · · − · · · < , which is absurd. Thus α > . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. Thus α = 2 and (7 · · | (2 D − . Thus D > and σ ( n ) n + dn < − · · · − · · < , which is impossible. CUI-FANG SUN AND ZHAO-CHENG HE
Subcase 4.4 p = 149 . Since ord (5) = ord (11) = ord (149) = 2 , we have β = 0 and | D . Thus D > . If D > , then σ ( n ) n + dn < − · · · · < , which is clearly false. Thus D = 1125 . Since ord (149) = 2 , ord (11) = 5 , we have | ( α + 1) and (11 − | (11 α +1 − . However, | (11 − , a contradiction. Subcase 4.5 p = 151 . If D > , then σ ( n ) n + dn < − · · · · < , which is clearly false. Thus D ∈ { , , } and β > . Noting that ord (5) =ord (11) = 75 , we have | ( α + 1) and (5 − | (5 α +1 − or | ( α + 1) and (11 − | (11 α +1 − . However, | (5 − and | (11 − , a contradiction. Subcase 4.6 p ∈ { , } . If D > , then σ ( n ) n + dn < − · · · · < , which is false. Thus D = 605 and β > . Since ord (11) = 5 and ord ( p ) = 4 , we have | ( α + 1) and (11 − | (11 α +1 − . However, | (11 − , a contradiction.This completes the proof of Lemma 2.1. (cid:3) Lemma 2.2.
There is no odd deficient-perfect number of the form n = 3 α α α p α .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If α > , then σ ( n ) n + dn > − · · − · · − · > , which is false. Thus α = 2 . If p , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. Thus p > . If D , then σ ( n ) n + dn > − · · − · · − · + 115 > , which is clearly false. Thus D > . By (1.1), we have(2.2) · α +1 − · α +1 − · p α +14 − p − · · α α p α − β β β p β . Now we divide into the following eight cases according to D . Case 1. D = 25 . If p > , then σ ( n ) n + dn < − · · · · < , which is false. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 125 > , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 5 which is also false. Thus p . It implies that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 125 > , which is impossible. Case 2. D ∈ { , , , } . Then p = D and σ ( n ) n + dn > − · · − · · − · · − · + 147 > , which is false. Case 3. D = 39 . If p > , then σ ( n ) n + dn < − · · · · < , which is a contradiction. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 139 > , which is also a contradiction. Thus p and α > . Since ord (13) = 4 , ord (5) = 6 and ord (13) = 2 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 139 > , which is impossible. Case 4. D = 45 . If p > , then σ ( n ) n + dn < − · · · · < , which is impossible. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 145 > , which is false. Thus p and α > . Since ord (13) = 4 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 145 > , which is impossible. Case 5. D ∈ { , , , , , , , } . Then p = D and β > . Noting that ord (13) =4 and ord ( p ) are all even, we deduce that the equality (2.2) can not hold. Case 6. D ∈ { , } . Then p = D and β > . Since ord (13) = 4 and ord ( p ) = 5 , wehave | ( α + 1) and ( p − | ( p α +14 − . However, | (61 − and | (71 − , acontradiction. Case 7. D ∈ { , } . If p > , then σ ( n ) n + dn < − · · · · < , which is false. Thus p and α > . Since ord (13) = 4 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 175 > , which is impossible. CUI-FANG SUN AND ZHAO-CHENG HE
Case 8. D > . If p > , then σ ( n ) n + dn < − · · · · < , which is false. Thus p . Subcase 8.1 p ∈ { , , } . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 and | (2 D − . Thus D > and σ ( n ) n + dn < − · · − · · · < , which is impossible. Subcase 8.2 p = 47 . Since ord (13) = ord (47) = 2 and ord (5) = ord (13) = 46 , wehave β = β = 0 , D > · and σ ( n ) n + dn < − · · · · · < , which is impossible. Subcase 8.3 p = 53 . If D > , then σ ( n ) n + dn < − · · · · < , which is false. Thus D ∈ { , } . Since ord (13) = ord (53) = 4 , we deduce that the equality(2.2) can not hold.This completes the proof of Lemma 2.2. (cid:3) Lemma 2.3.
There is no odd deficient-perfect number of the form n = 3 α α α p α .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If p = 19 , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus p > . If α = 2 , then | (2 D − . Thus D > . If p > , then σ ( n ) n + dn < − · · · · < , which is false. Thus p ∈ { , } and d = 1 , a contradiction. Thus α > . If p , then σ ( n ) n + dn > − · · − · · − · · − · > , which is clearly false. Thus p > . If D , then α > and σ ( n ) n + dn > − · · − · · − · + 137 > , which is absurd. Thus D > . By (1.1), we have(2.3) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . Now we divide into the following nine cases according to D . N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 7
Case 1. D ∈ { , , , , , , , , } . By (2.3), we have α > , α > and σ ( n ) n + dn > − · · − · · − · + 185 > , which is impossible. Case 2. D ∈ { , , , , , , , } . Then p = D, α > and σ ( n ) n + dn > − · · − · · − · · − · + 1113 > , which is impossible. Case 3. D ∈ { , , , , } . Noting that ord (5) = ord (17) = 2 , ord (3) =ord (17) = 4 and ord (3) = ord (5) = 16 , we have p ≡ . Thus α > , α > , α > and p > .If D = 125 , then σ ( n ) n + dn > − · · − · · − · + 1125 > , which is a contradiction.If D = 135 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 1135 > , which is also a contradiction.If D = 153 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 1153 > , which is also a contradiction.If D ∈ { , } , then σ ( n ) n + dn < · · · < , which is a contradiction. Case 4. D = p . Then α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 1359 > , which is impossible. Case 5. D = 225 . Since ord (17) = ord (5) = 2 and ord (3) = ord (5) = 16 , we have p ≡ . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p ∈ { , , } . Since ord (3) = ord (17) = 4 and ord ( p ) are alleven, we have α = 2 . Sine | (5 − , we deduce that the equality (2.3) can not hold. CUI-FANG SUN AND ZHAO-CHENG HE
Case 6. D ∈ { , , , , , , , , , , , } . Then p ∈ { , , , , , , , , , , , } , α > and σ ( n ) n + dn > − · · − · · − · · − · + 1339 > , which is impossible. Case 7. D = 289 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . Since ord (17) = ord (5) = 2 , ord (3) = ord (17) = 4 ,we have p ≡ . Thus p , α > , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 1289 > , which is impossible. Case 8. D ∈ { , , } . Then p ∈ { , , } , α > and σ ( n ) n + dn > − · · − · · − · · − · + 1365 > , which is impossible. Case 9. D > . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If p , then α = 2 and | (2 D − . Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is clearly false. If p , then σ ( n ) n + dn < · − · · · < , which is false. Thus p .If p , then α > . Otherwise, if α = 2 , then σ ( n ) n + dn < · − · · · < , which is absurd. If p , then α = 4 . Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. If p , then α > . Otherwise, if α = 4 , then σ ( n ) n + dn < − · · · · < , which is impossible. Thus p or p .Now we divide into the following fifteen subcases according to p . Subcase 9.1 p . If p , then α = 4 and | (2 D − . Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 9 which is clearly false. It follows that
D > and σ ( n ) n + dn < − · · − · · · < , which is impossible.If p = 83 , then β = β = β = 0 by ord (17) = ord (83) = 2 , ord (3) = ord (17) =ord (83) = 4 , ord (3) = 16 and ord (83) = 8 . Thus D > · · . If α , then σ ( n ) n + dn < − · · − · · · · · < , which is clearly false. Thus α > . If α = 2 , then σ ( n ) n + dn < · − · · − · · · · < , which is false. Thus α > and σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible.If p = 89 , then β = β = β = 0 by ord (17) = ord (89) = 2 , ord (3) = ord (17) =4 , ord (89) = 2 , ord (3) = 16 and ord (89) = 4 . Thus D > · · and σ ( n ) n + dn < · − · · · · · < , which is clearly false. Subcase 9.2 p ∈ { , , } . Since ord (3) = ord (17) = 4 , ord (3) = ord (5) = 16 , ord ( p ) and ord ( p ) are all even, we have β = β = 0 , D > · and σ ( n ) n + dn < − · · · · · < , which is clearly false. Subcase 9.3 p = 131 . Since ord (5) = ord (17) = ord (131) = 2 and ord (3) =ord (5) = ord (131) = 16 , we have β = β = 0 , D > · and σ ( n ) n + dn < − · · · · · < , which is clearly false. Subcase 9.4 p = 137 . Since ord (5) = ord (17) = ord (137) = 2 and ord (3) = ord (17) =ord (137) = 4 , we have β = β = 0 , D > · and σ ( n ) n + dn < − · · · · · < , which is clearly false. Subcase 9.5 p = 211 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is clearly false. Thus α = 6 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 4 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 . By (2.3), we have β = β = 0 and − (cid:18) (cid:19) = (cid:18) · · · · α (cid:19) = (cid:18) β β (cid:19) = 1 , which is a contradiction. Subcase 9.6 p = 223 . Since ord (3) = ord (17) = ord (223) = 4 , ord (3) = ord (5) =16 and ord (223) = 8 , we have β = β = 0 . If α = 6 , then − (cid:18) (cid:19) = (cid:18) · · α α α (cid:19) = (cid:18) β β (cid:19) = 1 , which is a contradiction. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is also a contradiction. Subcase 9.7 p = 227 . Since ord (3) = ord (5) = ord (227) = 16 and ord (3) =ord (17) = ord (227) = 4 , we have β = β = 0 . If α = 4 , then − (cid:18) (cid:19) = (cid:18) · α · · α α (cid:19) = (cid:18) β β (cid:19) = ( − β and (cid:18) (cid:19) = (cid:18) · α · · α α (cid:19) = (cid:18) β β (cid:19) = ( − β which is false. Thus α > . If α = 6 , then − (cid:18) (cid:19) = (cid:18) · · α α α (cid:19) = (cid:18) β β (cid:19) = 1 , which is a contradiction. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is also a contradiction. Subcase 9.8 p = 229 . Since ord (3) = ord (5) = 16 , ord (229) = 8 and ord (3) =ord (17) = 4 , ord (229) = 2 , we have β = β = 0 . If α = 4 , then − (cid:18) (cid:19) = (cid:18) · α · · α α (cid:19) = (cid:18) β β (cid:19) = 1 , which is false. Thus α > . If α = 6 , then − (cid:18) (cid:19) = (cid:18) · · α α α (cid:19) = (cid:18) β β (cid:19) = ( − β . Thus ∤ β . Noting that ord (17) = 19 , ord (5) = 114 , we have (cid:18) (cid:19) = (cid:18) · · α α α (cid:19) = (cid:18) β β (cid:19) = − . N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 11 which is a contradiction. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is also a contradiction. Subcase 9.9 p ∈ { , } . Since ord p i ( p j ) are all even for i = j , we have β = β = β = β = 0 and d = 1 , a contradiction. Subcase 9.10 p = 239 . Since ord (5) = ord (17) = ord (239) = 2 and ord (3) =ord (17) = 4 , ord (239) = 2 , we have β = β = 0 and D > · . If α = 4 , then σ ( n ) n + dn < · − · · · · < , which is false. Thus α > . If α = 6 , then σ ( n ) n + dn < − · · · · · < , which is a contradiction. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is also a contradiction. Subcase 9.11 p = 241 . Since ord (3) = ord (5) = ord (241) = 16 , ord (3) =120 , ord (5) = 40 , ord (17) = 80 , we have β = β = 0 and D > · . If α = 4 , then σ ( n ) n + dn < · − · · · · < , which is false. Thus α > . If α = 6 , then σ ( n ) n + dn < − · · · · · < , which is impossible. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 8 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. Thus α = 2 . However, σ ( n ) n + dn < − · · · − · · · < , which is also a contradiction. Subcase 9.12 p = 251 . Since ord (5) = ord (17) = ord (251) = 2 and ord (3) =ord (5) = 16 , ord (251) = 4 , we have β = β = 0 and D > · . If α = 4 , then σ ( n ) n + dn < · − · · · · < , which is false. Thus α > . If α = 6 , then σ ( n ) n + dn < − · · · · · < , which is impossible. Thus α > . If α = 2 , then σ ( n ) n + dn < · · − · · · < , which is absurd. Thus α > and σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. Subcase 9.13 p ∈ { , , , , , , , , , , , , , , , , } . Since ord (3) = ord (17) = 4 , ord (3) = ord (5) = 16 , ord ( p ) and ord ( p ) are all even, we have β = β = 0 , D > · and σ ( n ) n + dn < · · · · < , which is impossible. Subcase 9.14 p ∈ { , , , } . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Since ord (3) = ord (5) = 16 and ord ( p ) are all even, we have β = 0 .Thus D ∈ { , } and α = 2 . However, | (17 − , a contradiction. Subcase 9.15 p = 307 . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Since ord (3) = ord (17) = ord (307) = 4 , we have β = 0 , D = 625 and α = 4 . However, | (5 − , a contradiction.This completes the proof of Lemma 2.3. (cid:3) Lemma 2.4.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(2.4) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If α = 2 , then | (2 D − . Thus D > . If p > , then σ ( n ) n + dn < − · · · · < , which is false. Thus p = 23 . Since ord (3) = ord (23) = 4 , ord (19) = 2 , ord (5) =ord (19) = 22 , we have β = β = 0 , D > · and σ ( n ) n + dn < − · · · · · < , which is impossible. Thus α > . If p , then σ ( n ) n + dn > − · · − · · − · · − · > , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 13 which is absurd. Thus p > . If D , then σ ( n ) n + dn > − · · − · · − · + 121 > , which is false. Thus D > . Now we divide into the following nine cases according to D . Case 1. D ∈ { , , } . By (2.4), we have α > , α > and σ ( n ) n + dn > − · · − · · − · + 145 > , which is impossible. Case 2. D ∈ { , , , , , , , , , , , , , , , } . Then p = D . Noting that ord (3) = 4 , ord (19) = 2 and ord ( p ) are all even, we deduce that the equality(2.4) can not hold. Case 3. D = 57 . If p > σ ( n ) n + dn < · · · < , which is impossible. If | (5 α +1 − , then | ( α + 1) and (5 − | (5 α +1 − . However, | (5 − , a contradiction. Since ord (3) = 18 and ord (3) = 4 , ord (19) = 2 , we have p and | ( α + 1) . Thus α > , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 157 > , which is impossible. Case 4. D ∈ { , , , } . Then p = D . Noting that ord (3) = 4 , ord (19) = 2 , we have | ( α + 1) and ( p − | ( p α +14 − . However, | (61 − , | (71 − , | (101 − and | (131 − , a contradiction. Case 5. D = 75 . If p > σ ( n ) n + dn < · · · < , which is clearly false. By (2.4), we have α > , α > , p and σ ( n ) n + dn > − · · − · · − · · − · + 175 > , which is impossible. Case 6. D ∈ { , } . If p > σ ( n ) n + dn < · · · < , which is false. Thus p . If | (5 α +1 − , then | ( α + 1) and (5 − | (5 α +1 − .However, | (5 − , a contradiction. Since ord (3) = 18 and ord (3) = 4 , ord (19) = 2 , wehave p and | ( α + 1) . Thus α > , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 195 > , which is impossible. Case 7. D = 125 . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . By (2.4), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 1125 > , which is impossible. Case 8. D = 135 . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . By (2.4), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 1135 > , which is impossible. Case 9. D > . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p .If p , then α = 2 , | (2 D − and D > . Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible.If p , then α > . Otherwise, if α = 2 , then D > and σ ( n ) n + dn < · − · · · < , which is absurd. Subcase 9.1 p . If p = 47 , then α = 4 . Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Since ord (3) = ord (47) = 4 and ord (19) = 2 , we have β = 0 , D > and σ ( n ) n + dn < − · · − · · · < , which is impossible.If p = 53 , then α . Otherwise, if α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. If α = 4 , then σ ( n ) n + dn < − · · − · · · < , which is impossible. Thus α = 6 . Since ord (3) = ord (53) = 4 , ord (19) = 2 , ord (19) = 52 and ord (53) = 18 , we have β = β = β = 0 and − (cid:18) · · · α α (cid:19) = (cid:18) β (cid:19) = 1 , which is a contradiction. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 15 If p = 59 , then β = 0 by ord (3) = 4 and ord (19) = ord (59) = 2 . Thus D > and σ ( n ) n + dn < · − · · · < , which is impossible.If p = 61 , then D . Otherwise, if D > , then σ ( n ) n + dn < · − · · · < , which is impossible. Thus D = 171 . Since ord (3) = 4 , ord (19) = 2 and ord (61) = 5 , wehave | ( α + 1) and (61 − | (61 α +1 − . However, | (61 − , a contradiction. Subcase 9.2 p = 67 . By ord (3) = ord (67) = 18 , we have β = 0 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. If α = 4 , then − (cid:18) (cid:19) = (cid:18) · α α α (cid:19) = (cid:18) β β β (cid:19) = 1 , which is also a contradiction. Subcase 9.3 p = 71 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 4 and | (2 D − . Thus D > and σ ( n ) n + dn < − · · · · < , which is impossible. Subcase 9.4 p ∈ { , , } . Since ord (3) = 4 , ord (19) = 2 and ord ( p ) are all even, wehave β = 0 and D > . If α = 4 , then σ ( n ) n + dn < − · · · · < , which is a contradiction. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is also a contradiction. Subcase 9.5 p = 89 . Since ord (3) = 4 , ord (19) = ord (89) = 2 , we have β = 0 and D > . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. If α = 4 , then σ ( n ) n + dn < − · · · · < , which is also a contradiction. Thus α = 6 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. Thus α = 4 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 2 . By (2.4) and ord (89) = 18 , we have β = 1 , β = β = 0 and − (cid:18) (cid:19) = (cid:18) · · · · α (cid:19) = (cid:18) (cid:19) = 1 , which is absurd. Subcase 9.6 p = 97 . Since ord (3) = ord (97) = 4 , ord (19) = 2 , ord (3) = 48 , ord (5) =ord (19) = 96 , we have β = β = 0 , D > · and σ ( n ) n + dn < · · · · < , which is impossible. Subcase 9.7 p = 101 . If β > , then | ( α + 1) and (101 − | (101 α +1 − . Since | (101 − , we have | (2 D − . Thus D > and σ ( n ) n + dn < · · · D < , which is impossible. Thus β = 0 and | D . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D = 5 and α = 4 . However, we deduce that the equality (2.4) cannot hold. Subcase 9.8 p ∈ { , , , , , , } . Since ord (3) = 4 , ord (19) = 2 and ord ( p ) are all even, we have β = 0 . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D = 5 and α = 4 . However, we deduce that the equality (2.4) cannot hold. Subcase 9.9 p = 131 . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D = 171 and β > . Since ord (3) = 4 , ord (19) = 2 , ord (131) = 5 , wehave (131 − | (131 α +1 − . However, | (131 − , a contradiction.This completes the proof of Lemma 2.4. (cid:3) Lemma 2.5.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(2.5) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If α = 2 , then σ ( n ) n + dn < − · · · · < , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 17 which is false. Thus α > . If p , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus p > . Now we divide into the following five cases according to D . Case 1. D ∈ { , } . By (2.5), we have p ≡ , α > , α > and σ ( n ) n + dn > − · · − · · − · + 125 > , which is impossible. Case 2. D = 27 . If p > , then σ ( n ) n + dn < · · · < , which is impossible. Since ord (3) = ord (23) = 4 , ord (5) = ord (23) = 2 , we have p ≡ . Thus p . By (2.5), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 127 > , which is false. Case 3. D ∈ { , , , , , , } . Then p = D . By (2.5), we have α > and σ ( n ) n + dn > − · · − · · − · · − · + 161 > , which is impossible. Case 4. D = 45 . If p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (23) = 4 , ord (5) = ord (23) = 2 , we have p ≡ . Thus p . By (2.5), we have α > and σ ( n ) n + dn > − · · − · · − · · − · + 145 > , which is impossible. Case 5. D > . If p > , then σ ( n ) n + dn < · · · < , which is impossible. Thus p . Subcase 5.1 p = 37 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 2 and | (2 D − . Since ord (3) = ord (23) = ord (37) = 2 , wehave β = 0 . Thus D > and σ ( n ) n + dn < · − · · · < , which is impossible. Subcase 5.2 p = 41 . Since ord (5) = ord (23) = ord (41) = 2 , we have β = 0 and | D .If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 2 and | (2 D − . Thus D > and σ ( n ) n + dn < · − · · · < , which is impossible. Subcase 5.3 p = 47 . Since ord (5) = ord (23) = ord (47) = 2 , ord (3) = ord (23) =ord (47) = 4 , we have β = β = 0 and D > · . If α = 2 , then σ ( n ) n + dn < · − · · · · < , which is impossible. Thus α > . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is a contradiction. If α = 4 , then σ ( n ) n + dn < − · · · · · < , which is also a contradiction. Subcase 5.4 p ∈ { , } . Since ord (5) = ord (23) =ord ( p ) = 2 , ord (3) = ord (23) =4 and ord ( p ) are all even, we have β = β = 0 , D > · and σ ( n ) n + dn < · · · · < , which is impossible. Subcase 5.5 p = 61 . If α = 2 , then σ ( n ) n + dn < · − · · · < , which is impossible. Thus α > . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Thus D ∈ { , , , , , } . Since ord (3) = ord (23) = 4 and ord (61) = 5 , we have | ( α + 1) and (61 − | (61 α +1 − . Noting that | (61 − , wehave | (2 D − , a contradiction. Subcase 5.6 p = 71 . Since ord (5) = ord (23) = ord (71) = 2 , we have β = 0 . If α = 4 ,then σ ( n ) n + dn < − · · · · < , which is impossible. If α > , then D > and σ ( n ) n + dn < · · · < , which is false. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 19
Subcase 5.7 p ∈ { , , } . Since ord (3) = ord (23) = 4 and ord ( p ) are all even, wehave β = 0 . If α = 2 , then σ ( n ) n + dn < · − · · · < , which is impossible. If α > , then D > and σ ( n ) n + dn < · · · < , which is false.This completes the proof of Lemma 2.5. (cid:3) Lemma 2.6.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(2.6) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If α = 2 , then σ ( n ) n + dn < − · · · · < , which is false. Thus α > . Now we divide into the following four cases according to D . Case 1. D ∈ { , , } . By (2.6), we have α > and α = 4 . Since ord (5) = ord (29) =2 , ord (3) = 4 and ord (29) = 2 , we have p ≡ . If p > , then σ ( n ) n + dn < · · · < , which is impossible. Thus p . If α = 2 , then p = 31 and σ ( n ) n + dn > − · · − · · − · · − · + 129 > , which is false. If α > , then σ ( n ) n + dn > − · · − · · − · · − · + 129 > , which is impossible. Case 2. D ∈ { , } . Then p = D, α > and σ ( n ) n + dn > − · · − · · − · · − · + 137 > , which is impossible. Case 3. D = 43 . Then p = 43 , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 143 > , which is impossible. Case 4. D > . If p >
592 = σ ( n ) n + dn < · · · < , which is impossible. Thus p . Subcase 4.1 p = 31 . Since ord (3) = ord (31) = 28 , ord (5) = 14 , we have β = 0 and D > . If α = 4 , then − (cid:18) (cid:19) = (cid:18) · α α α α (cid:19) = (cid:18) β β β (cid:19) = 1 , which is false. Thus α > . If α = 2 , then σ ( n ) n + dn < · − · · · < , which is a contradiction. If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is also a contradiction. Subcase 4.2 p ∈ { , , , } . Since ord (3) =ord ( p ) = 28 , ord (5) = 14 , we have β = 0 , D > and σ ( n ) n + dn < · · · < , which is impossible. Subcase 4.3 p = 53 . By ord (5) = ord (29) = ord (53) = 2 , we have β = 0 , D > and σ ( n ) n + dn < · · · < , which is impossible.This completes the proof of Lemma 2.6. (cid:3) Lemma 2.7.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(2.7) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If α = 2 , then σ ( n ) n + dn < − · · · · < , which is clearly false. Thus α > . If D > , then σ ( n ) n + dn < · · · < , which is false. If D = p , then σ ( n ) n + dn < · · · < , which is impossible. Thus D ∈ { , , , , , , , , , , , , } . Case 1. D = 15 . By (2.7), we have α > . If α > , then σ ( n ) n + dn > − · · − · · − · + 115 > , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 21 which is impossible. Thus α = 2 . If p > , then σ ( n ) n + dn < · − · · · < , which is clearly false. Thus p . It follows that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible. Case 2. D = 25 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . It follows that α > , α > . If α > , then σ ( n ) n + dn > − · · − · · − · · − · + 125 > , which is impossible. Thus α = 2 . If p > , then σ ( n ) n + dn < · − · · · < , which is a contradiction. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 125 > , which is also a contradiction. Case 3. D ∈ { , , } . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p ∈ { , , , , , , , , , } and β > . Since ord (3) =ord (37) = ord (43) = ord (47) = ord (53) = ord (67) = ord (73) = 4 , ord (59) = 2 and ord (31) = ord (41) = ord (61) = ord (71) = 5 , we have (31 − | (31 α +1 − , (41 − | (41 α +1 − , (61 − | (61 α +1 − or (71 − | (71 α +1 − . However, | − , | − , | − and | − , a contradiction. Case 4. D ∈ { , } . Then p = D . Since ord (3) =ord ( p ) = 4 , ord (31) = 5 , we have | ( α + 1) and (31 − | (31 α +1 − . However, | (31 − , a contradiction. Case 5. D = { , } . Then p = D . Since ord (5) =ord ( p ) = 2 , ord (31) = 3 , we have | ( α + 1) and (31 − | (31 α +1 − . However, | (31 − , a contradiction. Case 6. D ∈ { , , , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p = 37 . If α = 2 , then σ ( n ) n + dn < · − · · · < , which is false. Thus α > and β > . Since ord (3) = ord (37) = 4 and ord (31) = 5 , wehave (31 − | (31 α +1 − . However, | (31 − , a contradiction.This completes the proof of Lemma 2.7. (cid:3) Lemma 2.8.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D ∈ { , , , } . Now we divide into the following four cases. Case 1. D = 15 . If p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (37) = 4 , we have p ≡ . Thus p , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible. Case 2. D = 25 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . It follows that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 125 > , which is impossible. Case 3. D = 27 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . It follows that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 127 > , which is impossible. Case 4. D = 37 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p = 41 . It follows that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 137 > , which is impossible.This completes the proof of Lemma 2.8. (cid:3) Lemma 2.9.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > . N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 23
Proof.
Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If D > , then σ ( n ) n + dn < · · · < , a contradiction. Thus D ∈ { , , } .If D = 15 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is clearly false. It follows that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible.If D ∈ { , } , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is clearly false. It follows that α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 127 > , which is impossible.This completes the proof of Lemma 2.9. (cid:3) Lemma 2.10.
There is no odd deficient-perfect number n of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If D > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus D ∈ { , } .If D = 15 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (43) = 4 , we have p ≡ . Thus p , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible.If D = 25 , then p = 47 . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 125 > , which is impossible.This completes the proof of Lemma 2.10. (cid:3) Lemma 2.11.
There is no odd deficient-perfect number of the form n = 3 α α p α p α with p .Proof. Assume that n = 3 α α p α p α is an odd deficient-perfect number with deficient divisor d = 3 β β p β p β and p . If D > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus D = 9 . By (1.1), we have(2.8) α +1 − · α +1 − · p α +13 − p − · p α +14 − p − · α − α p α p α . Thus α > and α > . If α = 6 , then p = 1093 . Since ord (3) = ord (5) = ord (1093) =16 , we have p ≡ . Thus p ∈ { , , } . If p = 239 , then σ ( n ) n + dn < · · · < , which is clearly false. If p ∈ { , } , then σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. By (2.8), we have α > . If p , then σ ( n ) n + dn > − · · − · · − · + 19 > , which is impossible. Thus p > . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (2.8), we have α > . Now we divide into thefollowing twelve cases according to p . Case 1. p = 137 . Since ord (5) = ord (137) = 2 and ord (3) = ord (137) = 4 , we have p ≡ . Noting that ord (137) = 17 , we have if | ( α + 1) and (137 − | (137 α +1 − , then | (137 − , a contradiction. Since ord (3) = ord (5) = 16 , we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 2. p = 139 . If p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (5) = ord (139) = 16 , we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 25
Case 3. p = 149 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Noting that ord (3) = ord (5) = 16 and ord (149) = 4 , we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 4. p = 151 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 5. p = 157 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 6. p ∈ { , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Since ord (3) = ord (5) = ord (163) = ord (167) = 16 , we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 7. p = 173 . If p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (5) = ord (173) = 16 , we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 8. p ∈ { , , , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Since ord (3) = ord (5) = 16 and ord ( p ) are even, we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 9. p ∈ { , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Since ord (3) = ord (5) = 16 and ord ( p ) are even, we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 10. p ∈ { , , , , , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Since ord (3) = ord (5) = 16 and ord ( p ) are even, we have p ≡ . Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 11. p = 239 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p , α > and σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is impossible. Case 12. p ∈ { , , , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . Since ord (3) = ord (5) = 16 and ord ( p ) , ord ( p ) are even, we deduce that the equality (2.8) can not hold.This completes the proof of Lemma 2.11. (cid:3)
3. T
HE CASE OF p = 7 In this section, we consider the case of n = p α p α p α p α with p = 3 and p = 7 . Lemma 3.1. If n = 3 α α α p α is an odd deficient-perfect number with D > , then n =3 · · · with deficient divisor d = 3 · · . N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 27
Proof.
By (1.1), we have(3.1) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . We will divide into the following seven cases according to D . Case 1. D = 7 . If α > , then σ ( n ) n + dn > − · · − · · − · + 17 > , which is clearly false. Thus α = 2 . If p , then σ ( n ) n + dn > − · · − · · − · · − · + 17 > , which is impossible. If p > , then σ ( n ) n + dn < · · · < , which is also impossible. Thus p . By (3.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 17 > , which is false. Case 2. D ∈ { , } . If α > , then σ ( n ) n + dn > − · · − · · − · + 111 > , which is clearly false. Thus α = 2 , p = 13 and σ ( n ) n + dn > − · · − · · − · · − · + 111 > , which is impossible. Case 3. D ∈ { , , } . Then p = D and σ ( n ) n + dn > − · · − · · − · · p − p − · p + 1 p > , which is impossible. Case 4. D = 21 . If α = 2 , then p = 13 and σ ( n ) n + dn > − · · − · · − · · − · + 121 > , which is impossible. Thus α > . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 121 > , which is false. Thus p . By (3.1), we have α > and α > . If p , then σ ( n ) n + dn > − · · − · · − · · − · + 121 > , which is absurd. Thus p ∈ { , , , } . Since ord (3) = 8 , ord (7) = ord (11) =ord (67) = ord (71) = 40 , ord (61) = 20 and ord (59) = 5 , we have p = 59 , | ( α + 1) and (59 − | (59 α +1 − . However, | (59 − , a contradiction. Case 5. D ∈ { , , } . Then p = D, α > and σ ( n ) n + dn > − · · − · · − · · p − p − · p + 1 p > , which is impossible. Case 6. D = 27 . Then α > . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 127 > , which is false. Thus p ∈ { , , } . Since ord (3) = ord (47) = 6 , ord (41) = 2 , ord (11) =3 and ord (43) = 7 , we have | ( α + 1) and (11 − | (11 α +1 − or | ( α + 1) and (43 − | (43 α +1 − . However, | (11 − and | (43 − , a contradiction. Case 7. D > . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . Subcase 7.1 p = 13 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 2 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is impossible. Thus α = 2 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is false. Thus α = 2 . By (3.1), we have n = 3 · · · and d = 3 · · . Subcase 7.2 p ∈ { , } . If α > , then σ ( n ) n + dn > − · · − · · − · · p − p − · p > , which is clearly false. Thus α = 2 and σ ( n ) n + dn < − · · · · < , which is impossible. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 29
Subcase 7.3 p = 23 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is clearly false. If α = 2 , then σ ( n ) n + dn < − · · · · < , which is impossible. Thus α = 4 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 . If D , then σ ( n ) n + dn > − · · − · · − · · − · + 1303 > , which is a contradiction. If D > , then σ ( n ) n + dn < − · · − · · · < , which is also a contradiction. Thus D ∈ { , , , , , } . However, | (7 − and ∤ (2 D − d , which contradicts with (3.1). Subcase 7.4 p = 29 . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D ∈ { , , , , , , , , } . Since ord (3) = ord (11) = 28 and ord (7) = 7 , we have | ( α + 1) and (7 − | (7 α +1 − . However, | (7 − , acontradiction. Subcase 7.5 p = 31 . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D ∈ { , , , , } . Since ord (3) = ord (11) = 30 and ord (7) =15 , we have | ( α + 1) and (7 − | (7 α +1 − . However, | (7 − , a contradiction. Subcase 7.6 p = 37 . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D ∈ { , } . Noting that ord (3) = 6 , ord (11) = ord (37) = 3 , we have | ( α + 1) and (11 − | (11 α +1 − or | ( α + 1) and (37 − | (37 α +1 − . However, | (11 − and | (37 − , a contradiction. Subcase 7.7 p = 41 . If D > , then σ ( n ) n + dn < · · · < , which is false. Thus D = 33 . Noting that ord (3) = 6 , ord (41) = 2 and ord (11) = 3 , we have | ( α + 1) and (11 − | (11 α +1 − . However, | (11 − , a contradiction.This completes the proof of Lemma 3.1. (cid:3) Lemma 3.2.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > . Proof.
Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(3.2) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . We will divide into the following six cases according to D . Case 1. D = 7 . If α > , then σ ( n ) n + dn > − · · − · · − · · + 17 > , which is clearly false. Thus α = 2 . If p > , then σ ( n ) n + dn < − · · · · < , which is a contradiction. If p , then σ ( n ) n + dn > − · · − · · − · · − · + 17 > , which is also a contradiction. Thus p = 53 . Since ord (13) = 2 and ord (53) = 3 , we have | ( α + 1) and (53 − | (53 α +1 − . However, | (53 − , a contradiction. Case 2. D = 9 . If p , then σ ( n ) n + dn > − · · − · · − · · − · + 19 > , which is clearly false. Thus p > . If α > , then σ ( n ) n + dn > − · · − · · − · + 19 > , which is impossible. Since ord (7) = 16 , ord (13) = 4 , we have p ≡ . Thus p > and α = 2 . However, σ ( n ) n + dn < − · · · · < , a contradiction. Case 3. D = 13 . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . Noting that ord (3) = ord (7) = ord (13) = 4 , we have p ∈{ , , } , | ( α + 1) and ( p − | ( p α +14 − . However, | (31 − , | (41 − , | (61 − , a contradiction. Case 4. D = 17 . Then p = 17 . Since ord (3) = ord (7) = 16 and ord (13) = 4 , wededuce that equality (3.2) cannot hold. Case 5. D = 19 . Then p = 19 . Since ord (3) = 18 , ord (13) = ord (19) = 36 and ord (7) = 9 , we have | ( α + 1) and (7 − | (7 α +1 − . However, | (7 − , acontradiction. Case 6. D > . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 31
Subcase 6.1 p = 17 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is clearly false. If α = 2 , then σ ( n ) n + dn < − · · · · < , which is impossible. Thus α = 4 . If α > , then σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Thus α = 2 . Noting that ord (13) = 2 , ord (17) = 6 , ord (17) = 6 and ord (13) = 16 , we have β = β = β = 0 and − (cid:18) (cid:19) = (cid:18) · · · α α (cid:19) = (cid:18) β (cid:19) = 1 , which is a contradiction. Subcase 6.2 p = 19 . Noting that ord (3) = ord (19) = 6 and ord (13) = 2 , we have β = 0 and D > .If α = 2 , then σ ( n ) n + dn < − · · · · < , which is clearly false. If α = 4 , then | (2 D − . Thus D > and σ ( n ) n + dn < − · · · · < , which is absurd. If α = 6 , then − (cid:18) (cid:19) = (cid:18) · · α α α (cid:19) = (cid:18) β β β (cid:19) = 1 , which is a contradiction. Thus α > .If α = 2 , then − (cid:18) (cid:19) = (cid:18) · α α · · α (cid:19) = (cid:18) β β β (cid:19) = 1 , which is a contradiction. Thus α > .If α = 2 , then D . Otherwise, if D > , then σ ( n ) n + dn < · − · · · < , which is clearly false. If D , then σ ( n ) n + dn > − · · − · · − · · − · + 1197 > , which is impossible. Thus D . However, the case cannot hold since | D . Thus α > and σ ( n ) n + dn > − · · − · · − · · − · > , which is absurd. Subcase 6.3 p = 23 . If D > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus D ∈ { , , , , } . Since ord (3) = 6 , ord (13) = 2 and ord (23) = 3 , we have if | (2 D − d , then | ( α + 1) and (23 − | (23 α +1 − . However, | (23 − , a contradiction. Thus D = 49 , α = 2 and σ ( n ) n + dn < · − · · · < , which is also a contradiction. Subcase 6.4 p = 29 . If D > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus D ∈ { , } . Since ord (3) = 8 and ord (7) = ord (13) =ord (29) = 40 , we have D = 27 and α > . Since ord (29) = 2 , we have | ( α + 1) and (7 − | (7 α +1 − or | ( α + 1) and (13 − | (13 α +1 − . However, | (7 − and | (13 − , a contradiction. Subcase 6.5 p = 31 . If D > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus D = 21 . Noting that ord (3) = ord (31) = 6 and ord (13) = 2 , wededuce that the equality (3.2) cannot hold.This completes the proof of Lemma 3.2. (cid:3) Lemma 3.3.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(3.3) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Thus D ∈ { , , , , , } . Case 1. D = 7 . Since ord (3) = ord (7) = 16 , we have p > . If α = 2 , then σ ( n ) n + dn < − · · · · < , which is false. Thus α > , α > and σ ( n ) n + dn > − · · − · · − · + 17 > , which is a contradiction. Case 2. D ∈ { , , , , } . If p > , then σ ( n ) n + dn < · · · < , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 33 which is absurd. Thus p . Since ord (3) = ord (7) = 16 and ord ( p ) are all even, wededuce that the equality (3.3) cannot hold.This completes the proof of Lemma 3.3. (cid:3) Lemma 3.4.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(3.4) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Thus D ∈ { , } . Case 1. D = 7 . If α = 2 , then p = 23 . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is absurd. Since ord (19) = 6 and ord (23) = 3 , we have | ( α + 1) and (23 − | (23 α +1 − . However, | (23 − , a contradiction. By (3.4), we have α > . If p ,then σ ( n ) n + dn > − · · − · · − · · − · + 17 > , which is false. Since ord (3) = ord (19) = 6 , we have p ∈ { , , , , } . If p = 127 , then | ( α + 1) and (127 − | (127 α +1 − , However, | (127 − , acontradiction. If p ∈ { , , , } , then | ( α + 1) and ( p − | ( p α +14 − . However, | (137 − , | (149 − , | (151 − , | (163 − , which is also a contradiction. Case 2. D = 9 . Since ord (3) = ord (7) = 16 , ord (19) = 8 and ord (3) = ord (19) = 6 ,we have p = 137 . Thus | ( α +1) and (137 − | (137 α +1 − . However, | (137 − ,a contradiction.This completes the proof of Lemma 3.4. (cid:3) Lemma 3.5.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . By (1.1), we have(3.5) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α α p α − β β β p β . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p ∈ { , , , , , , , , , } . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Thus D ∈ { , } . Case 1. D = 7 . If α = 2 , then σ ( n ) n + dn < − · · · · < , which is absurd. By (3.5), we have α > , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 17 > , which is a contradiction. Case 2. D = 9 . Since ord (3) = ord (7) = ord (23) = 16 and ord ( p ) are all even, wededuce that the equality (3.5) cannot hold.This completes the proof of Lemma 3.5. (cid:3) Lemma 3.6.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p ∈ { , , } . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Thus D = 7 . By (1.1), we have(3.6) α +1 − · α +1 − · α +1 − · p α +14 − p − · α α − α p α . If α = 2 , then σ ( n ) n + dn < − · · · · < , which is absurd. By (3.6), we have α > , α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 17 > , which is a contradiction.This completes the proof of Lemma 3.6. (cid:3) Lemma 3.7.
There is no odd deficient-perfect number of the form n = 3 α α α p α with D > .Proof. Assume that n = 3 α α α p α is an odd deficient-perfect number with deficient divisor d = 3 β β β p β . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p = 37 . If D > , then σ ( n ) n + dn < · · · < , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 35 which is impossible. Thus D = 7 . By (1.1), we have α +1 − · α +1 − · α +1 − · α +1 −
136 = 13 · α · α − · α · α . Since ord (3) = ord (31) = 6 and ord (37) = 3 , we have | ( α +1) and (37 − | (37 α +1 − .However, | (37 α +1 − , a contradiction.This completes the proof of Lemma 3.7. (cid:3)
4. T
HE CASE OF p ∈ { , } In this section, we study the case of n = p α p α p α p α with p = 3 and p ∈ { , } . Lemma 4.1.
There is no odd deficient-perfect number of the form n = 3 α α p α p α .Proof. Assume that n = 3 α α p α p α is an odd deficient-perfect number with deficient divisor d = 3 β β p β p β . By (1.1), we have(4.1) α +1 − · α +1 − · p α +13 − p − · p α +14 − p − · α α p α p α − β β p β p β . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . Case 1. p = 13 . If D > , then σ ( n ) n + dn < · · · < , which is false. If D = 3 , then σ ( n ) n + dn > − · · − · · − · + 13 > , which is absurd. Thus D = 9 . If p > , then σ ( n ) n + dn < · · · < , which is impossible Thus p = 17 . Since ord (3) = ord (11) = 16 and ord (13) = 4 , weknow that the equality (4.1) cannot hold. Case 2. p > . If D > , then σ ( n ) n + dn < · · · < , which is impossible. Thus D = 3 . By (4.1), we have α > and α > . If p , then σ ( n ) n + dn > − · · − · · − · + 13 > , which is clearly false. Thus p > .If α = 6 , then p = 1093 and p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction. Thus α = 6 . By (4.1), we have α = 8 and α = 10 . If p > ,then α = 4 . Otherwise, if α = 4 , then σ ( n ) n + dn < − · · · · < , which is impossible. Subcase 2.1 p ∈ { , , , , } . If α > , then σ ( n ) n + dn > − · · − · · − · + 13 > , which is clearly false. Thus α = 4 .If p = 73 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. By (4.1), we have α > . If α = 2 , then p = 1801 and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Thus α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction.If p = 79 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. By (4.1), we have α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction.If p = 83 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. By (4.1), we have α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction.If p = 89 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. By (4.1), we have α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 37 If p = 97 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.2 p = 101 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. By (4.1), we have α > . If α = 4 , then p = 3221 and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. Thus α > . Noting that | (101 − , we have α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , a contradiction. Subcase 2.3 p = 103 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. Noting that ord (3) = ord (103) = 4 and | (11 − , we have α > and p ≡ . Thus p and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > and α > . If α = 4 , then p = 3221 and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Thus α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction. Subcase 2.4 p = 107 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.5 p = 109 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.6 p = 113 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 39 which is a contradiction.
Subcase 2.7 p = 127 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.8 p = 131 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.9 p = 137 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p .If α = 4 , then p = 139 . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible.It follows that σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is absurd. By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.10 p = 139 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.11 p ∈ { , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.12 p = 157 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.13 p = 163 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.14 p = 167 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.15 p = 173 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 41 which is a contradiction.
Subcase 2.16 p = 179 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.17 p = 181 . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . By (4.1), we have α > , α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is a contradiction. Subcase 2.18 p ∈ { , , } . If p > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus p . Noting that ord (3) = 4 and ord ( p ) are all even, we have | ( α + 1) and (11 − | (11 α +1 − or | ( α + 1) and (191 − | (191 α +1 − . However, | (11 − and | (191 − , a contradiction.This completes the proof of Lemma 4.1. (cid:3) Lemma 4.2.
There is no odd deficient perfect number of the form n = 3 α α p α p α .Proof. Assume that n = 3 α α p α p α is an odd deficient-perfect number with deficient divisor d = 3 β β p β p β . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . If D > , then σ ( n ) n + dn < · · · < , which is absurd. Thus D = 3 . By (1.1), we have(4.2) α +1 − · α +1 − · p α +13 − p − · p α +14 − p − · α − α p α p α . Since | (3 − , we have α = 4 . If p > , then α > . Otherwise, if α = 2 , then σ ( n ) n + dn < − · · · · < , which is impossible. If p > , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. Now we divide into the following seven cases according to p . Case 1. p ∈ { , , , , , } . If α > , then σ ( n ) n + dn > − · · − · · − · + 13 > , which is impossible. Thus α = 2 and p ∈ { , , , } .If p = 17 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is impossible. Since ord (13) = 4 and ord (13) = ord (17) = 4 , we have p ≡ . Thus p > , a contradiction.If p = 19 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is false. Since ord (13) = 4 and ord (19) = 2 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is impossible.If p = 23 , then p . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is false. Since ord (13) = ord (23) = 4 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is impossible.If p = 29 , then p = 31 . Otherwise, if p > , then σ ( n ) n + dn < − · · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction. Case 2. p = 41 . If p > , then σ ( n ) n + dn < · · · < , which is impossible. Since ord (3) = ord (13) = 4 and | (41 − , we have p ≡ and | ( α + 1) . Thus p and α = 4 . By (4.2), we have α > and α > .If α = 2 , then p = 61 and (61 − | (61 α +1 − . However, | (61 − , a contradiction.Thus α > and σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction. N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 43
Case 3. p ∈ { , , , } . Since ord (3) = ord (13) and ord ( p ) are all even, we have p ≡ and | ( α + 1) . By (4.2), we have α > and α > . If α = 2 , then p = 61 and (61 − | (61 α +1 − . However, | (61 − , a contradiction. Thus α > .If p = 43 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction.If p = 47 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction.If p = 53 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction.If p = 59 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is a contradiction. However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is also a contradiction. Case 4. p = 61 . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . Since ord (3) = ord (13) = 4 , ord (61) = ord (71) =ord (101) = 5 , we have (61 − | (61 α +1 − or (71 − | (71 α +1 − or (101 − | (101 α +1 − . However, | (61 − , | (71 − , | (101 − , a contradiction. Case 5. p = 67 . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . Observing that ord (3) = ord (13) = ord (67) = 4 , we have p ≡ and | ( α + 1) . Thus p = 71 and (71 − | (71 α +1 − . However, | (71 − , a contradiction. Case 6. p = 71 . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . Since ord (3) = ord (13) = 4 and ord ( p ) are all even, we have | ( α + 1) and (71 − | (71 α +1 − . However, | (71 − , a contradiction. Case 7. p = 73 . If p > , then σ ( n ) n + dn < · · · < , a contradiction. Thus p ∈ { , } . Observing that ord (3) = ord (13) = ord (73) = 4 and ord ( p ) are all even, we see that the case cannot hold.This completes the proof of Lemma 4.2. (cid:3)
5. P
ROOFS
Proof of Theorem 1.
Assume that n = p α p α p α p α is an odd deficient-perfect number with defi-cient divisor d = p β p β p β p β . By [6], we need to consider p = 3 and p ∈ { , , , , } . Case 1. p = 5 . If D = 3 , then σ ( n ) n + dn > − · · − · + 13 > , which is clearly false. If D = 5 , then by (1.1), we have α +1 − · α +1 − · p α +13 − p − · p α +14 − p − α +2 α − p α p α . If α > , then σ ( n ) n + dn > − · · − · + 15 > , which is absurd. Thus α = 2 . If α > , then σ ( n ) n + dn > − · · − · + 15 > , which is impossible. Thus α = 2 , p = 13 and p = 31 . However, σ ( n ) n + dn > − · · − · · − · · − · + 15 > , a contradiction. If p = 7 , then σ ( n ) n + dn > − · · − · · − · > , which is a contradiction. Thus D > . If p > , then σ ( n ) n + dn < · · · < , which is false. Thus p . If p , then D > . Otherwise, if D = 9 , then α > and σ ( n ) n + dn > − · · − · · − · + 19 > , which is impossible. Now we divide into the following five subcases according to p . Subcase 1.1 p ∈ { , , } . By Lemma 2.1-Lemma 2.3, there is no odd deficient-perfectnumber of the form n = 3 α α p α p α . N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 45
Subcase 1.2 p ∈ { , , } . If D = 15 , then α > . If p = 31 , then σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is a contradiction. If p > , then α > and σ ( n ) n + dn > − · · − · · − · + 115 > , which is also a contradiction. Thus D > . By Lemma 2.4-Lemma 2.6, there is no odd deficient-perfect number of the form n = 3 α α p α p α . Subcase 1.3 p ∈ { , , , } . By Lemma 2.7-Lemma 2.10, there is no odd deficient-perfectnumber of the form n = 3 α α p α p α . Subcase 1.4 p ∈ { , , , } . If D > , then σ ( n ) n + dn < · · · < , which is clearly false. Thus D = 15 , α > and α > .If p = 47 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (47) = 4 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible.If p = 53 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = ord (53) = 4 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible.If p = 59 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is false. Since ord (3) = 4 , ord (59) = 2 , we have p and σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible.If p = 61 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is clearly false. However, σ ( n ) n + dn > − · · − · · − · · − · + 115 > , which is impossible. Subcase 1.5 p . By Lemma 2.11, there is no odd deficient-perfect number of theform n = 3 α α p α p α . Case 2. p = 7 . If D = 3 , then σ ( n ) n + dn > − · · − · + 13 > , which is clearly false. Thus D > . If p > , then σ ( n ) n + dn < · · · < , which is impossible. Thus p . By Lemma 3.1-Lemma 3.7, we have n = 3 · · · with deficient divisor d = 3 · · . Case 3. p ∈ { , } . By Lemma 4.1-Lemma 4.2, there is no odd deficient-perfect number ofthe form n = 3 α p α p α p α . Case 4. p = 17 . If D > , then σ ( n ) n + dn < · · · < , which is absurd. Thus D = 3 . If p > , then σ ( n ) n + dn < · · · < , which is false. If p = 19 , then α > and σ ( n ) n + dn > − · · − · · − · + 13 > , which is impossible. Thus p ∈ { , , , , , } . By (1.1), we have α +1 − · α +1 − · p α +13 − p − · p α +14 − p − · α − α p α p α and α > . Since ord (3) = 16 and ord ( p ) are all even, we have p ≡ .If p = 23 , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is false. Noting that ord (3) = ord (17) = ord (23) = 4 , we have p ≡ and α > . Noting that ord (17) = ord (23) = 2 , we have p ≡ . Thus p ∈{ , , , } . It follows that α > , α > and α > . However, σ ( n ) n + dn > − · · − · · − · · − · + 13 > , which is impossible.If p ∈ { , , , , } , then p . Otherwise, if p > , then σ ( n ) n + dn < · · · < , which is false. Thus p = 103 and p = 29 . Since ord (3) = ord (17) = ord (103) = 4 and ord (29) = 2 , we know that the case can not occur.This completes the proof of Theorem 1. (cid:3) N ODD DEFICIENT-PERFECT NUMBERS WITH FOUR DISTINCT PRIME DIVISORS 47 R EFERENCES [1] J. Ahn, Y.S. Shin, The minimal free resolution of a star-configuration in P n and the weak-Lefschetz property.J. Korean Math. Soc. (2012), 405–417.[2] A. Anavi, P. Pollack and C. Pomerance, On congruences of the form σ ( n ) ≡ a (mod n ) , Int. J. NumberTheory 9 (2013), 115124.[3] G. L. Cohen, On odd perfect numbers (II), multiperfect numbers and quasiperfect numbers , J. Aust. Math. Soc.29 (1980), 369384.[4] P. Hagis and G. L. Cohen,
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