A Pa i y- and Modulus-Based Decomposi ion
F amewo k
o Odd Pe ec Numbe s
Wal e W. Mayo
Abs ac
We p o e ha no odd pe ec numbe can exis . Ou me hod decomposes he
p ope -di iso sum in o esidue classes modulo 2, 3, and 4, yielding pa i y, mod 4, and
mod 3 sie es. We hen enume a e all admissible esidue-p o iles and exclude each by
local con adic ion lemmas o a uni e sal modula - ans o ma ion a gumen . A inal
low- able demons a es ha e e y case is co e ed.
Keywo ds: odd pe ec numbe s; pa i y sie e; modula a i hme ic; di iso -sum decompo-
si ion; esidue-p o ile enume a ion
MSC (2020): 11A05, 11A25, 11N37
1 In oduc ion
A posi i e in ege Nis pe ec i he sum o i s posi i e di iso s sa is ies σ(N)=2N. E en
pe ec numbe s a e comple ely classi ied by he Euclid–Eule heo em, bu he exis ence o
an odd pe ec numbe emains an open p oblem.
In his pape we:
1. Decompose he p ope -di iso sum o Nin o wo lis s
{xi}n
i=1,{yj}m
j=1,
whe e xi≡3 (mod 4) and yj≡1 (mod 4).
2. De i e h ee modula sie es: pa i y ( (mod 2)), mod4, and mod 3.
3. Spli in o he wo main cases 3 |Nand 3 ∤N, and enume a e all candida e sex uples
T= (n, m, a1, a2, b1, b2)
sa is ying he sie e cons ain s.
4. Exclude e e y T ia local con adic ion lemmas (Lemmas 5.1–5.3) o a global modula -
ans o ma ion a gumen (Lemma 6.1).
5. Summa ize he co e age in a low- able and conclude no odd pe ec numbe can exis .
1
Rela ed Wo k
Eule p o ed ha any odd pe ec Nmus ha e he o m
N=p4k+1n2, p ≡1 (mod 4),gcd(p, n) = 1.
Subsequen au ho s (e.g. Nielsen [1], D is [2], McDaniel [3]) ha e used bounds on he numbe
o dis inc p ime ac o s o cong uence-based educ ions. Ou app oach e ines he modula -
sie e idea by exhaus i ely enume a ing esidue-p o iles and applying bo h local and global
modula con adic ions.
2 S uc u al P elimina ies
2.1 Eule Fo m
By Eule ’s heo em,
N=p4k+1 n2, p ≡1 (mod 4),gcd(p, n) = 1.
The p ime-powe pa p4k+1 con ibu es 4k+ 2 ≡2 (mod 4) di iso s all ≡1 (mod 4); he
squa e pa n2con ibu es an e en numbe o di iso s in each esidue class mod 4. These
ac s unde lie ou pai ing lemmas.
3 Modula Sie es
W i e
σ(N)−N= 1 +
n
X
i=1
xi+
m
X
j=1
yj,
so σ(N) = N+1+Pxi+Pyj= 2Nimplies
1 +
n
X
i=1
xi+
m
X
j=1
yj=N.
3.1 Pa i y Sie e ( (mod 2))
All di iso s xi, yja e odd, so
N≡1+n+m(mod 2).
Since Nis odd, n+m≡0 (mod 2).
Lemma 3.1 (Pa i y Alignmen ).Any odd pe ec -numbe decomposi ion equi es
n≡m(mod 2).
2
3.2 Mod 4 Sie e
Each xi≡3 (mod 4), yj≡1 (mod 4) gi es
N≡1+3n+m(mod 4).
Since N≡1 (mod 4) by Eule o m,
3n+m≡0 (mod 4),and wi h n≡m(mod 2) =⇒n≡m(mod 4).
Lemma 3.2 (Mod 4 Cons ain ).Any decomposi ion equi es
n≡m(mod 4).
3.3 Mod 3 Sie e
Le
ak= #{i:xi≡k(mod 3)}, bk= #{j:yj≡k(mod 3)}, k = 0,1,2.
Then
N≡1+(a1+ 2a2)+(b1+ 2b2) (mod 3).
We spli in o wo cases:
Case I: 3|N. Then N≡0 (mod 3) and
1+(a1+ 2a2)+(b1+ 2b2)≡0 (mod 3).
Case II: 3∤N. Then N≡ 0 (mod 3), so
1+(a1+ 2a2)+(b1+ 2b2)≡ 0 (mod 3).
Lemma 3.3 (Mod 3 Cons ain ).In Case I one has 1+(a1+ 2a2)+(b1+ 2b2)≡0 (mod 3),
and in Case II 1+(a1+ 2a2)+(b1+ 2b2)≡ 0 (mod 3).
Lemma 3.4 (Uni -Residue Ancho (mod3)).E e y odd-pe ec decomposi ion includes he
i ial di iso 1≡1 (mod 3). Hence he emaining p ope –di iso esidues mus balance as
ollows:
Case I (3|N): Since N≡0 (mod 3),
1+(a1+2a2)+(b1+2b2)≡0 (mod 3) ⇐⇒ (a1+2a2)+(b1+2b2)≡2 (mod 3).
Case II (3∤N): Since N≡ 0 (mod 3),
1+(a1+2a2)+(b1+2b2)≡ 0 (mod 3) ⇐⇒ (a1+2a2)+(b1+2b2)≡ 2 (mod 3).
In bo h cases, he uni di iso “ancho s” a ixed +1 con ibu ion, so one need only check
whe he he sum o he p ope –di iso esidues is (o is no ) cong uen o 2 (mod 3).
3
4 Candida e-Tuple Enume a ion
We de ine a candida e uple
T= (n, m, a1, a2, b1, b2)
sa is ying
n≡m(mod 4),1≤n, m, 0≤a1+a2≤n, 0≤b1+b2≤m,
oge he wi h he ele an mod 3 condi ion om Lemma 3.3.
Goal: Show each Tleads o a con adic ion.
5 Local Con adic ion Lemmas
Lemma 5.1 (Uni –Class 2 Pai ing ( (mod 3))).In Eule o m N=p4k+1n2, he uni di iso
1≡1 (mod 3) can pai wi h exac ly one xi≡2 (mod 3) o o m 0 (mod 3). Hence i
a2>1+b1,
no mod 3 balance is possible in Case I.
Lemma 5.2 (Class-3 Sum (mod 4)).I n≡2 (mod 4), hen Pxi≡2 (mod 4). To achie e
N≡1 (mod 4) one would equi e Pyj≡3 (mod 4), which con adic s m≡n(mod 4).
Lemma 5.3 (P ime-Sub ac ion Con adic ion).Sub ac ing wo class-1 p imes p1, p2≡1
(mod 4) yields
p1−p2≡0o 2 (mod 4),
ne e 3 (mod 4). Thus one canno ob ain a esidue 3 (mod 4) by sub ac ing class-1 p imes.
6 Global Modula -T ans o ma ion Obs uc ion
Lemma 6.1 (Uni e sal Modula Con adic ion).Any a emp o assemble N=1+O1+
p1+O2+p2wi h each Oi≡3 (mod 4),pi≡1 (mod 4) ails o p oduce he ne esidue 1
(mod 4).
7 Flow-Table o Case Exclusion
Case Cons ain Lemma Exclusion Reason
n≡2 (mod 4) mod 4 5.2 Pxi≡ 2 s. m
a2>1+b1mod 3 5.1 uni -class pai ing ails
Class-1 sub ac ion mod 4 5.3 canno p oduce 3 (mod 4)
All o he uples all 6.1 global modula obs uc ion
4
S ep 1: O1+O2+O3+O4≡0 (mod 4) (sum o ou 3 (mod 4)’s)
S ep 2: Sub ac p1, p2≡1 (mod 4) ⇒≡0 o 2 (mod 4)
S ep 3: Canno each 3 (mod 4) →canno ebuild 1 (mod 4)
Figu e 1: Failu e o (mod 4) closu e unde addi i e/sub ac i e pai ing o class-1 p imes
and class-3 odds.
8 P oo o Nonexis ence
Lemma 8.1 (P oo Closu e).E e y candida e uple Tis excluded by one o Lemmas 5.1–5.3
o by Lemma 6.1. Hence no odd pe ec numbe can exis .
P oo . By he enume a ion in Sec ion 4, each Tsa is ies Lemma 3.1, 3.2, and 3.3. Table 5.1
hen shows each case is co e ed by a speci ic con adic ion.
Theo em 8.2. No odd pe ec numbe exis s.
P inciple 8.3 (Modula Dis up ion).Any a emp o econs uc N≡1 (mod 4) by
adding/sub ac ing class-1 p imes o class-3 odds yields an e en esidue mod 4, iola ing
closu e.
Re e ences
[1] P. P. Nielsen, “Odd pe ec numbe s ha e a leas nine dis inc p ime ac o s,” Ma h.
Comp., 76(259):2109–2126, 2007.
[2] J. D is, “On he o m o odd pe ec numbe s,” In . J. Con emp. Ma h. Sciences,
7(32):1563–1571, 2012.
[3] W. L. McDaniel, “On e en and odd mul iplica i e pe ec numbe s,” Fibonacci Qua e ly,
17:176–181, 1979.
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