On the Additive Uniqueness of Generalized Nonagonal Numbers for Multiplicative Functions

#A99 INTEGERS 25 (2025)
ON THE ADDITIVE UNIQUENESS OF GENERALIZED
NONAGONAL NUMBERS FOR MULTIPLICATIVE FUNCTIONS
Poo-Sung Pa k1
Depa men o Ma hema ics Educa ion, Kyungnam Uni e si y, Changwon,
Republic o Ko ea
[email p o ec ed]
Recei ed: 2/28/25, Re ised: 6/30/25, Accep ed: 10/10/25, Published: 11/5/25
Abs ac
We p o e ha he se No all nonze o gene alized nonagonal numbe s is an addi i e
uniqueness se . I a mul iplica i e unc ion sa is ies he equa ion
(a+b) = (a) + (b)
o all a, b ∈ N, hen is he iden i y unc ion.
1. In oduc ion
In 1992, C. Spi o [11] in oduced he no ion o an addi i e uniqueness se , b ie ly
AU se , E⊂No a subse So a i hme ic unc ions, which means ∈Sis uniquely
de e mined by he condi ion (a+b) = (a) + (b) o all a, b ∈E. She showed
ha he se o p imes is an addi i e uniqueness se o he se
S={ | is mul iplica i e and (p0)= 0 o some p ime p0},
whe e is mul iplica i e i (1) = 1 and (ab) = (a) (b) o all aand bwi h
gcd(a, b) = 1. Since he pape was published, many ma hema icians ha e been
s udying he k-addi i e uniqueness, b ie ly k-AU, o a ious se s o na u al numbe s
wi h he condi ion
(a1+a2+···+ak) = (a1) + (a2) + ···+ (ak).
In 1999, Chung and Phong [2] showed ha he se o iangula numbe s is an
AU se o he se o mul iplica i e unc ions. This se is also a k-AU se wi h k≥3
DOI: 10.5281/zenodo.17535282
1This wo k was suppo ed by he Na ional Resea ch Founda ion o Ko ea (NRF) g an unded
by he Ko ean go e nmen (MSIT) (RS-2021-NR058832).
INTEGERS: 25 (2025) 2
[9]. Howe e , he se o squa es is no a 2-AU se o mul iplica i e unc ions [1]
and i is a k-AU se o k≥3 [8].
Le us conside he addi i e uniqueness o polygonal numbe s o mul iplica i e
unc ions o gene alize he abo e esul s. The au ho and colleagues [6] p o ed ha
he se P={n(3n−2)
2|n∈Z, n = 0}o gene alized pen agonal numbe s is an AU
se o mul iplica i e unc ions. This is also a k-AU se o k≥3 [3, 10].
I is much mo e di icul o conside he se P+={n(3n−2)
2|n∈Z, n ≥1}o
o dina y pen agonal numbe s. This se is also a 2-AU se , which was p o ed in [7].
The se H+={n(2n−1) |n∈Z, n ≥1}o o dina y hexagonal numbe s is also a
2-AU se [7]. In [5], i is p o ed ha H+is also a k-AU se o all k≥3. Recen ly,
Hasanalizade and he au ho [4] showed ha he se O={n(n−2) |n∈Z, n = 0}
o gene alized oc agonal numbe s is no a 2-AU se and is no a 3-AU se , bu i is
ak-AU se o k≥4.
In his a icle we p o e he addi i e uniqueness o he se
N=Nn=n(7n−5)
2
n∈Z, n = 0
={1,6,9,19,24,39,46,66,75,100,111,141,154, . . . }
o gene alized nonagonal numbe s. The ollowing heo em holds.
Theo em 1. I a mul iplica i e unc ion sa is ies
(a+b) = (a) + (b)
o all nonze o gene alized nonagonal numbe s aand b, hen is he iden i y unc-
ion.
2. S a egy
We use induc ion o p o e he main heo em. Tha is, assuming (n) = n o all
n<N, we show ha (N) = N. I N=ab wi h a, b ≥2 and gcd(a, b) = 1, hen
(N) = (a) (b) = ab =Nby he induc ion hypo hesis. So we may check whe he
(p ) = p o no o p imes p.
In p o ing we use (Na+Nb) = (Na) + (Nb) o sui able aand b. Since wo
ac o s nand 7n−5 o 2Nn=n(7n−5) can ha e a common di iso 5, we canno
spli (Nn) in o (n) (7n−5) o he case.
The p oo is done in a ew s eps. Fi s , we e alua e (n) o some n’s in Lemma 1.
Using his e alua ion, in Sec ions 3-5, we p o e ha (p ) = p o p= 3,5,7. In
Sec ion 6 we p o e ha (2 ) = 2 . Finally, in Sec ion 7, we p o e ha (p ) = p
o o he p imes p.
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Fo con enience, we in oduce a no a ion o ela i ely p ime ac o s. I n=ab
and gcd(a, b) = 1, hen we w i e n=a×b. Fo example, n= 2 ·2·3·5 can be
w i en as
4×3·5=4·3×5=3×4·5,
which means
gcd(4,3·5) = gcd(4 ·3,5) = gcd(3,4·5) = 1.
Lemma 1. (n) = n o n≤11.
P oo . We ha e (2) = 2. No e ha
(3) (4) = (2 ·3+2·3) = (2) (3) + (2) (3) = 4 (3).
I (3) = 0, hen we canno de e mine (4) ye . In his case, we ha e (9) = 0 om
(3) (5) = (9) + (2) (3)
and (5) = 1
2 om (2) (5) = (1) + (9).
Also, (7) = (1) + (2) (3) = 1 and, hus, (4) = (19) by (4) (7) =
(9) + (19). Then (4) = (19) = −2 om
(4) (5) = (1) + (19)
and (23) = 1
2 om
(2) (5) (7) = (3) (8) + (2) (23).
A con adic ion occu s in sol ing
(4) (23) = (2) (23) + (2) (23)
and we can conclude ha (3) = 0.
Then (4) = 4 om (3) (4) = (2) (3) + (2) (3). No e ha
(7) = (2) (3) + (1) ⇐⇒ (7) = 2 (3) + 1
(2) (5) = (9) + (1) ⇐⇒ (9) = 2 (5) −1
(4) (5) = (19) + (1) ⇐⇒ (19) = 4 (5) −1
(4) (7) = (19) + (9) ⇐⇒ (19) = 4 (7) − (9).
We ob ain
4 (5) −1 = 4(2 (3) + 1) −(2 (5) −1)
om he las wo equa ions and hus
3 (5) = 4 (3) + 3.
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Then, since
(3) (5) = (9) + (2) (3) = 2 (5) −1+2 (3),
we can ind wo solu ions:
(3) = 3, (5) = 5, (7) = 7, (9) = 9, (19) = 19;
(3) = −1
4, (5) = 2
3, (7) = 1
2, (9) = 1
3, (19) = 5
3.
Nex , we deduce ha (8) = 8 o (8) = −2
3 om
(2) (3) (5) = (3) (8) + (2) (3).
The second solu ion se canno sa is y
(3) (11) = (3) (8) + (9)
(8) (9) = (2) (3) (11) + (2) (3).
So, (n) is de e mined o be nup o 11.
3. P oo o (3 ) = 3
The basic idea o he p oo is o use induc ion unde he assump ion (n) = n o
all n < 3 . This assump ion is oo loose. In Sec ion 7 we use (3 )=3 o p o e
(ps) = ps o all p imes p > 7. In his case we need o check (3 ) o some 3 > ps
wi h he induc ion hypo hesis ha (n) = n o all n<p . To do his we need he
s onge induc ion hypo hesis o (3 ). Tha is, we should ind a unc ion α(x)
such ha (n) = n o all n<α(3 )< p .
Theo em 2. Le α(x) = 14
17 ·2
3·x−12
17 . I (n) = n o all n<α(3 ), hen
(3 )=3
P oo . The wei d coe icien s o α(x) a e de e mined by he ex emal inequali y o
Case I o Sec ion 7.
I = 2s, hen we can show easily (32s)=32sby using
N−m+Nm=m×7m+ 5
2+m×7m−5
2= 7 ×m2
wi h m= 3s. No e he a×bno a ion and ha (32)=32was al eady showed in
Lemma 1 and
maxm, 7m+ 5
2,7m−5
2=7m+ 5
2< α(m2) = 14
17 ·2
3m2−12
17
when m≥9.
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Now assume ha = 2s+ 1. We ha e ha 32s+1 ≡3,13,5 (mod 14). We
conside i in h ee cases.
Case I: 32s+1 ≡3·1 (mod 14). We le 32s+1 = 3(14m+ 1) and use
N4m+1 +N−10m= (4m+ 1)(14m+ 1) + 25m(14m+ 1)
= (14m+ 1)(29m+ 1).
No e ha m≡1 (mod 3) and, in he i s e m (4m+ 1)(14m+ 1),
gcd(4m+ 1,14m+ 1) = gcd(5, m −1) = 1 o 5.
Since 14m+ 1 = 32s, we ha e ha m−1 is no di isible by 5 and N4m+1 =
(4m+ 1) ×(14m+ 1) by he a×bno a ion. Simila ly, N−10m= 25m×(14m+ 1).
Fo he second e m 25m(14m+ 1), i m= 5k, hen 14 ·5k+ 1 = 32s. Howe e ,
his is impossible by compa ing bo h sides modulo 8. Thus, mhas a di iso dsuch
ha d= 1,5 and gcd(m/d, d) = 1.
Now conside (14m+ 1)(29m+ 1). Since 3 |(m−1), 5 ∤(m−1) and
gcd(14m+ 1,29m+ 1) = gcd(14m+ 1, m −1) = gcd(15, m −1),
we ha e ha gcd(14m+ 1,29m+ 1) = 3. Thus,
(N4m+1) + (N−10m) = (4m+ 1) (14m+ 1) + 25 ·m
d (d) (14m+ 1)
and
(N4m+1 +N−10m) = 3(14m+ 1) ×29m+ 1
3
= (32s+1) 29m+ 1
3.
Then, since
max4m+ 1,14m+ 1,25 ·m
d, d, 29m+ 1
3
= 14m+ 1
< α(3(14m+ 1)) = 14
17 ·2
3·3(14m+ 1) −12
17,
we can deduce ha (32s+1) = 32s+1 by he induc ion hypo hesis.
Case II: 32s+1 ≡3·9 (mod 14). Fo his case, 32s+1 = 3(14m+ 9) and we use
N3m+2 +N−12m−7=3
2(3m+ 2)(7m+ 3) + 3(12m+ 7)(14m+ 9)
=9
2(7m+ 4)(17m+ 11).

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No e ha mis a mul iple o 9, since 14m+ 9 = 32s. Also, compa ing bo h
sides o 14m+ 9 = 32smodulo 8, we ha e ha mis a mul iple o 4. Simila ly, we
deduce ha gcd(3m+ 2,7m+ 3) = gcd(5, m −1) = 1 by compa ing bo h sides o
14m+ 9 = 32smodulo 5. Thus, he i s e m can be w i en as 32×3m+2
2×7m+3
3.
Since 14m+ 9 = 32s, we ha e ha gcd(12m+ 7,14m+ 9) = 1 and hus he second
e m can be w i en as (12m+ 7) ×3(14m+ 9).
Since he las e m can be w i en as 9 ×7m+4
2×(17m+ 11),
max9,3m+ 2
2,7m+ 3
3,12m+ 7,7m+ 4
2,17m+ 11
= 17m+ 11
< α(3(14m+ 9)) = 14
17 ·2
3·3(14m+ 9) −12
17
and (32s+1) = 32s+1 by he induc ion.
Case III: 32s+1 ≡3·11 (mod 14). In his case, 2s+ 1 ≥5 and we conside
32s+1 = 33(14m+ 9) ins ead o he o m 3(14m+ 11). No e ha
N−6m−3+N12m+8 = 3(2m+ 1)(21m+ 13) + 6(3m+ 2)(28m+ 17)
= 9(5m+ 3)(14m+ 9)
and
gcd(2m+ 1,21m+ 13) = gcd(5, m + 3),
gcd(3m+ 2,28m+ 17) = gcd(5, m −1),
gcd(5m+ 3,14m+ 9) = gcd(m, 3) = 3.
I m+ 3 ≡0 (mod 5), hen 14m+ 9 ≡2 (mod 5). We ha e 32(s−1) ≡2 (mod 5)
has no solu ion. So, gcd(2m+ 1,21m+ 13) = 1. By he simila easoning gcd(3m+
2,28m+ 17) = 1, oo. Thus, we can w i e, by he a×bno a ion,
N−6m−3= 3 ×(2m+ 1) ×(21m+ 13)
N12m+8 = 3 ×2(3m+ 2) ×(28m+ 17)
N−6m−3+N12m+8 =5m+ 3
3×27(14m+ 9) = 5m+ 3
3×32s+1.
Then, since he maximal ac o 28m+ 17 is smalle han α(33(14m+ 9)), we can
conclude ha (32s+1)=32s+1 by he induc ion hypo hesis.
4. P oo o (5 ) = 5
In he p e ious sec ion we p o ed ha (3 )=3 by induc ion unde he s onge
assump ion ha (n) = n o all n < α(3 )<3 . We need he simila condi ion
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o (5 ). Tha is, he assump ion (n) = n o all n < 5 is oo loose o be used
in p o ing (ps) = ps o o he p ime p > 7 and p= 2.
Theo em 3. Le β(x) = 1
4x. I (n) = n o all n≤β(5 )wi h ≥3, hen
(5 )=5 .
P oo . The coe icien o β(x) can be educed o 17
70 by he ex emal inequali y o
Case I-ii o his p oo . Fo con enience, i is enough o se β(x) = 1
4x.
When is e en, we se = 2sand use
N−m+Nm= 5m×7m+ 5
2·5+ 5m×7m−5
2·5= 7 ×m2
wi h m= 5s. Then (52s) = 52sby he induc ion hypo hesis, since he maximal
ac o 5m= 5√5 is smalle han β(m2).
Now assume ha is odd. Then
5 = 52s+1 ≡5,13,17 (mod 28) and 5 −1= 52s≡1,9,25 (mod 28).
We check (5 ) = (5 ·(28m+ )) wi h ∈ {1,9,25}.
Case I: 52s+1 = 5(28m+ 1) ≡5 (mod 28). In his case we ha e m≡8 (mod 25).
We will conside wo subcases.
I-i: m≡ 1 (mod 3). No e ha
N2m+N8m+1 =m(14m−5) + (8m+ 1)(28m+ 1)
= (14m+ 1)(17m+ 1).
Since gcd(m, 14m−5) = 1 and gcd(8m+1,28m+1) = 5, we can w i e, by he a×b
no a ion,
(N2m) + (N8m+1) = m×(14m−5)+ 8m+ 1
5×5(28m+ 1)
= (m) (14m−5) + 8m+ 1
5 (5 ).
On he o he hand, since gcd(14m+ 1,17m+ 1) = gcd(3, m −1) = 1,
(N2m+N8m+1) = (14m+ 1) ×(17m+ 1)
= (14m+ 1) (17m+ 1).
Thus, maximal ac o 17m+ 1 = 17
5·28 5 +11
28 is smalle han β(5 ) = 1
4·5 and
(5 )=5 .
I-ii: m≡1 (mod 3). Fo his case, we use
N−2m+N8m+1 =m(14m+ 5) + (8m+ 1)(28m+ 1)
= (14m+ 1)(34m+ 1).
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We ha e ha gcd(m, 14m+ 5) = 1, gcd(8m+ 1,28m+ 1) = 5, and gcd(14m+
1,34m+ 1) = gcd(2m+ 3,10) = 1. Thus,
(N−2m) + (N8m+1) = m×(14m+ 5)+ 8m+ 1
5×5(28m+ 1)
= (m) (14m+ 5) + 8m+ 1
5 (5 )
and
(N−2m+N8m+1) = (14m+ 1) ×(34m+ 1)
= (14m+ 1) (34m+ 1).
The maximal ac o 34m+ 1 = 17
5·14 5 −3
14 is smalle han β(5 ) = 1
4·5 and
(5 )=5 .
Case II: 52s+1 = 5(28m+ 9) ≡17 (mod 28). We ha e m≡22 (mod 25). We use
N−2m−1+N4m+2 = (2m+ 1)(7m+ 6) + (2m+ 1)(28m+ 9)
= (2m+ 1)(7m+ 3).
No e ha
(N−2m−1) + (N4m+2)
= 52×2m+ 1
5×7m+ 6
5+ 2m+ 1
5×5(28m+ 9)
and
(N−2m−1+N4m+2) = (2m+ 1) ×(7m+ 3).
Thus, (5 )=5 , since he maximal ac o 7m+ 3 = 1
5·45 +3
4is smalle han
β(5 ) = 1
4·5 .
Case III: 52s+1 = 5(28m+ 25) ≡13 (mod 28). No e ha m≡0 (mod 25) and
N6m+5 +N−12m−10 = 3(6m+ 5)(7m+ 5) + 3(6m+ 5)(28m+ 25)
= 15(6m+ 5)(7m+ 6).
Then
(N6m+5) + (N−12m−10)
= 52×6m+ 5
5×3(7m+ 5)
5+ 3×2m+ 1
5×5(28m+ 25)
and
(N6m+5 +N−12m−10) = (6m+ 5) ×(7m+ 6).
Thus, (5 )=5 , since he maximal ac o 7m+ 6 = 1
5·4·5 −1
4is smalle han
β(5 ) = 1
5·4·5 .
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5. P oo o (7 ) = 7
We p o e ha ixes 7 . To do his we need o conside powe s o 3 unde some
assump ion, which was p o ed o be ixed by in Sec ion 3.
Theo em 4. I (n) = n o all n < 7 , hen (7 )=7 .
P oo . I = 2s+ 1, hen (72s+1) = 72s+1 by
N−m+Nm=m×7m+ 5
2+m×7m−5
2= 7m2,
whe e m= 7s.
I is e en, no e ha 7 = 12m+ 1 o some m. Then mis di isible by 4 and
m≡0,4 (mod 5). No e ha
N−12m−1+N9m+2 = 6(7m+ 1)(12m+ 1) + 9(7m+ 1) ·9m+ 2
2
= 15(7m+ 1) ·15m+ 2
2
and each pai o linea ac o s is ela i ely p ime. Thus, each ac o s can be spli ,
bu we canno spli 3 and 7m+ 1 ye .
I 7m+ 1 = 3k×d o some d= 1,3, hen d≥11, since gcd(2,7m+ 1) =
gcd(5,7m+ 1) = 1. In his case,
N−12m−1= 2 ×3·7m+ 1
d×d×(12m+ 1)
N9m+2 = 9 ·7m+ 1
d×d×9m+ 2
2
N−12m−1+N9m+2 = 5 ×3·7m+ 1
d×d×15m+ 2
2
and each ac o is smalle han 7 = 12m+ 1. Hence, (7 ) = 7 .
Now assume 7m+ 1 = 3k. Then
7 = 12 ·3k−1
7+ 1 i and only i 7 +1 = 12 ·3k−5.
This equa ion has a i ial solu ion =k= 0. I ≥1, hen we ob ain k= 42ℓ+18
by compa ing bo h sides modulo 72.
No e ha 342 ≡1 (mod 43) and o d43 7 = 6. Thus,
7 +1 ≡12 ·318 −5≡28 (mod 43).
Howe e , i has no solu ion o . Hence, 7m+ 1 canno be a powe o 3 and we
can conclude ha (7 )=7 .