On the Additive Uniqueness of Generalized Nonagonal Numbers for Multiplicative Functions
Full text
#A99 INTEGERS 25 (2025) ON THE ADDITIVE UNIQUENESS OF GENERALIZED NONAGONAL NUMBERS FOR MULTIPLICATIVE FUNCTIONS Poo-Sung Park1 Department of Mathematics Education, Kyungnam University, Changwon, Republic of Korea [email protected] Received: 2/28/25, Revised: 6/30/25, Accepted: 10/10/25, Published: 11/5/25 Abstract We prove that the set Nof all nonzero generalized nonagonal numbers is an additive uniqueness set. If a multiplicative function fsatisfies the equation f(a+b) = f(a) + f(b) for all a, b ∈ N, then fis the identity function. 1. Introduction In 1992, C. Spiro [11] introduced the notion of an additive uniqueness set, briefly AU set, E⊂Nof a subset Sof arithmetic functions, which means f∈Sis uniquely determined by the condition f(a+b) = f(a) + f(b) for all a, b ∈E. She showed that the set of primes is an additive uniqueness set for the set S={f|fis multiplicative and f(p0)= 0 for some prime p0}, where fis multiplicative if f(1) = 1 and f(ab) = f(a)f(b) for all aand bwith gcd(a, b) = 1. Since her paper was published, many mathematicians have been studying the k-additive uniqueness, briefly k-AU, of various sets of natural numbers with the condition f(a1+a2+···+ak) = f(a1) + f(a2) + ···+f(ak). In 1999, Chung and Phong [2] showed that the set of triangular numbers is an AU set for the set of multiplicative functions. This set is also a k-AU set with k≥3 DOI: 10.5281/zenodo.17535282 1This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MSIT) (RS-2021-NR058832).
INTEGERS: 25 (2025) 2 [9]. However, the set of squares is not a 2-AU set for multiplicative functions [1] and it is a k-AU set for k≥3 [8]. Let us consider the additive uniqueness of polygonal numbers for multiplicative functions to generalize the above results. The author and colleagues [6] proved that the set P={n(3n−2) 2|n∈Z, n = 0}of generalized pentagonal numbers is an AU set for multiplicative functions. This is also a k-AU set for k≥3 [3, 10]. It is much more difficult to consider the set P+={n(3n−2) 2|n∈Z, n ≥1}of ordinary pentagonal numbers. This set is also a 2-AU set, which was proved in [7]. The set H+={n(2n−1) |n∈Z, n ≥1}of ordinary hexagonal numbers is also a 2-AU set [7]. In [5], it is proved that H+is also a k-AU set for all k≥3. Recently, Hasanalizade and the author [4] showed that the set O={n(n−2) |n∈Z, n = 0} of generalized octagonal numbers is not a 2-AU set and is not a 3-AU set, but it is ak-AU set for k≥4. In this article we prove the additive uniqueness of the set N=Nn=n(7n−5) 2 n∈Z, n = 0 ={1,6,9,19,24,39,46,66,75,100,111,141,154, . . . } of generalized nonagonal numbers. The following theorem holds. Theorem 1. If a multiplicative function fsatisfies f(a+b) = f(a) + f(b) for all nonzero generalized nonagonal numbers aand b, then fis the identity function. 2. Strategy We use induction to prove the main theorem. That is, assuming f(n) = nfor all n<N, we show that f(N) = N. If N=ab with a, b ≥2 and gcd(a, b) = 1, then f(N) = f(a)f(b) = ab =Nby the induction hypothesis. So we may check whether f(pr) = pror not for primes p. In proving we use f(Na+Nb) = f(Na) + f(Nb) for suitable aand b. Since two factors nand 7n−5 of 2Nn=n(7n−5) can have a common divisor 5, we cannot split f(Nn) into f(n)f(7n−5) for the case. The proof is done in a few steps. First, we evaluate f(n) for some n’s in Lemma 1. Using this evaluation, in Sections 3-5, we prove that f(pr) = prfor p= 3,5,7. In Section 6 we prove that f(2r) = 2r. Finally, in Section 7, we prove that f(pr) = pr for other primes p.
INTEGERS: 25 (2025) 3 For convenience, we introduce a notation for relatively prime factors. If n=ab and gcd(a, b) = 1, then we write n=a×b. For example, n= 2 ·2·3·5 can be written 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. f(n) = nfor n≤11. Proof. We have f(2) = 2. Note that f(3) f(4) = f(2 ·3+2·3) = f(2) f(3) + f(2) f(3) = 4f(3). If f(3) = 0, then we cannot determine f(4) yet. In this case, we have f(9) = 0 from f(3) f(5) = f(9) + f(2) f(3) and f(5) = 1 2from f(2) f(5) = f(1) + f(9). Also, f(7) = f(1) + f(2) f(3) = 1 and, thus, f(4) = f(19) by f(4) f(7) = f(9) + f(19). Then f(4) = f(19) = −2 from f(4) f(5) = f(1) + f(19) and f(23) = 1 2from f(2) f(5) f(7) = f(3) f(8) + f(2) f(23). A contradiction occurs in solving f(4) f(23) = f(2) f(23) + f(2) f(23) and we can conclude that f(3) = 0. Then f(4) = 4 from f(3) f(4) = f(2) f(3) + f(2) f(3). Note that f(7) = f(2) f(3) + f(1) ⇐⇒ f(7) = 2f(3) + 1 f(2) f(5) = f(9) + f(1) ⇐⇒ f(9) = 2f(5) −1 f(4) f(5) = f(19) + f(1) ⇐⇒ f(19) = 4f(5) −1 f(4) f(7) = f(19) + f(9) ⇐⇒ f(19) = 4f(7) −f(9). We obtain 4f(5) −1 = 4(2f(3) + 1) −(2f(5) −1) from the last two equations and thus 3f(5) = 4f(3) + 3.
INTEGERS: 25 (2025) 4 Then, since f(3) f(5) = f(9) + f(2) f(3) = 2f(5) −1+2f(3), we can find two solutions: f(3) = 3, f(5) = 5, f(7) = 7, f(9) = 9, f(19) = 19; f(3) = −1 4, f(5) = 2 3, f(7) = 1 2, f(9) = 1 3, f(19) = 5 3. Next, we deduce that f(8) = 8 or f(8) = −2 3from f(2) f(3) f(5) = f(3) f(8) + f(2) f(3). The second solution set cannot satisfy f(3) f(11) = f(3) f(8) + f(9) f(8) f(9) = f(2) f(3) f(11) + f(2) f(3). So, f(n) is determined to be nup to 11. 3. Proof for f(3r) = 3r The basic idea of the proof is to use induction under the assumption f(n) = nfor all n < 3r. This assumption is too loose. In Section 7 we use f(3r)=3rto prove f(ps) = psfor all primes p > 7. In this case we need to check f(3r) for some 3r> ps with the induction hypothesis that f(n) = nfor all n<pr. To do this we need the stronger induction hypothesis for f(3r). That is, we should find a function α(x) such that f(n) = nfor all n<α(3r)< pr. Theorem 2. Let α(x) = 14 17 ·2 3·x−12 17 . If f(n) = nfor all n<α(3r), then f(3r)=3r Proof. The weird coefficients of α(x) are determined by the extremal inequality of Case I of Section 7. If r= 2s, then we can show easily f(32s)=32sby using N−m+Nm=m×7m+ 5 2+m×7m−5 2= 7 ×m2 with m= 3s. Note the a×bnotation and that f(32)=32was already showed in Lemma 1 and maxm, 7m+ 5 2,7m−5 2=7m+ 5 2< α(m2) = 14 17 ·2 3m2−12 17 when m≥9.
INTEGERS: 25 (2025) 5 Now assume that r= 2s+ 1. We have that 32s+1 ≡3,13,5 (mod 14). We consider it in three cases. Case I: 32s+1 ≡3·1 (mod 14). We let 32s+1 = 3(14m+ 1) and use N4m+1 +N−10m= (4m+ 1)(14m+ 1) + 25m(14m+ 1) = (14m+ 1)(29m+ 1). Note that m≡1 (mod 3) and, in the first term (4m+ 1)(14m+ 1), gcd(4m+ 1,14m+ 1) = gcd(5, m −1) = 1 or 5. Since 14m+ 1 = 32s, we have that m−1 is not divisible by 5 and N4m+1 = (4m+ 1) ×(14m+ 1) by the a×bnotation. Similarly, N−10m= 25m×(14m+ 1). For the second term 25m(14m+ 1), if m= 5k, then 14 ·5k+ 1 = 32s. However, this is impossible by comparing both sides modulo 8. Thus, mhas a divisor dsuch that d= 1,5 and gcd(m/d, d) = 1. Now consider (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 have that gcd(14m+ 1,29m+ 1) = 3. Thus, f(N4m+1) + f(N−10m) = f(4m+ 1) f(14m+ 1) + f25 ·m df(d)f(14m+ 1) and f(N4m+1 +N−10m) = f3(14m+ 1) ×29m+ 1 3 =f(32s+1)f29m+ 1 3. Then, since max4m+ 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 that f(32s+1) = 32s+1 by the induction hypothesis. Case II: 32s+1 ≡3·9 (mod 14). For this 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).
INTEGERS: 25 (2025) 6 Note that mis a multiple of 9, since 14m+ 9 = 32s. Also, comparing both sides of 14m+ 9 = 32smodulo 8, we have that mis a multiple of 4. Similarly, we deduce that gcd(3m+ 2,7m+ 3) = gcd(5, m −1) = 1 by comparing both sides of 14m+ 9 = 32smodulo 5. Thus, the first term can be written as 32×3m+2 2×7m+3 3. Since 14m+ 9 = 32s, we have that gcd(12m+ 7,14m+ 9) = 1 and thus the second term can be written as (12m+ 7) ×3(14m+ 9). Since the last term can be written as 9 ×7m+4 2×(17m+ 11), max9,3m+ 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 f(32s+1) = 32s+1 by the induction. Case III: 32s+1 ≡3·11 (mod 14). In this case, 2s+ 1 ≥5 and we consider 32s+1 = 33(14m+ 9) instead of the form 3(14m+ 11). Note that 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. If m+ 3 ≡0 (mod 5), then 14m+ 9 ≡2 (mod 5). We have 32(s−1) ≡2 (mod 5) has no solution. So, gcd(2m+ 1,21m+ 13) = 1. By the similar reasoning gcd(3m+ 2,28m+ 17) = 1, too. Thus, we can write, by the a×bnotation, 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 the maximal factor 28m+ 17 is smaller than α(33(14m+ 9)), we can conclude that f(32s+1)=32s+1 by the induction hypothesis. 4. Proof for f(5r) = 5r In the previous section we proved that f(3r)=3rby induction under the stronger assumption that f(n) = nfor all n < α(3r)<3r. We need the similar condition
INTEGERS: 25 (2025) 7 for f(5r). That is, the assumption f(n) = nfor all n < 5ris too loose to be used in proving f(ps) = psfor other prime p > 7 and p= 2. Theorem 3. Let β(x) = 1 4x. If f(n) = nfor all n≤β(5r)with r≥3, then f(5r)=5r. Proof. The coefficient of β(x) can be reduced to 17 70 by the extremal inequality of Case I-ii of this proof. For convenience, it is enough to set β(x) = 1 4x. When ris even, we set r= 2sand use N−m+Nm= 5m×7m+ 5 2·5+ 5m×7m−5 2·5= 7 ×m2 with m= 5s. Then f(52s) = 52sby the induction hypothesis, since the maximal factor 5m= 5√5ris smaller than β(m2). Now assume that ris odd. Then 5r= 52s+1 ≡5,13,17 (mod 28) and 5r−1= 52s≡1,9,25 (mod 28). We check f(5r) = f(5 ·(28m+t)) with t∈ {1,9,25}. Case I: 52s+1 = 5(28m+ 1) ≡5 (mod 28). In this case we have m≡8 (mod 25). We will consider two subcases. I-i: m≡ 1 (mod 3). Note that 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 write, by the a×b notation, f(N2m) + f(N8m+1) = fm×(14m−5)+f8m+ 1 5×5(28m+ 1) =f(m)f(14m−5) + f8m+ 1 5f(5r). On the other hand, since gcd(14m+ 1,17m+ 1) = gcd(3, m −1) = 1, f(N2m+N8m+1) = f(14m+ 1) ×(17m+ 1) =f(14m+ 1) f(17m+ 1). Thus, maximal factor 17m+ 1 = 17 5·28 5r+11 28 is smaller than β(5r) = 1 4·5rand f(5r)=5r. I-ii: m≡1 (mod 3). For this case, we use N−2m+N8m+1 =m(14m+ 5) + (8m+ 1)(28m+ 1) = (14m+ 1)(34m+ 1).
INTEGERS: 25 (2025) 8 We have that gcd(m, 14m+ 5) = 1, gcd(8m+ 1,28m+ 1) = 5, and gcd(14m+ 1,34m+ 1) = gcd(2m+ 3,10) = 1. Thus, f(N−2m) + f(N8m+1) = fm×(14m+ 5)+f8m+ 1 5×5(28m+ 1) =f(m)f(14m+ 5) + f8m+ 1 5f(5r) and f(N−2m+N8m+1) = f(14m+ 1) ×(34m+ 1) =f(14m+ 1) f(34m+ 1). The maximal factor 34m+ 1 = 17 5·14 5r−3 14 is smaller than β(5r) = 1 4·5rand f(5r)=5r. Case II: 52s+1 = 5(28m+ 9) ≡17 (mod 28). We have m≡22 (mod 25). We use N−2m−1+N4m+2 = (2m+ 1)(7m+ 6) + (2m+ 1)(28m+ 9) = (2m+ 1)(7m+ 3). Note that f(N−2m−1) + f(N4m+2) =f52×2m+ 1 5×7m+ 6 5+f2m+ 1 5×5(28m+ 9) and f(N−2m−1+N4m+2) = f(2m+ 1) ×(7m+ 3). Thus, f(5r)=5r, since the maximal factor 7m+ 3 = 1 5·45r+3 4is smaller than β(5r) = 1 4·5r. Case III: 52s+1 = 5(28m+ 25) ≡13 (mod 28). Note that 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 f(N6m+5) + f(N−12m−10) =f52×6m+ 5 5×3(7m+ 5) 5+f3×2m+ 1 5×5(28m+ 25) and f(N6m+5 +N−12m−10) = f(6m+ 5) ×(7m+ 6). Thus, f(5r)=5r, since the maximal factor 7m+ 6 = 1 5·4·5r−1 4is smaller than β(5r) = 1 5·4·5r.
INTEGERS: 25 (2025) 9 5. Proof for f(7r) = 7r We prove that ffixes 7r. To do this we need to consider powers of 3 under some assumption, which was proved to be fixed by fin Section 3. Theorem 4. If f(n) = nfor all n < 7r, then f(7r)=7r. Proof. If r= 2s+ 1, then f(72s+1) = 72s+1 by N−m+Nm=m×7m+ 5 2+m×7m−5 2= 7m2, where m= 7s. If ris even, note that 7r= 12m+ 1 for some m. Then mis divisible by 4 and m≡0,4 (mod 5). Note that N−12m−1+N9m+2 = 6(7m+ 1)(12m+ 1) + 9(7m+ 1) ·9m+ 2 2 = 15(7m+ 1) ·15m+ 2 2 and each pair of linear factors is relatively prime. Thus, each factors can be split, but we cannot split 3 and 7m+ 1 yet. If 7m+ 1 = 3k×dfor some d= 1,3, then d≥11, since gcd(2,7m+ 1) = gcd(5,7m+ 1) = 1. In this 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 factor is smaller than 7r= 12m+ 1. Hence, f(7r) = 7r. Now assume 7m+ 1 = 3k. Then 7r= 12 ·3k−1 7+ 1 if and only if 7r+1 = 12 ·3k−5. This equation has a trivial solution r=k= 0. If r≥1, then we obtain k= 42ℓ+18 by comparing both sides modulo 72. Note that 342 ≡1 (mod 43) and ord43 7 = 6. Thus, 7r+1 ≡12 ·318 −5≡28 (mod 43). However, it has no solution for r. Hence, 7m+ 1 cannot be a power of 3 and we can conclude that f(7r)=7r.