1. Introduction
An arithmetic function
$f:\mathbb {N}\to \mathbb {C}$
is called multiplicative if
$f(1)=1$
and
$f(mn)=f(m)f(n)$
whenever
$\text {gcd}(m,n)=1$
. Let
$\mathcal {M}$
denote the set of complex-valued multiplicative functions.
A set
$E\subseteq \mathbb {N}$
is called an additive uniqueness set of a set of arithmetic functions
$\mathcal {F}$
if
$f\in \mathcal {F}$
is uniquely determined under the condition
$$ \begin{align} f(m+n)=f(m)+f(n) \quad \text{for all } m,n\in E. \end{align} $$
For example,
$\mathbb {N}$
and
$\{1\}\cup 2\mathbb {N}$
are trivially additive uniqueness sets of
$\mathcal {M}$
. This concept was first introduced by Spiro [Reference Spiro12] in 1992. She proved that the set of primes is an additive uniqueness set of
$\mathcal {M}_0=\{f\in \mathcal {M} \mid f(p_0)\neq 0 \ \text {for some prime}\ p_0\}$
. Spiro’s work has been extended in many directions.
Let
$k\geq 2$
be a fixed integer. If there is only one function
$f\in \mathcal {F}$
satisfying
$f(x_1+x_2+\cdots +x_k)=f(x_1)+f(x_2)+\cdots +f(x_k)$
for arbitrary
$x_i\in E$
,
$i\in \{1,2,\ldots ,k\}$
, then E is called a k-additive uniqueness set of
$\mathcal {F}$
.
In 2010, Fang [Reference Fang4] proved that the set of primes is a 3-additive uniqueness set of
$\mathcal {M}_0$
. In 2013, Dubickas and Šarka [Reference Dubickas and Šarka3] generalised Fang’s result to sums of an arbitrary number of primes. In 1999, Chung and Phong [Reference Chung and Phong2] showed that the set of positive triangular numbers
$T_n=\tfrac 12n(n+1)$
,
$n\in \mathbb {N}$
, and the set of positive tetrahedral numbers
$ \textit {Te}_n=\tfrac 16n(n+1)(n+2)$
,
$n\in \mathbb {N}$
, are new additive uniqueness sets for
$\mathcal {M}$
and Park [Reference Park10] extended their work to sums of k triangular numbers for
$k\geq 3$
. Park [Reference Park9] proved that the set of nonzero squares is a k-additive uniqueness set of
$\mathcal {M}$
for every
$k\geq 3$
, although it is not a 2-additive uniqueness set (see [Reference Chung1]). Recently, in [Reference Park11], he showed that the set of positive cubes is a k-additive uniqueness set, like the set of positive squares, for multiplicative functions when
$k\ge 3$
.
In 2018, Kim et al. [Reference Kim, Kim, Lee and Park6] proved that the set of nonzero generalised pentagonal numbers
$\mathcal {P}=\{\tfrac 12n(3n-1) \mid n\in \mathbb {Z}, n \ne 0\}$
, is an additive uniqueness set for
$\mathcal {M}$
and the first author [Reference Hasanalizade5] extended their result to sums of k nonzero pentagonal numbers for
$k\geq 3$
. Later, Kim et al. [Reference Kim, Kim, Lee and Park7] showed that the set of positive pentagonal numbers and the set of positive hexagonal numbers
$\mathcal {H}=\{n(2n-1)\,|\, n\in \mathbb {N}\}$
are new additive uniqueness sets for
$\mathcal {M}$
. They also conjectured that among the sets of s-gonal numbers, only the sets of triangular, pentagonal and hexagonal numbers are additive uniqueness sets for
$\mathcal {M}$
.
Let
$\mathcal {O}=\{\mathcal {O}_m=m(3m-2) \mid m\in \mathbb {Z},m\neq 0\}$
be the set of nonzero generalised octagonal numbers. That is,
$$ \begin{align*} \mathcal{O}=\{1,5,8,16,21,33,40,56,65,85,\ldots\}. \end{align*} $$
The main result of this paper is the following theorem.
Theorem 1.1. Fix
$k\ge 4$
. The set
$\mathcal {O}$
of nonzero generalised octagonal numbers is a k-additive uniqueness set of
$\mathcal {M}$
; that is, if a multiplicative function
$f_k$
satisfies
$$ \begin{align*} f_k(x_1+x_2+\cdots+x_k)=f_k(x_1)+f_k(x_2)+\cdots+f_k(x_k) \end{align*} $$
for arbitrary
$x_1,\ldots ,x_k\in \mathcal {O}$
, then
$f_k$
is the identity function.
In Sections 2 and 3, we will also show that
$\mathcal {O}$
is neither an additive nor a
$3$
-additive uniqueness set of
$\mathcal {M}$
.
2. Non
$2$
-additive uniqueness
Theorem 2.1. If a multiplicative function
$f_2$
satisfies
$$ \begin{align*} f_2(a+b) = f_2(a)+f_2(b) \quad\text{ for }a,b \in \mathcal{O}, \end{align*} $$
then
$f_2(2) = 2$
and
$f_2(4)$
can be an arbitrary number. For other n, there are two possibilities for
$f_2$
: either
$f_2(n) = n$
or
$$ \begin{align*} f_2(n) = \begin{cases} 1 & \text{when}\ n\ \text{is odd},\\ 0 & \text{when}\ 8\mid n. \end{cases} \end{align*} $$
Proof. Trivially,
$f_2(2) = 2$
.
Since
$8$
and
$16$
are generalised octagonal numbers and
$$ \begin{align*} f_2(3) \cdot f_2(8) = f_2(8) + f_2(16) = f_2(8) + f_2(8) + f_2(8), \end{align*} $$
if
$f_2(8) \ne 0$
, then
$f_2(3) = 3$
. Then,
$f_2(5) = 5$
from
$$ \begin{align*} f_2(2)\cdot f_2(3) = f_2(1) + f_2(5) = 1+f_2(5). \end{align*} $$
Note that
$$ \begin{align*} f_2(13) &= f_2(5) + f_2(8) = 5 + f_2(8), \\ f_2(3) \cdot f_2(7) &= f_2(5) + f_2(16) = 5 + 2f_2(8), \\ f_2(2) \cdot f_2(11) &= f_2(1) + f_2(3) \cdot f_2(7) = 6 + 2f_2(8). \end{align*} $$
Thus,
$f_2(8) = 8$
by solving
$$ \begin{align*} f_2(2) \cdot f_2(3) \cdot f_2(11) = f_2(1) + f_2(5) \cdot f_2(13) \Longleftrightarrow 3(6+2f_2(8)) = 1 + 5(5+f_2(8)). \end{align*} $$
Now assume that
$f_2(8) = 0$
. Then,
$$ \begin{align*} f_2(13) = f_2(5 + 8) = f_2(5)\quad\text{and}\quad f_2(3) \cdot f_2(7) = f_2(5+16) = f_2(5). \end{align*} $$
Since
$$ \begin{align*} f_2(2) \cdot f_2(11) = f_2(1) + f_2(3) \cdot f_2(7) = 1 + f_2(5), \end{align*} $$
we obtain an equation
$f_2(3)(1+f_2(5)) = 1 + f_2(5)^2$
from
$$ \begin{align*} f_2(2) \cdot f_2(3) \cdot f_2(11) = f_2(1) + f_2(5) \cdot f_2(13). \end{align*} $$
However, since
$f_2(2) \cdot f_2(3) = 1+f_2(5)$
, we can conclude that
$$ \begin{align*} f_2(3) = f_2(5) = 1. \end{align*} $$
Note that if a generalised octagonal number is even, then it is divisible by
$8$
. Also, the sum of two odd generalised octagonal numbers is singly even. So we cannot determine
$f_2(4)$
. It can be an arbitrary value.
Now, we have two cases for the values of
$f_2(n)$
for
$n \le 8$
except
$n=4$
:
$$ \begin{align*} &f_2(1) = 1, f_2(2) = 2, f_2(3) = 3, f_2(5) = 5, f_2(6) = 6, f_2(7) = 7, f_2(8) = 8;\\ &f_2(1) = 1, f_2(2) = 2, f_2(3) = 1, f_2(5) = 1, f_2(6) = 2, f_2(7) = 1, f_2(8) = 0. \end{align*} $$
Consider the first case. That is, let us show that
$f_2(n)=n$
for all n except for
$4$
. We use induction on n. Suppose that
$f_2(n)=n$
for
$4\neq n<N$
. If N is not a prime power, then
$N=ab$
with
$\text {gcd}(a,b)=1$
and
$f_2(N) = f_2(a) \cdot f_2(b) = ab =N$
by the multiplicativity of
$f_2$
and the induction hypothesis. Thus, we may assume that
$N=p^k$
is a prime power.
If
$N = 2^r$
with
$r \ge 4$
, then
$f_2(N) = N$
from the equations
$$ \begin{align*} f_2(\mathcal{O}_{2^{r-2}} + \mathcal{O}_{2^{r-2}}) &= f_2( 2^{r-1}(3\cdot2^{r-3} - 1) + 2^{r-1}(3\cdot2^{r-3} - 1) ) \\ &= f_2(2^r) \cdot f_2(3\cdot2^{r-3} - 1) \\ &= f_2(2^r)\,(3\cdot2^{r-3} - 1) \end{align*} $$
and
$$ \begin{align*}f_2(\mathcal{O}_{2^{r-2}}) + f_2(\mathcal{O}_{2^{r-2}}) = 2 f_2( 2^{r-1} ) \cdot f_2(3\cdot2^{r-3} - 1) = 2^r (3\cdot2^{r-3} - 1),\end{align*} $$
since
$3\cdot 2^{r-3}-1 < 2^r$
.
Now, suppose
$N = 3^{2r+1}$
with
$r \ge 1$
. In this case,
$$ \begin{align*} \mathcal{O}_{3^r} + \mathcal{O}_{-3^r} = 3^r(3\cdot3^r-2) + 3^r(3\cdot3^r+2) = 2 \cdot 3^{2r+1} \end{align*} $$
gives
$f_2(3^{2r+1}) \kern1.3pt{=}\kern1.3pt 3^{2r+1}$
. Next, let us consider
$N \kern1.3pt{=}\kern1.3pt 3^{2r}$
with
$r \kern1.3pt{\ge}\kern1.3pt 2$
. Since
${3^2 \kern1.3pt{\equiv}\kern1.3pt -1 \pmod {5}}$
, we have
$3^{2r} = 5m\pm 1$
with m even. Note that
$$ \begin{align*} \mathcal{O}_{-2m} + \mathcal{O}_{m+1} &= 4m(3m+1) + (m+1)(3m+1) = (3m+1)(5m+1), \\ \mathcal{O}_{-m+1} + \mathcal{O}_{2m} &= (m-1)(3m-1) + 4m(3m-1) = (3m-1)(5m-1). \end{align*} $$
Since
$\gcd (m,3m+1) = \gcd (m+1,3m+1) = \gcd (3m+1,5m+1) = 1$
, and both of
$4m$
and
$3m+1$
are less than
$5m+1$
, we can conclude that
$f_2(5m+1) = 5m+1$
. Also,
$f_2(5m-1) = 5m-1$
by the same method. Thus,
$f_2(3^{2r}) = 3^{2r}$
.
The final case is
$N = p^r$
with
$p \ne 2,3$
. Note that
$p^r = 6m\pm 1$
with
$m \ge 2$
and
$$ \begin{align} \begin{aligned} \mathcal{O}_{-4m} + \mathcal{O}_{-2m} = 8m(6m+1) + 4m(3m+1) = 12m(5m+1), \\ \mathcal{O}_{2m} + \mathcal{O}_{4m} = 8m(6m-1) + 4m(3m-1) = 12m(5m-1). \end{aligned} \end{align} $$
If
$\gcd (8,m) = \gcd (4,3m\pm 1) = \gcd (4,5m\pm 1) = \gcd (3,5m\pm 1)$
, then (*) yields
$f_2(6m\pm 1) = 6m\pm 1$
. However, otherwise, we must split each factor carefully. This part is complicated, so we break it into several cases. The proofs for
$6m+1$
and
$6m-1$
are similar, so we only treat
$6m+1$
.
Case (i).
$8m$
is a power of
$2$
. In this case, letting
$m = 2^r$
,
$8m = 2^{r+3}$
is greater than
$6m+1$
. So, we cannot apply the induction hypothesis to
$f_2(8m)$
. However, note that
$$ \begin{align*} \mathcal{O}_{2^{r+1}} + \mathcal{O}_{2^{r+1}} = 2^{r+2}(3\cdot2^r-1) + 2^{r+2}(3\cdot2^r-1) = 2^{r+3}(3\cdot2^r-1) \end{align*} $$
and
$$ \begin{align*} 2^{r+2} = 4m < 6m+1 \quad\text{and}\quad 3\cdot2^r-1 = 3m-1 < 6m+1. \end{align*} $$
Thus,
$f_2$
fixes
$2^{r+3} = 8m$
.
Case (ii).
$4(3m+1)$
is a power of
$2$
. If we let
$3m+1 = 2^r$
, then
$4(3m+1) = 2^{r+2}$
and
$$ \begin{align*} \mathcal{O}_{2^r} + \mathcal{O}_{2^r} = 2^{r+1}(3\cdot2^{r-1}-1) + 2^{r+1}(3\cdot2^{r-1}-1) = 2^{r+2}(3\cdot2^{r-1}-1). \end{align*} $$
However, in this equation,
$2^{r+1} = 2(3m+1)$
is not less than
$6m+1$
. So, we apply this method again to verify
$f_2(2^{r+1}) = 2^{r+1}$
.
Case (iii). The divisor
$4(5m+1)$
of
$12m(5m+1)$
is a power of
$2$
. We can show that
$f_2(4(5m+1)) = 4(5m+1)$
by applying the previous method twice.
Case (iv). The divisor
$3(5m+1)$
of
$12m(5m+1)$
is a power of
$3$
. Since
$\operatorname {ord}_5{3} = 4$
, we have
$3(5m+1) = 3^{4r+1}$
. Thus,
$$ \begin{align*} \mathcal{O}_{3^{2r}} + \mathcal{O}_{-3^{2r}} = 3^{2r}(3\cdot3^{2r}-2) + 3^{2r}(3\cdot3^{2r}+2) = 2 \cdot 3^{4r+1}. \end{align*} $$
We can conclude that
$f_2$
fixes
$3(5m+1)$
.
Case (v).
$4(5m+1) = 2^{a+2}\cdot 3$
with
$a \ge 1$
. Since
$\tfrac 134(5m+1)> 6m+1$
, we should check this case. Note that
$5m+1$
is divisible by
$2\cdot 3$
when
$m = 6n+1$
with
$n \ge 1$
. Then,
$$ \begin{align*} 5(6n+1)+1 = 2^a\cdot3 \Longleftrightarrow 5n+1 = 2^{a-1}. \end{align*} $$
If we take both sides modulo
$5$
, we deduce that
$a-1 = 4k$
. That is,
$5n = 16^k-1$
with
$k \ge 1$
. Calculating
$6m+1$
:
$$ \begin{align*} 6m+1 &= 6(6n+1)+1 = 6^2 \times \frac{16^k-1}{5} + 7 \\ &= \frac{(6\cdot4^k)^2-1}{5} = \frac{1}{5}(6\cdot4^k+1)(6\cdot4^k-1). \end{align*} $$
Thus,
$6m+1$
cannot be a power of a prime.
Case (vi).
$3(5m+1) = 2\cdot 3^{b+1}$
with
$b \ge 1$
. Note that
$\tfrac 123(5m+1)> 6m+1$
. However,
$m = 6n+1$
with
$n \ge 1$
. Then,
$$ \begin{align*} 5(6n+1)+1 = 2\cdot3^b \Longleftrightarrow 5n+1 = 3^{b-1} \end{align*} $$
and thus
$5n = 3^{4k} - 1$
with
$k \ge 1$
. Thus,
$$ \begin{align*} 6m+1 = \tfrac{1}{5}(6\cdot9^k+1)(6\cdot9^k-1) \end{align*} $$
and this cannot be a power of p.
From (*) and Cases (i)–(vi), we can conclude that
$f_2(6m+1) = 6m+1$
. Similarly,
$f_2$
also fixes
$6m-1$
. Hence,
$f_2(n) = n$
except for
$n=4$
.
By the same process, we can deduce the alternative with
$f_2(2^r) = 0$
for
$r \ge 3$
and
$f_2(n) = 1$
if n is odd.
3. Non
$3$
-additive uniqueness
Theorem 3.1. If a multiplicative function
$f_3$
satisfies
$$ \begin{align*} f_3(a+b+c)=f_3(a)+f_3(b)+f_3(c) \end{align*} $$
for
$a,b,c\in \mathcal {O}$
, then
$f_3(n)=n$
for
$n\neq 4$
, and
$f_3(4)$
can be an arbitrary number.
Proof of Theorem 3.1.
Trivially,
$f_3(3) = 3$
. Note that
$$ \begin{align*} f_3(7) &= f_3(5+1+1) = f_3(5) + 2,\\ f_3(11) &= f_3(5+5+1) = 2f_3(5)+1,\\ f_3(5)\cdot f_3(7) &= f_3(3)\cdot f_3(11) + f_3(1)+f_3(1). \end{align*} $$
If we set
$a = f_3(5)$
, then
$a(a+2) = 3(2a+1)+2$
. We obtain two solutions
$a = -1$
and
$a = 5$
.
If
$f_3(5) = -1$
, then
$f_3(8) = 2$
from
$$ \begin{align*} f_3(3) \cdot f_3(7) = 3 \cdot 1 = f_3(8+8+5) = 2f_3(8) -1. \end{align*} $$
However, this leads to a contradiction:
$$ \begin{align*} f_3(2) \cdot (-1) = f_3(2) \cdot f_3(5) &= f_3(8+1+1) = f_3(8) + 2 = 4, \\ f_3(2) = f_3(2) \cdot f_3(7) &= f_3(8+5+1) = f_3(8) = 2. \end{align*} $$
Thus, we can conclude that
$f_3(5)=5$
. Then
$f_3(n)=n$
for
$n\le 8$
except
$n=4$
. We can also find
$f(9) = 9$
and
$f(16) = 16$
from
$$ \begin{align*} f(2)\cdot f(9) = f(8+5+5) = 18, \quad f(2)\cdot f(9) = f(16+1+1) = f(16) + 2. \end{align*} $$
Let us show that
$f_3(n)=n$
for all n except
$n=4$
. We use induction on n. Suppose that
$f_3(n)=n$
for
$4\neq n<N$
. If N is not a prime power, then
$N=ab$
with
$\text {gcd}(a,b)=1$
and
$f_3(N)=N$
by the multiplicativity of
$f_3$
and the induction hypothesis. Thus, we may assume that N is a prime power.
If
$N=3^r$
, then from the equalities
$$ \begin{align*} f_3(3\mathcal{O}_{3^{r-1}})&=3f_3(\mathcal{O}_{3^{r-1}})=3f_3(3^{r-1}(3^r-2))=3f_3(3^{r-1}) \cdot f_3(3^r-2)\\ &=f_3(3^r(3^r-2))=f_3(3^r) \cdot f_3(3^r-2), \end{align*} $$
we conclude that
$f_3(3^r)=3^r$
since
$f_3(3^r-2)=3^r-2$
by the induction hypothesis.
Now suppose that
$N=2^{2s+1}$
with
$s\ge 2$
. In this case,
$$ \begin{align*} \mathcal{O}_{2^s}+\mathcal{O}_{2^s}+\mathcal{O}_{-2^{s+1}}=2^{s+1}(3\cdot2^{s-1}-1)+2^{s+1}(3\cdot2^{s-1}-1)+2^{s+2}(3\cdot2^s+1)=9\cdot2^{2s+1} \end{align*} $$
yields
$f_3(2^{2s+1})=2^{2s+1}$
since
$3\cdot 2^{s-1}-1<2^{2s+1}$
,
$3\cdot 2^s+1<2^{2s+1}$
. Next, assume that s is even and
$N=2^{s+2}$
with
$s\ge 4$
. In this case, from
$$ \begin{align*} \mathcal{O}_{2^{s-1}}+\mathcal{O}_{-2^{s-1}}+\mathcal{O}_{2^{s+1}} &=2^s(3\cdot2^{s-2}-1)+2^s(3\cdot2^{s-2}+1)+2^{s+2}(3\cdot2^s-1)\\ &=2^{s+2}(27\cdot2^{s-3}-1), \end{align*} $$
we have
$f_3(2^{s+2})=2^{s+2}$
since
$3\cdot 2^{s-2}\pm 1<2^{s+2}$
,
$3\cdot 2^s-1<2^{s+2}$
and
$27\cdot 2^{s-3}-1<2^{s+2}$
.
Now consider
$N = p^r$
with
$p \ne 2,3$
. Assume
$N = p^r = 6m-1$
with
$m \ge 1$
and
$f_3(n) = n$
for all
$n < N$
.
Note that
$$ \begin{align*} \mathcal{O}_m + \mathcal{O}_m + \mathcal{O}_{4m} &= m(3m-2) + m(3m-2) + 8m(6m-1) = 6m(9m-2), \\ \mathcal{O}_{2m} + \mathcal{O}_{2m} + \mathcal{O}_{4m} &= 4m(3m-1) + 4m(3m-1) + 8m(6m-1) = 8m(9m-2). \end{align*} $$
To calculate
$f_3(6m-1)$
, we consider two cases.
Case (i). m is odd. In this case,
$\gcd (m,3m-2) = \gcd (2,m) = \gcd (m,9m-2) = 1$
and every factor of the above two equations is less than
$6m-1$
except
$9m-2$
or a power of
$2$
or
$3$
. So, letting
$x = f_3(6m-1)$
and
$y = f_3(9m-2)$
, we can write
$$ \begin{align*} m(3m-2) + m(3m-2) + 8m\,x = 6m\,y, \\ 4m(3m-1) + 4m(3m-1) + 8m\,x = 8m\,y, \end{align*} $$
and these equations yield
$x = f_3(6m-1) = 6m-1$
and
$y = f_3(9m-2) = 9m-2$
.
Case (ii). m is even. Let
$m = 2n$
. Then, we should check
$f_3(12n-1)$
. Note that
$$ \begin{align*} \mathcal{O}_m + \mathcal{O}_m + \mathcal{O}_{4m} = 4n(3n-1) + 4n(3n-1) + 16n(12n-1) = 12n(9n-1) \end{align*} $$
and
$\gcd (n,3n-1) \kern1.4pt{=}\kern1.4pt \gcd (2,12n-1) \kern1.4pt{=}\kern1.4pt \gcd (n,12n-1) \kern1.4pt{=}\kern1.4pt \gcd (n,9n-1) \kern1.4pt{=}\kern1.4pt 1$
, and every factor is less than
$12n-1$
or a power of
$2$
or
$3$
. However, we have an exceptional case to check:
$16n$
is a power of
$2$
greater than
$12n-1$
. In this case, we can use the equalities for
$f_3(2^r)$
. So,
$f_3$
fixes every factor. Thus, if we rearrange the equality
$$ \begin{align*} 4n(3n-1) + 4n(3n-1) + 16n\,f_3(12n-1) = 12n(9n-1), \end{align*} $$
we obtain
$f_3(6m-1) = f_3(12n-1) = 12n-1$
.
For the case
$N = p^r = 6m+1$
, we can use
$\mathcal {O}_{-m} = m(3m+2)$
.
4.
$4$
-additive uniqueness
Theorem 4.1. If a multiplicative function
$f_4$
satisfies
$$ \begin{align*} f_4(a+b+c+d)=f_4(a)+f_4(b)+f_4(c)+f_4(d) \end{align*} $$
for
$a,b,c,d\in \mathcal {O}$
, then
$f_4$
is the identity function.
We will need the following lemma.
Lemma 4.2 (Kim et al. [Reference Kim, Lee and Oh8, Theorem 4.7]).
For any integer
$k\ge 4$
, any positive integer n is a sum of k nonzero generalised octagonal numbers, except for
$n=1,2,\ldots ,k-1$
and
$k+b$
, where
$b\in B=\{1,2,3,5,6,9,10,13,17\}$
.
Proof of Theorem 4.1.
By Lemma 4.2, every positive integer can be represented as a sum of four nonzero generalised octagonal numbers except for
$1,2,3,5,6,7,9,10,13$
,
$14,17,21$
. Note that it is sufficient to determine
$f_4(n)$
for
$n=2,3,5,7,9,13,17$
.
Trivially,
$f_4(1)=1$
and
$f_4(4)=4$
. Note that
$f_4(12)=4f_4(3)=f_4(5+5+1+1)=2f_4(5)+2$
. Thus,
$f_4(3)=\tfrac 12(f_4(5)+1)$
. Moreover,
$$ \begin{align*} f_4(8) &=f_4(5+1+1+1)=f_4(5)+3,\\ f_4(11) &=f_4(8+1+1+1)=f_4(5)+6. \end{align*} $$
Now observe that
$$ \begin{align*} f_4(40)&=f_4(8) \cdot f_4(5)=(f_4(5)+3) \cdot f_4(5)\\ &=f_4(33+5+1+1)=f_4(3) \cdot f_4(11)+f_4(5)+2=f_4(3)(f_4(5)+6)+f_4(5)+2. \end{align*} $$
After simplification,
$$ \begin{align*} f_4(5)^2+2f_4(5)=f_4(3)(f_4(5)+6)+2. \end{align*} $$
For convenience, let
$f_4(5)=a$
. Then
$2a^2+4a=(a+1)(a+6)+4$
and we obtain two solutions
$f_4(5)=-2$
and
$f_4(5)=5$
. The first solution yields
$f_4(3)=-\tfrac 12$
and
$$ \begin{align*} f_4(16) &=3f_4(5)+1=-5,\\ f_4(21) &=f_4(3)(f_4(5)+3)-3=-\tfrac{7}{2}. \end{align*} $$
However, this leads to a contradiction:
$$ \begin{align*} f_4(48)& =f_4(21+21+5+1)=2f_4(21)+f_4(5)+1=-8\\ & =f_4(16) \cdot f_4(3)=\tfrac{5}{2}. \end{align*} $$
Thus, we can conclude that
$f_4(3)=3$
,
$f_4(5)=5$
and
$f_4(7)=7$
. Since
$$ \begin{align*} f_4(30) &=f_4(2)\cdot f_4(3)\cdot f_4(5)=f_4(1+5+8+16),\\ f_4(18) &=f_4(2)\cdot f_4(9)=f_4(8+8+1+1),\\ f_4(26) &=f_4(13)\cdot f_4(2)=f_4(16+8+1+1),\\ f_4(34) &=f_4(17)\cdot f_4(2)=f_4(16+16+1+1), \end{align*} $$
we find
$f_4(2)=2$
,
$f_4(9)=9$
,
$f_4(13)=13$
and
$f_4(17)=17$
.
Since other numbers can be expressed as sums of four positive generalised octagonal numbers, we can conclude that
$f_4$
is the identity function by induction.
5.
$5$
-additive uniqueness
Theorem 5.1. If a multiplicative function
$f_5$
satisfies
$$ \begin{align*} f_5(a+b+c+d+e)=f_5(a)+f_5(b)+f_5(c)+f_5(d)+f_5(e) \end{align*} $$
for
$a,b,c,d,e\in \mathcal {O}$
, then
$f_5$
is the identity function.
Proof of Theorem 5.1.
By Lemma 4.2, the exceptional set is
$$ \begin{align*}\{1,2,3,4\}\cup\{6,7,8,10,11,14,15,18,22\}.\end{align*} $$
Note that
$f_5(5)=5$
and
$f_5(9)=f_5(5+1+1+1+1)=f_5(5)+4=9$
. So it is sufficient to determine
$f_5(n)$
for
$n=2,3,4,7,8,11$
. Note that
$$ \begin{align*} f_5(16)=f_5(8+5+1+1+1)=f_5(8)+8. \end{align*} $$
Then
$$ \begin{align*} f_5(72)=9f_5(8)=f_5(16+16+16+16+8)=4f_5(16)+f_5(8)=5f_5(8)+32. \end{align*} $$
Therefore,
$f_5(8)=8$
and from
$f_5(24)=f_5(3) \cdot f_5(8)=f_5(8+5+5+5+1)= 16+f_5(8)$
, we get
$f_5(3)=3$
. Similarly, from
$$ \begin{align*} f_5(30)&=5f_5(2) \cdot f_5(3)=f_5(8+8+8+5+1)=3f_5(8)+6,\\ f_5(36)&=9f_5(4)=f_5(5+5+5+5+16)=20+f_5(16),\\ f_5(21)&=f_5(3)\cdot f_5(7)=f_5(5+5+5+5+1)=21,\\ f_5(33)&=f_5(3)\cdot f_5(11)=f_5(8+8+8+8+1)=4f_5(8)+1, \end{align*} $$
we deduce that
$f_5(2)=2$
,
$f_5(4)=4$
,
$f_5(7)=7$
and
$f_5(11)=11$
, respectively.
Since other numbers can be represented as sums of five positive generalised octagonal numbers, we can conclude that
$f_5$
is the identity function by induction.
6. Proof for k-additivity with
$k\ge 6$
Let
$k\geq 6$
and
$f_k$
be k-additive on positive generalised octagonal numbers. That is,
$f_k(x_1+x_2+\dotsb +x_k)=f_k(x_1)+f_k(x_2)+\dotsb +f_k(x_k)$
for arbitrary
$x_1,\ldots ,x_k\in \mathcal {O}$
. Trivially,
$f_k(1)=1$
and
$f_k(k)=k$
.
Note that
$$ \begin{align*} (k-3)+18&=(k-3)\cdot1+5+5+8\\ &=(k-3)\cdot1+1+1+16,\\ (k-3)+37&=(k-3)\cdot1+8+8+21\\ &=(k-3)\cdot1+5+16+16,\\ (k-3)+42&=(k-3)\cdot1+5+16+21\\ &=(k-3)\cdot1+40+1+1,\\ (k-5)+64&=(k-5)\cdot1+8+8+16+16+16\\ &=(k-5)\cdot1+1+1+1+21+40. \end{align*} $$
Let
$x=f_k(5)$
,
$y=f_k(8)$
,
$z=f_k(16)$
and
$w=f_k(21)$
. The above equalities give rise to the system of equations
$$ \begin{align*} \begin{cases} 2x+y=2+z,\\ 2y+w=x+2z,\\ x+z+w=2+xy,\\ 2y+3z=3+xy+w. \end{cases} \end{align*} $$
The solutions are
$$ \begin{gather*} f_k(5)=f_k(8)=f_k(16)=f_k(21)=1,\\ f_k(5)=5, f_k(8)=8, f_k(16)=16, f_k(21)=21. \end{gather*} $$
Consider the first solution set
$f_k(5)=f_k(8)=f_k(16)=f_k(21)=1$
. Arrange the positive generalised octagonal numbers into an increasing sequence and let
$O_n$
denote the nth term. Then
$f_k(O_1)=f_k(O_2)=f_k(O_3)=f_k(O_4)=f_k(O_5)=1$
. As seen in Section 5, every
$O_n$
with
$n\geq 6$
can be written as a sum of five positive generalised octagonal numbers. From the equality
$$ \begin{align} (k-6)+1+1+1+16+21+O_f =(k-6)+40+O_a+O_b+O_c+O_d+O_e \end{align} $$
with
$a,b,c,d,e<f$
, we infer that
$f_k(O_n)=1$
for all
$n\geq 6$
inductively. However, for sufficiently large n,
$O_n$
can be represented as a sum of k positive generalised octagonal numbers. So
$f_k(O_n)=k$
, which is a contradiction.
Hence, we conclude that
$f_k(5)=5$
,
$f_k(8)=8$
,
$f_k(16)=16$
and
$f_k(21)=21$
. Moreover, (6.1) yields
$f_k(O_n)=O_n$
for every
$n\geq 1$
.
If N is a sum of k positive generalised octagonal numbers, then
$f_k(N)=N$
. Otherwise, choose an integer
$M>k+17$
such that
$\text {gcd}(M,N)=1$
. Then M and
$MN$
can be represented as sums of k positive generalised octagonal numbers by Lemma 4.2. By the multiplicativity of
$f_k$
,
$Mf_k(N)=f_k(M) \cdot f_k(N)=f_k(MN)=MN$
. Therefore,
$f_k(N)=N$
and this completes the proof.
