Subfield: Combinatorics


Question: Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.

Solution:
Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\mathbb{Z}_{>0}$ denote the set of all positive integers.

Suppose, to the contrary, that there exist only finitely many positive integers that are not clean. Therefore, there exists a positive integer $N$ such that every integer $n>N$ has exactly one odd representation.

Clearly, $S$ is infinite. We now claim the following properties of odd and even representations.

Property 1. Any positive integer $n$ has at most one odd and at most one even representation. Proof. We first show that every integer $n$ has at most one even representation. Since $S$ is infinite, there exists $x \in S$ such that $x>\max \{n, N\}$. Then, the number $n+x$ must be clean, and $x$ does not appear in any even representation of $n$. If $n$ has more than one even representation, then we obtain two distinct odd representations of $n+x$ by adding $x$ to the even representations of $n$, which is impossible. Therefore, $n$ can have at most one even representation.

Similarly, there exist two distinct elements $y, z \in S$ such that $y, z>\max \{n, N\}$. If $n$ has more than one odd representation, then we obtain two distinct odd representations of $n+y+z$ by adding $y$ and $z$ to the odd representations of $n$. This is again a contradiction.

Property 2. Fix $s \in S$. Suppose that a number $n>N$ has no even representation. Then $n+2$ as has an even representation containing $s$ for all integers $a \geqslant 1$.

Proof. It is sufficient to prove the following statement: If $n$ has no even representation without $s$, then $n+2 s$ has an even representation containing $s$ (and hence no even representation without $s$ by Property 1).

Notice that the odd representation of $n+s$ does not contain $s$; otherwise, we have an even representation of $n$ without $s$. Then, adding $s$ to this odd representation of $n+s$, we get that $n+2 s$ has an even representation containing $s$, as desired.

Property 3. Every sufficiently large integer has an even representation.

Proof. Fix any $s \in S$, and let $r$ be an arbitrary element in $\{1,2, \ldots, 2 s\}$. Then, Property 2 implies that the set $Z_{r}=\{r+2 a s: a \geqslant 0\}$ contains at most one number exceeding $N$ with no even representation. Therefore, $Z_{r}$ contains finitely many positive integers with no even representation, and so does $\mathbb{Z}_{>0}=\bigcup_{r=1}^{2 s} Z_{r}$.

In view of Properties 1 and 3, we may assume that $N$ is chosen such that every $n>N$ has exactly one odd and exactly one even representation. In particular, each element $s>N$ of $S$ has an even representation.

Property 4. For any $s, t \in S$ with $N<s<t$, the even representation of $t$ contains $s$.

Proof. Suppose the contrary. Then, $s+t$ has at least two odd representations: one obtained by adding $s$ to the even representation of $t$ and one obtained by adding $t$ to the even representation of $s$. Since the latter does not contain $s$, these two odd representations of $s+t$ are distinct, a contradiction.

Let $s_{1}<s_{2}<\cdots$ be all the elements of $S$, and set $\sigma_{n}=\sum_{i=1}^{n} s_{i}$ for each nonnegative integer $n$. Fix an integer $k$ such that $s_{k}>N$. Then, Property 4 implies that for every $i>k$ the even representation of $s_{i}$ contains all the numbers $s_{k}, s_{k+1}, \ldots, s_{i-1}$. Therefore,

$$
s_{i}=s_{k}+s_{k+1}+\cdots+s_{i-1}+R_{i}=\sigma_{i-1}-\sigma_{k-1}+R_{i},\tag{1}
$$

where $R_{i}$ is a sum of some of $s_{1}, \ldots, s_{k-1}$. In particular, $0 \leqslant R_{i} \leqslant s_{1}+\cdots+s_{k-1}=\sigma_{k-1}$.



Let $j_{0}$ be an integer satisfying $j_{0}>k$ and $\sigma_{j_{0}}>2 \sigma_{k-1}$. Then (1) shows that, for every $j>j_{0}$,

$$
s_{j+1} \geqslant \sigma_{j}-\sigma_{k-1}>\sigma_{j} / 2\tag{2}
$$

Next, let $p>j_{0}$ be an index such that $R_{p}=\min _{i>j_{0}} R_{i}$. Then,

$$
s_{p+1}=s_{k}+s_{k+1}+\cdots+s_{p}+R_{p+1}=\left(s_{p}-R_{p}\right)+s_{p}+R_{p+1} \geqslant 2 s_{p}
$$

Therefore, there is no element of $S$ larger than $s_{p}$ but smaller than $2 s_{p}$. It follows that the even representation $\tau$ of $2 s_{p}$ does not contain any element larger than $s_{p}$. On the other hand, inequality (2) yields $2 s_{p}>s_{1}+\cdots+s_{p-1}$, so $\tau$ must contain a term larger than $s_{p-1}$. Thus, it must contain $s_{p}$. After removing $s_{p}$ from $\tau$, we have that $s_{p}$ has an odd representation not containing $s_{p}$, which contradicts Property 1 since $s_{p}$ itself also forms an odd representation of $s_{p}$.