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I want to prove the following:

Let $A$ be a set of $n$ positive integers such that for any subset $M$, $M$ neither empty nor equal to $A$, the sum of elements of $M$ is not divisible by $n$. Prove that the elements of $A$ are all congruent $\pmod n.$

The $n=2$ case is obvious (both numbers must be odd). For $n=3$, the only possible residues are 1,2, and it is obvious that if both residues are present then their sum is divisible by 3, contradiction. $n=4$ is again easily provable by casework (since you can't have both 1 and 3 as residues in the same set.)

However, i do not know how one would generalize this. For starters, if $r$ is a residue $\pmod n$, you cannot include both $r$ and $n-r$ as residues in the same set, thus obtaining that such a set has at most $\left \lfloor \dfrac{n+1}{2} \right \rfloor$ distinct residues.( $\lfloor a\rfloor$ is the integer part of $a$)

How can I finish the proof?

Caleb Stanford
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Tanny Sieben
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1 Answers1

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List out the $n$ elements of $A$ as $a_1, a_2, \ldots, a_n$. We will show that either a proper, nonempty subset of them sums to $0$ mod $n$, or else they are all congruent. Consider the partial sums $s_i = a_1 + a_2 + \cdots + a_i$, for $0 \le i \le n$.

By the pigeonhole principle, two of these $n+1$ partial sums fall into the same residue class mod $n$, say $s_i = s_j$. Unless $i = 0$ and $j = n$, we are done because we get $a_{i+1} + a_{i+2} + \cdots + a_{j} \equiv 0 \pmod{n}$. Therefore, assume that $s_0 \equiv s_n \equiv 0$ but that the $n-1$ other partial sums are exactly the $n-1$ nonzero residues mod $n$.

Now for any $1 \le i \le n-1$, consider switching $a_{i}$ and $a_{i+1}$, forming new partial sums $s_0, s_1, s_2, \ldots, s_{i}', s_{i+1}', \ldots, s_n$. If this causes two of the new partial sums to coincide mod $n$, we are again done as before. So we may assume that the collection of partial sums of the new sequence are still all distinct, specifically: $\{s_i', s_{i+1}'\} \equiv \{s_i, s_{i+1}\}$. Subtracting $s_{i-1}$, $\{s_i' - s_{i-1}, s_{i+1}' - s_{i-1}\} \equiv \{s_i - s_{i-1}, s_{i+1} - s_{i-1}\}$. That is, $$ \{a_{i+1}, a_{i+1} + a_i\} \equiv \{a_i, a_i + a_{i+1}\}. $$ In general, if $\{q,r\} = \{s,t\}$, either $q = s$ and $r = t$ or $q = t$ and $r = s$. In this case, either $a_{i+1} \equiv a_i + a_{i+1}$ or $a_{i+1} \equiv a_i$. If $a_{i+1} \equiv a_i + a_{i+1}$, then $a_{i} \equiv 0$, thus $\{a_{i}\}$ is a one-element set that sums to $0$ mod $n$. Otherwise, $a_i \equiv a_{i+1} \pmod{n}$.

Applying this for all $i$, we get that $a_1 \equiv a_2 \equiv \cdots \equiv a_n$. (We may further notice that this common residue must be relatively prime to $n$.) $\square$

Caleb Stanford
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