Problem 1481

To prove that given a set $M$ of $1985$ distinct positive integers, none of which has a prime divisor greater than $23$, contains a subset of $4$ elements whose product is the $4$th power of an integer, we will use a combinatorial approach involving vectors in $\mathbb{Z}_4^9$.

1. Representation of Numbers:
Each number in $M$ can be represented in the form $n = 2^{a_1} \cdot 3^{a_2} \cdot \ldots \cdot 23^{a_9}$, where $a_i$ are non-negative integers. We can map each number to a vector $v = (a_1 \pmod 4, a_2 \pmod 4, \ldots, a_9 \pmod 4)$ in $\mathbb{Z}_4^9$.

2. **Vector Sum in $\mathbb{Z}_4^9$**:
We need to show that among $1985$ such vectors, there exist $4$ vectors whose sum is $0$ in $\mathbb{Z}_4^9$. This is equivalent to finding $4$ numbers in $M$ whose exponents, when summed component-wise and taken modulo $4$, result in a vector of all zeros.

3. Reduction to Combinatorial Problem:
We need to prove that given any $1985$ elements of $\mathbb{Z}_4^9$, we can find $4$ which sum to $0$. This is a specific case of a more general combinatorial problem.

4. **General Assertion $P_k(n)$**:
Consider the assertion $P_k(n)$, which states: Given $2^k n - (2^k - 1)$ elements of $\mathbb{Z}_n^k$, we can find $n$ elements which sum to $0$.

5. Proof by Induction:
We will prove that $P_k(a)$ and $P_k(b)$ imply $P_k(ab)$. This will help us establish $P_9(4)$.

- Base Case: $P_k(2)$ is a simple application of the pigeonhole principle.
- Inductive Step: Assume $P_k(a)$ and $P_k(b)$ hold. Consider $2^k ab - (2^k - 1)$ elements in $\mathbb{Z}_{ab}^k$. Group these elements into $2^k a - (2^k - 1)$ subsets, each containing $a$ elements. By $P_k(a)$, each subset has a sum divisible by $a$. Now, consider these sums modulo $a$ as elements in $\mathbb{Z}_b^k$. By $P_k(b)$, we can find $b$ such sums that add up to $0$ modulo $b$. Multiplying back by $a$, we get $ab$ elements whose sum is $0$ in $\mathbb{Z}_{ab}^k$.

6. Application to Our Problem:
Since $P_1(n)$ is the Erdős–Ginzburg–Ziv theorem, and $P_2(n)$ has been proven by C. Reiher, we can use these results to establish $P_9(4)$. Given $2^9 \cdot 4 - (2^9 - 1) = 1537$ elements in $\mathbb{Z}_4^9$, we can find $4$ elements whose sum is $0$.

7. Conclusion:
Since $1985 > 1537$, the given set $M$ of $1985$ elements must contain a subset of $4$ elements whose product is the $4$th power of an integer.

$\blacksquare$