Problem 1573

Proof (1) We prove that there exist $n$ consecutive positive integers, each of which has at least one prime factor that appears exactly once in its prime factorization, making the number not a power.

Since there are infinitely many primes, we can take $n$ distinct primes $p_{1}, \cdots, p_{n}$. Consider the system of congruences
$$x \equiv -i + p_{i} \left(\bmod p_{i}^{2}\right), \quad i=1,2, \cdots, n$$

Since $p_{1}^{2}, p_{2}^{2}, \cdots, p_{n}^{2}$ are pairwise coprime, by the Chinese Remainder Theorem, the above system of congruences has a positive integer solution $x$. For $1 \leqslant i \leqslant n$, since $x + i \equiv p_{i} \left(\bmod p_{i}^{2}\right)$, it follows that $p_{i} \mid (x + i)$; but by (1), $p_{i}^{2} \nmid (x + i)$, meaning $p_{i}$ appears exactly once in the prime factorization of $x + i$, hence $x + 1, x + 2, \cdots, x + n$ are not powers.

(2) We construct inductively an infinite sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, all of which are not powers, such that for each $n$, the sum of any subset of $a_{1}, \cdots, a_{n}$ is not a power, thereby proving the conclusion in (2).

First, $a_{1}$ can be any non-power number, for example, take $a_{1} = 2$. Suppose $a_{1}, \cdots, a_{n}$ have been determined, we prove that we can choose $a_{n+1}$ to be a non-power number, $a_{n+1} > a_{n}$, and $a_{n+1}$ plus any subset of $a_{1}, \cdots, a_{n}$ is not a power.

Let $s_{1}, \cdots, s_{m}$ be all the distinct sums of any subset of $a_{1}, \cdots, a_{n}$, where $m = 2^{n} - 1$. Since there are infinitely many primes, we can take $m + 1$ distinct primes $p, p_{1}, \cdots, p_{m}$, and consider the system of congruences
$$x \equiv p \left(\bmod p^{2}\right), \quad x \equiv -s_{i} + p_{i} \left(\bmod p_{i}^{2}\right), \quad i=1, \cdots, m$$

Since $p^{2}, p_{1}^{2}, \cdots, p_{m}^{2}$ are pairwise coprime, the system of congruences (2) has infinitely many positive integer solutions $x$. Take any solution greater than $a_{n}$, and denote it as $a_{n+1}$. Then $a_{n+1} \equiv p \left(\bmod p^{2}\right)$ implies that $a_{n+1}$ is divisible by $p$ but not by $p^{2}$, hence $a_{n+1}$ is not a power. Also, $a_{n+1} \equiv -s_{i} + p_{i} \left(\bmod p_{i}^{2}\right)$ implies that $a_{n+1} + s_{i}$ is divisible by $p_{i}$ but not by $p_{i}^{2}$, thus for each $i = 1, \cdots, m$, the number $a_{n+1} + s_{i}$ is not a power. This recursively constructs an infinite sequence $a_{1}, a_{2}, \cdots$ that meets the aforementioned requirements. Q.E.D.