Consecutive Numbers pt. 2

Let $p_1, p_2, \dots, p_{2n}$ be distinct prime numbers. To show that there exists $n$ consecutive integers that are not perfect powers of any prime, we only need to show that the $i$ th number in the sequence is divisible by $p_{2i - 1}p_{2i},$ because the extra prime factor makes it so that the number cannot possibly be a power of one prime.

Thus, we must find a positive integer $x$ such that $x + i \equiv 0 \! \pmod{p_{2i - 1}p_{2i}}$ for $i = 1, 2, \dots, n.$ By the Chinese Remainder Theorem, there exists a solution $x$ to the system. This choice of $x$ guarantees that all of $x + 1, x + 2, \dots, x + n$ are not the power of any prime, so we are done. $\blacksquare$

Addendum: This problem is from the 1989 IMO, #5 .

Consider the following set $S$ of $n$ consecutive integers $S= \{((n+1)!)^2+j, j=2,3, \ldots, n+1\}.$ Any element from the set $S$ may be factorized as follows $((n+1)!)^2+j=j(1+cj),$ for some integer $c$ . Clearly, $\textrm{GCD}(j, 1+cj)=1$ . Hence, it can not be expressed as $p^k$ for any prime $p$ and any integer $k> 1. \blacksquare$