I have a solution but I am confused why this book uses more abstruse language or I am missing something (as in are my solutions not rigorous enough):
Find the least positive integer n such that any set of $n$ pairwise relatively prime integers greater than $1$ and less than $2005$ contains at least one prime number. The solution from Putnam and Beyond uses contradiction and least/maximums of prime factorizations. I think the solutions are basically the same, but the Putnam one has me a bit confused.
Since the set of positive integers is relatively prime, each positive integer is comprised of the primes $2,\dots,43$. Considering a set of no primes, we can use powers of these primes to construct the set, the largest set of which being the squares of the primes from $2,...,43$ which has $14$ elements.
In order for the set to remain pairwise relatively prime any new integer appended to the set must have be divisible by a prime greater than $43$. The smallest being $47$. Since the set must be comprised of composite numbers, the smallest number that satisfies this property is $47^2$ which is greater than $2005$. So any such set with at least $n = 15$ elements must have at least one prime number and from above it is seen that we can construct sets that have pairwise relatively prime integers and are composite for $n < 15$ by considering n squares of primes. Hence $n = 15$.