>>11161961This is ez, rtard
finite polynomials are not bounded as you make n arbitrarily large, so, you are guaranteed to find f(n) that isn’t 1, -1, or 0. You can take a prime factor of this number, and set it to p. Then, because p divides f(n), then p divides f(n + k*p), where k is any integer, as f is a finite polynomial with integer coefficients. k is any integer, and, because of the archimedian property, you found infinitely many positive integers to satisfy the condition, hence, that p should be in the set. Producing infinitely many different p is something any dumbass could do. Suppose every f(n) is either 0, or a product of some finite list of prime numbers (the primes may have any non-negative multiplicity), with the inclusion of -1. Obviously, the constant term must be a product of the terms from the list, or 0. If the constant term is 0, then the case is trivial, so, suppose the constant term is nonzero. Take n to be the product of every prime number on that list, each with multiplicity 1, along with the constant term. f(n) can be expanded into the sum of the constant term, and non-constant terms. The non-constant terms can be reverse-distributed, showing that the sum is a multiple of n. The constant term can be reverse distributed with the sum of the non-constant terms, showing that f(n) is the product of the constant term with 1 plus a multiple of the product of the prime numbers from the list. Obviously, 1 plus a multiple of the product of the prime numbers from the list must have prime factors from outside the list, a contradiction, therefore, there is an infinitely long list of prime numbers that divide f(n) for some n. This proves the statement.
