SOLUTION: Let P(x) be a nonconstant polynomial, where all the coefficients are nonnegative integers. Prove that there exist infinitely many positive integers n such that P(n) is composite. I

Question 1173733: Let P(x) be a nonconstant polynomial, where all the coefficients are nonnegative integers. Prove that there exist infinitely many positive integers n such that P(n) is composite. I know that this was already solved on the forum, but I'm curious to know if there is another way to solve it. Thank you!
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.

                                        For your info:

I found (thanks to GOOGLE) that post in wwww.algebra.com
to which you missed to refer, relevant to this problem.

https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1160694.html

I read that solution and came to the conclusion that it is INCORRECT, unfortunately.

            That solution would work for polynomials with the constant term different from 1,

            but DOES NOT work for polynomials with the constant term equal to 1.

            From the other side, only such polynomials with the constant term 1 are of interest ---
            --- all the others are OUT OF the INTEREST, because for them the statement is TRIVIAL.

I also looked with this request to popular "people" web-sites, and noticed, that they repeat the same error.

/\/\/\/\/\/\/\/

After some time break, I browsed in the Internet again, looking for appropriate sources.

I found this article online

http://www.cs.umd.edu/~gasarch/BLOGPAPERS/polyprimes.pdf

where the statement is proved

    Let f(x) ∈ Z[x]. There are an infinite number of y ∈ Z such that f(y) is not prime.

See Colollary 2.2 in this article.

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