Prove that if f(x) is a nonconstant polynomial with integer coefficients, then there is an integer y such that f(y) is composite. [Hint: Assume thai f(x0) = p is prime. Show that p divides f(x0 + kp) for all integers k. Obtain a contradiction of the fact that a polynomial of degree n, where n > 1, takes on each value at most n times.]

Solution:Step-1:In this problem we need to prove that if f(x) is a nonconstant polynomial with integer coefficients , then there is an integer y such that f(y) is composite.[Given hint: Assume that is a prime. .We need to show that p divides for all integers k.Obtain a contradiction of the fact that a polynomial of degree n, where n>1, takes on each value at most n times].Step-2:Given: is a prime number and f(x) is a non constant polynomial.Let us consider, with and integer coefficients.At then the value of ,where p is prime number and k is an integer. We know that for any x , y and n for some C.Because highest degree of that function is n.So, can be written as : since Therefore, .Step-3:p(1+C) is not a constant value.Because for all values of C we have infinite number of values.So, p(1+C) is not a prime number………….(1)But the given is function takes only prime values………..(2)So, from (1) and (2) it is a contradiction.Therefore, there exist an integer y such that f(y) is composite.