Prove that the polynomial is irreductible over for all integers .
Miklos Schweitzer 2009 Problem 4
A generalization of this problem was proposed by myself in the Romanian TST 2010, the solution being similar to the one below. The generalization said to prove that if is a prime and are positive integers with then is irreductible over .
Solution: Say . It’s easy to see that . Therefore, if we consider , then this polynomial satisfies the hypothesis of the following Problem so all the roots of have modulus strictly greater than 1, or has a root which is the root of unity of some order different of 1. Then, since the minimal polynomial of is and is a root of which has integer coefficients, it follows that divides . Suppose are the remainders of modulo , and suppose they are non zero. Then which shows that at least one of is greater or equal to , which is a contradiction, because . Therefore, one of is divisible by . But we know that , so divides the other one too. Therefore , which is a contradiction, because is different from 1 since 1 is not a root for . Therefore has only roots which have modulus greater than 1.
Take a root of . Then is a root for , so . This implies , so has all roots of modulus strictly greater than 1.
Suppose . Then . This implies that one of is 1. Suppose . But . This is a contradiction proving that our assumption was false. Therefore is irreductible.