# Solution June 22, 2007

### Problem

Does there exist a a sequence in , such that for each , and for each n, the polynomial is irreducible in ?

### Solution

Yes.

**Lemma**: Let with . Then, any complex root of *f*(*x*) has absolute value less than 1.

**Proof**: Assume . Then, by the triangle inequality,

so is not a root of . The Lemma follows.

Now define and where denotes the smallest prime number larger than . This sequence satisfies the conditions of the problem.

**Proof**: Assume there is a *n* such that is reducible, i.e.

with non-constant polynomials and . Because the leading coefficient of is prime, one of the polynomials and must be monic (have leading coefficient ). Assume without loss of generality that is monic.

By the Fundamentel Theorem of Algebra (*http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra*) we can write

where are the complex roots of . By our lemma, for all .

All roots of are also roots of . The constant term of is the product of these roots. The absolute value of that product is less than , so the constant term of - being an integer - must be . But then , a contradidiction because 0 is not a root of - indeed, .

So must be irreducible as claimed.