## Vojtech Jarnik Competition 2015 – Category 1 Problems

**Problem 1.** Let be differentiable on . Prove that there exists such that

**Problem 2.** Consider the infinite chessboard whose rows and columns are indexed by positive integers. Is it possible to put a single positive rational number into each cell of the chessboard so that each positive rational number appears exactly once and the sum of every row and of every column is finite?

**Problem 3.** Let and . Determine for each of the polynomials and whether it is a divisor of some nonzero polynomial whose coefficients are all in the set .

**Problem 4.** Let be a positive integer and let be a prime divisor of . Suppose that the complex polynomial with and is divisible by the cyclotomic polynomial . Prove that there at least non-zero coefficients .

The cyclotomic polynomial is the monic polynomial whose roots are the -th primitive complex roots of unity. Euler’s totient function denotes the number of positive integers less than or eual to which are coprime to .

## Every finite field is commutative – Wedderbrun’s Theorem

**Wedderbrun’s Theorem** Every finite field is commutative.

*Proof:* The statement of the theorem is very simple to understand, yet it is hard (if you’re not an expert) to imagine why the axioms of a field and the finiteness assumption allow us to deduce that the field is commutative.

We will prove the theorem by induction over the number of elements of our field. Denote the field in question. If has elements then which is commutative. Suppose now that any field with fewer elements than is commutative. The following lemma will be useful:

**Lemma** If is a finite field and is a subfield of which is commutative then there exists a positive integer such that . (where we denote by the number of elements of )

## Kronecker’s Theorem regarding Cyclotomic Polynomials

Suppose is irreducible, monic and all roots of have absolute value at most . Then is a cyclotomic polynomial.

*Proof:* It suffices to prove that every root of is a root of unity. If

consider the family of polynomials

The coefficients of are algebraic integers, since they are calculated using multiplications and additions starting from . On the other hand, the coefficients of are symmetric polynomials in , so they are rational, and therefore integers.

Furthermore, since then the coefficient of in satisfies

This bounds the coefficients of independently of . Since it follows that we can only have a finite number of polynomials in the family .

## Approximation by polynomials of a function defined on the entire real line

Suppose is a function which can be uniformly approximated by polynomials on . Then is also a polynomial.

Note the striking difference between this result and the Weierstrass approximation theorem.

## Irreductible polynomial with paired roots

Suppose is an irreductible polynomial which has a complex root such that is also a root for . Prove that for any other root of , is also a root for .

## Useful Polynomial Roots position Lemma

Denote by an open disk in the complex plane. If is a polynomial with roots , such that for all and is a complex number in , then all roots of are not in .

*Source: mathlinks.ro, user: andrew*

## Geometric mean for n complex numbers

Let be a closed disc in the complex plane. Prove that for all positive integers , and for all complex numbers there exists a such that .

*Romanian TST 2004*