**Vector Calculus and the Topology of Domains in 3-space**

by Jason Cantarella, Dennis DeTurck, and Herman Gluck

cantarel@math.uga.edu, deturck@math.upenn.edu, gluck@math.upenn.edu

Suppose you have a vector field defined on a bounded domain in 3-space. How can you tell whether your vector field is the gradient of some function? Or the curl of another vector field? Can you find a nonzero field on your domain that is divergence-free, curl-free, and tangent to the boundary? How about a nonzero field that is divergence-free, curl-free, and orthogonal to the boundary?

To answer these questions, you need to understand the relationship between the calculus of vector fields and the topology of their domains of definition. The Hodge Decomposition Theorem provides the key by decomposing the space of vector fields on a given domain into five mutually orthogonal subspaces that are topologically and analytically meaningful. This decomposition is useful not only in mathematics, but also in fluid dynamics, electrodynamics, and plasma physics. Furthermore, carrying out the proof provides a pleasant introduction to homology and cohomology theory in a familiar setting, and a chance to see both the general Hodge theorem and the deRham isomorphism theorem in action. In this paper we give an elementary exposition of these ideas.

**The Fermat-Steiner Problem **

by Shay Gueron and Ran Tessler

shay@math.Haifa.ac.il

Given a triangle ABC, how can we find a point P for which PA + PB + PC is minimal? This problem was originally proposed by Fermat some 300 years ago, and since then has reappeared in the literature with different variations, solutions, and credits. In one reincarnation it became known as Steiner's Problem, and we therefore call it the Fermat-Steiner Problem. In this paper we explore the weighted Fermat-Steiner Problem, a 200 years old generalization of the original problem.

**Prime Numbers and Irreducible Polynomials**

by M. Ram Murty

murty@mast.queensu.ca

A classical result of A. Cohen states that, if we express a prime *p* in base 10 as *p* = a_{m} 10^{m} + a_{m-1}10^{m-1} + Â… + *a*_{1} 10 + *a*_{0}, then the polynomial *f*(*x*) = *a _{m}*

**Squares from Products of Consecutive Integers**

by Alfred J. van der Poorten and Gerhard J. Woeginger

alf@math.mq.edu.au, g.j.woeginger@math.utwente.nl

Notice that 1 · 2 · 3 · 4 + 1 = 5*2*, 2 · 3 · 4 · 5 + 1 = 11^{2}, 3 · 4 · 5 · 6 + 1 = 19^{2},... . Indeed, it is well-known that the product of any four consecutive integers differs by 1 from a perfect square. However, a little experimentation readily leads one to guess that there is no integer *n*, other than four, so that the product of any *n* consecutive integers differs from a perfect square by some integer *c* = *c*(*n*) depending only on *n*. We explain the apparently special status of four and we show that of course there can be no *n* larger than four so that ... .

**Problems and Solutions**

**Notes**

**The Cycloidal Pendulum**

By Jeff Brooks and Satha Push

j.Brooks@latrobe.edu.au

**Galois Theory of Reciprocal Polynomials**

By Paulo Viana and Paula Murgel Veloso

paup@mat.puc-rio.br

**When is a Group the Union of Proper Normal Subgroups?**

By Mira Bhargava

matmzb@magic.hofstra.edu

**A Short Proof of Hall's Theorem on SDRs**

By Gregory F. Bachelis

greg@math.wayne.edu

**Reviews**

**The Math Gene: How Mathematical Thinking Evolved and Why Numbers Are Like Gossip**

By Keith Devlin

Reviewed by Marion D. Cohen

mathwoman199436@aol.com