Alexei Kolesnikov

Fermat's equation in falling powers

This project grew out of an independent study course in combinatorics. The course was based on "Concrete Mathematics" by Graham, Knuth, and Patashnik -- an outstanding text for the geeky at heart. The student, Matthew Green, grew interested in the version of power operation that is called a "falling power" or "factorial power". It is defined by the formula \[x^{\underline{n}} := x(x-1)(x-2)\cdots(x-n+1),\] pronounced "\(x\) to the \(n\) falling".

After experimenting to see whether the familiar properties (such as \( (x^n)^m=x^{n\cdot m}\)) also hold for the factorial powers (they don't), we turned our attention to the equation \(x^\underline{2}+y^\underline{2}=z^\underline{2}\). Can one descibe all the integer solutions? In the case of the usual power, the integers satisfying \(x^2+y^2=z^2\) are called Pythagorean triples. They have the property that if \((x,y,z)\) is a Pythagorean triple, then so is \((kx,ky,kz)\) for any integer \(k\). The "falling" Pythagorean triples do not have that property; this also means that the usual trick of describing Pythagorean triples via rational points on the unit circle will not work for the "falling" triples.

Matt managed to obtain a complete description of the triples for power 2 falling. It turns out that, for each integer \(x\), the number of pairs \((y,z)\) such that \(x^\underline{2}+y^\underline{2}=z^\underline{2}\) is finite and it is controlled by the number of odd divisors of \(x(x-1)\): for each odd divisor, there are 4 pairs.

Now what about the higher falling powers? We know that the equations \(x^n+y^n=z^n\) do not have integer solutions for \(n\ge 3\). But it turns out that there are infinitely many non-trivial solutions for the power 3 falling; and that the set of powers for which there is a non-trivial solution is unbounded. However, we do not have a complete description of solutions for powers \(n\ge 3\).

In particular, here's one interesting question. For power 4 falling, the equation has trivial solutions: \(7^\underline{4} +7^\underline{4}=8^\underline{4}\) and the solutions such that either \(x^\underline{4}=0\) or \(y^\underline{4}=0\). It also has the following non-trivial solution: \[132^\underline{4}+190^\underline{4}=200^\underline{4}.\] Are there more non-trivial solutions for the power 4 falling?

Senior seminar projects

Senior Seminar is a capstone course for mathematics majors at Towson. I taught the course in the Fall of 2011 and chose to run it as a seminar, with students giving most of the presentations. At the beginning of the semester, the topics were taken from the book "Mathematical Connections: a Capstone Course" by Conway. This text worked very well for the class.

The majority of the projects were expository; but two of the projects went beyond exposition. One of the projects, "The sum of two squares" Ben Vogel explored the following question. When can an element of a field (\(\mathbb{Q}\) or an algebraic extension of \(\mathbb{Q}\)) be written as the sum of squares of two other elements in that field? The answer is obtained for \(\mathbb{Q}\) (the upshot is that a rational number \(p/q\) is the sum of squares of rational numbers if and only if the integer \(p\cdot q\) is the sum of squares of integers; and the answer to the latter problem is classically known). The answer for \(\mathbb{Q}(i)\) is easy (everything is a sum of squares). But we do not know what happens for other extensions of \(\mathbb{Q}\).

"Mathematics Behind Reflective Anamorphosis" by Kimberly Rausch dealt with reflections in cylindrical and conical mirrors. Imagine a mirror that has the form of a cylindrical column. The mirror is placed on a table with some image. What will you see as the reflection? Suppose that you want to see a specific image as the reflection. What do you need to draw on the table to see the desired image in the mirror? Kim obtained explicit formulas that describe the bijection between the points on the table and their reflections in the cylindrical mirror. She obtained similar description for the mirror shaped as a cone. The formulas were then tested: a rectangular grid was transformed, printed, and it was verified that the reflection of the transformed grid is the original rectangular grid.