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The following “polynomial-logarithmic” algebraic identity that one encounters on many occasions turns out to have a rather useful set of applications!

POLYNOMIAL-LOGARITHMIC IDENTITY:If is a polynomial of degree with roots , then .

PROOF: This one is left as a simple exercise. (*Hint*: Logarithms!)

A nice application of the above identity is found in one of the exercises from the chapter titled *Analysis* (p120) in Proofs from the Book by Aigner, Ziegler and Hofmann.

EXERCISE:Let be a non-constant polynomial with only real zeros. Show that for all .

SOLUTION: If is a zero of , then the right hand side of the above inequality equals zero, and we are done. So, suppose is not a root of . Then, differentiating the above identity w.r.t. , we obtain , and we are done.

It turns out that the above identity can also used to prove the well-known *Gauss-Lucas* theorem.

GAUSS-LUCAS:If is a non-constant polynomial, then the zeros of lie in the convex hull of the roots of .

PROOF: See this.

HISTORY: The well-known Russian author V.V. Prasolov in his book *Polynomials* offers a brief and interesting historical background of the theorem, in which he points out that Gauss’ original proof (in 1836) of a variant of the theorem was motivated by physical concepts, and it was only in 1874 that F. Lucas, a French Engineer, formulated and proved the above theorem. (Note that the Gauss-Lucas theorem can also be thought of as some sort of a generalization (at least, in spirit!) of Rolle’s theorem.)

Even though I knew the aforesaid identity before, it was once again brought to my attention through a nice (and elementary) article, titled On an Algebraic Identity by Roberto Bosch Cabrera, available at Mathematical Reflections. In particular, Cabrera offers a simple solution, based on an application of the given identity, to the following problem (posed in the 2006 4*th* issue of *Mathematical Reflections*), the solution to which had either escaped regular problem solvers or required knowledge of some tedious (albeit elementary) technique.

PROBLEM:Evaluate the sum . (proposed byDorin Andrica and Mihai Piticari.)

SOLUTION: (Read Cabrera’s article.)

There is yet another problem which has a nice solution based again on our beloved identity!

PROBLEM:(Putnam A3/2005) Let be a polynomial of degree , all of whose zeros have absolute value 1 in the complex plane. Put . Show that all zeros of have absolute value 1.

SOLUTION: (Again, read Cabrera’s article.)

In an earlier issue of *Mathematical Reflections, *Iurie Boreico (from Harvard) proposed the following problem.

Problem: A polynomial is called a “mirror” if . Let and consider polynomials such that , and . Prove that is a mirror iff is a mirror.

Two solutions – the pdf file size is around 1 Mb – to the above problem were proposed, and the one by the author is very close to the one I had worked out myself (partially) earlier but never really got around finishing it. So, I will post my solution here but at a slightly later time. In the meantime, you might be interested in finding a solution yourself.

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