A masterpiece argument to establish the infamous pentagonal identity due to Euler. It is believed that the proof discussed in this blog post is due to Euler.

# Recent Updates

# Course materials

Hello everyone. Recently, I decided to upload some course materials, which I have prepared based on lectures I have been a part of, on the course materials webpage .

These notes will be updated regularly.

In some instances, I may also share some sections from my unpublished book, or other course related documents I have prepared.

# Waring’s problem and an introduction to the circle method

This blog post aims to discuss about Waringâ€™s problem, its formulation and a proof in terms of the circle method (which is also known as Hardy-Littlewood method).

Let for any positive integers we denote the number of solutions to the equation

with as

In 1770, Edward Waring posed the following problem :

For every positive integer there exists some positive integer such that, for every positive integer

It remained unanswered for a long time, until building up on a partial progress due to Hurwitz, David Hilbert proved the assertion to be true, thereby solving the infamous Waringâ€™s problem. We shall not discuss Hilbertâ€™s proof in this blog post, since it happens to be well-known and can easily be found on the internet.

Consider the number In following the circle method, we write this number of solutions as an integral.

Here, I shall be demonstrating this basic instance of the circle method by writing the number as an integral in this case.

We begin by recalling the notation where as usual is the imaginary unit, and is Eulerâ€™s infamous number.

For any suppose

,

where is a positive integer.

We may write as

where represents the number of solutions to with This can be seen by multiplying the expression of (in ) many times, and observing that in the power of exponential we would get sum of th powers, and then taking it as .

If then, would be the same as

Now, multiplying both sides of equation with and integrating it over (and by exploiting the orthogonality of complex exponential integral) yields

The expression in above allows us to write the number of solutions to the equation as an integral.

Now, we make a claim the proof of which is left as an easy exercise for the reader :

(Hint : Try to prove this combinatorially)

Exercise :Prove that in the asymptotic formula given for above, the main term is of order

This observation allows us to neglect all those which contribute an amount of order strictly lower than to the integral in

We consider and regard the absolute value of the integrand as

Then, an appeal to Huaâ€™s inequality allows us to neglect all those for which where as usual denotes the Vinogradov asymptotic symbol.

A general plan in working with Waringâ€™s problem and other similar problems is to divide the values of into two sets : the *major arcs* whose contribution shapes the main term, and the *minor arcs* the contribution of which goes to the error term. The distinction between these sets depend very much on the available results about the main term.

Loosely speaking, while powerful (although involved) methods are available for dealing with the major arcs, the crux of the problem lies with the minor arcs. Once a method to deal with minor arcs is known, one tries to extend the method as much as the setup of the problem permits, to ease the work of dealing with major arcs (though dealing with major arcs are usually more straightforward).

Around every rational number (in lowest form), we put an interval given by the conditions

and we do this for and

It is easy to notice that the distance between the centres of these intervals is and this being greater than their length implies that these intervals do not overlap. Moreover by rounding and translation, we can assume that these intervals are contained in

Invoking Weylâ€™s inequality, one can see that these intervals gives the major arcs, and their complement relative to give the minor arcs, the totality of which we denote by

Lemma :If then we havewhere is a positive number depending upon

*Proof : *Invoking a classical result due to Dirichlet on rational approximations, we get that : for every one can find a rational number satisfying and

Furthermore, if then we shall have : Since gives a stronger criteria than so it follows that : we should have if In other words, this means for every we must have

Since, so applying Weylâ€™s inequality to the exponential sum and using the fact that we get :

where

Now, applying Huaâ€™s inequality to this we get :

where and depends on

This concludes the proof of the above lemma.

We now turn our attention to the major arcs

Lemma: For we havewhere,

and

*Proof* : To begin with, we collect those values of in the sum defining which are in the same residue class modulo To do this, we write with Here, runs through an interval depending upon corresponding to Now, doing this and making use of the fact that for any integer we conclude that :

Now, we attempt to replace the discrete variable by a continuous variable in a quest to replace the by an integral over After this, we can make a change of variable from to This would allow us to replace the summation over by the integral

So, all that remains now, is to estimate the error in going from the summation to the integral.

(Hint : Use Lagrangeâ€™s Mean Value Theorem)

Exercise: If is a differentiable function, then prove that for we must have

From the above exercise, we see that if we divide any interval into intervals of length together with two possibly broken intervals, then we obtain :

In this case, we have : and hence and

Also, in this case Furthermore, from the very definition of we would have :

Hence the error in replacing the sum by the integral is

Finally we notice that and hence multiplying by from outside gives us the error bound we had in

This completes the proof of the lemma.

Remark: The above proof of the lemma demonstrates a way to replace summation by integrals. There are other effective ways to do this. Such techniques of replacing summations by integrals and bounding the error term play a crucial role when one deals with asymptotic behaviour of terms as we are doing here.

In a previous lemma, we dealt with the minor arcs. Now, we consider dealing with the major arcs.

Now, we define the *singular series* for Waringâ€™s problem as :

Lemma: If denotes the totality of the major arcs, thenwhere

We notice that, the above quantities allows us to deal with

In fact we shall have :

Theorem: If thenfor some fixed and where

We shall gloss over the proof of this theorem, because it is quite technical in nature. One may refer to a paper by Landau for the proof.

Finally we notice that, the above theorem would give us what we aimed to prove.

There are several other equations, where the circle method finds a prominent place. For a survey on the remarkable circle method, one may refer to the book on Hardy-Littlewood method authored by Vaughan. The interested reader is suggested to take a look at this wonderful blog post on circle methods and its limitations (heuristics) authored by Prof. Terence Tao.

# A note on Combinatorial Nullstellensatz

Hi everyone !

Last year around this time I was introduced to the immensely powerful technique of combinatorial nullstellensatz (devised by prof. Noga Alon) by prof. Fedor Petrov.

I wanted to share a note on this remarkable technique with the audience here. I had two options : make a blog post and use the LaTeX feature of wordpress to type everything out here, or upload a file for the audience to read. I chose the latter one, since in my opinion it is more convenient.

Hope you enjoy reading this note. Feel free to comment below.

# Dilworth’s, Mirsky’s and ErdÅ‘sâ€“Szekeres’s theorems for partially ordered sets

Just today morning, after solving a problem here using the infamous ErdÅ‘sâ€“Szekeres theorem, I felt like posting some details of it over here, which I wrote earlier, and edited recently to add some examples.

So, I wrote this post, to share the handout I prepared on these contents. This is a part of the write-up on the notes from the advanced combinatorics course at St. Petersburg State University delivered during the 2019-2020 session.

For viewing the images, please click on the image to view it better.