I would suggest using the less eponymous term "Shatter function lemma" (that Matousek uses in his Lectures on Discrete Geometry) in addition to "Sauer-Shelah lemma". I personally have known this lemma for a long time, but always under the more descriptive name "Shatter function lemma".

It was actually Waclaw Sierpinski who obtained the exponent 1/3 in the Gauss circle problem, as part of his doctoral dissertation, and who published the result in 1906 (in Polish). His adviser Georgy Voronoi had obtained the same exponent in the Dirichlet divisor problem in 1903. There is an article by Andrzej Schinzel about Sierpinski's papers in number theory in volume XXI of Acta Mathematica (1972). Also there is supposed to exist (but which I have never seen) a description of Sierpinski's proof in German, possibly in some Jahrbuch der Deutsche Mathematikerverein from those years. Schinzel states that the proof is by Voronoi's geometric method. That would mean that the circle is approximated by a polygon, and Euler-Maclaurin summation applied on each piece.

Very good but it would be nice if variable seperation method would be included.

This article is unintelligible.

How can it be possible to transform the method of completing the square into such a "charabia".

By the way, a(x+x_0)^2 = ax^2 + 2ax x_0 + x_0^2 is false.

This describes one of the most fundamental and general mathematical techniques there is, one so basic that probably most math majors find it second nature already. As such, the description of it as a general technique may be most useful for lower-level pedagogy; failing to grasp this general technique can make math very hard for primary and secondary school students. However, I would love to have examples of its use at all levels and fields, since it is so general-purpose.

It's a Wiki. Usual advice is to rewrite the style yourself. :-)

For consistency I've now decided to use \chi throughout that section.

I don't think one really needs to apply the axiom of choice in this context: Since the multifunction is taking values in positive integers, we could just define the corresponding function by taking its value to be the least member of the set of values of the multifunciton. (This process will then, I guess, secretly take us back to the original function construction, where we write each rational in lowest terms — but we don't need to pay attention to this if we don't want to.)

I like how you snuck in the Axiom of Choice here, and hid it under the rug since its use is "clear".

There is a typo in the second paragraph of section Example 1, continued: one should cancel the symbol in the two occurrences of

the expected value of given that

And, reading the last paragraph of this section, I wondered why you picked as an example of extremely unlikely deviation. First, would already yield a rather respectable (i.e., small) bound, namely, something like . Second, there is nothing specific about here, although the formulation seems to indicate otherwise: you might want to add a for example somewhere.

Thank you for this article.

What is a "smart exhaust"?

Is it clear to the intended reader that none of those numbers can be factored any further? Of course one does something like trial division, but it's not obvious what "something like trial division" is.

One possible proof that 2, 3, 1 + \sqrt{-5}, 1 - \sqrt{-5} can't be factored any further is as follows. Let N(a+b\sqrt{-5}) = a^2 + 5b^2 (I use N because this is a "norm".) This norm is multiplicative, since it's the square of the complex absolute value. Then if, say, there exist non-unit elements of Z(\sqrt{-5}), a + b\sqrt{-5} and c + d\sqrt{-5}, such that

(a + b \sqrt{-5}) (c + d \sqrt{-5}) = 3

then they have norms either 1 and 9 or 3 and 3, respectively. There are no elements of the ring with norm 3, and the only elements of norm 1 are units. Similar proofs work for 2, 1 + \sqrt{-5}, 1 - \sqrt{-5} – in general, if an element of Z(\sqrt{-5}) can be factored, then its norm can be factored, and the factorization of a + b\sqrt{-5} will be a "lifting" of the factorization of the norm a^2 + 5b^2.

("Take norms" seems like its own trick, and perhaps this is better placed in an article on that trick.)