Roth first proved that any subset of the integers with positive density contains a three term arithmetic progression in 1953. Since then, many other proofs have emerged (I can think of eight off the top of my head). 


A lot of attention has gone into the bounds in Roth's theorem, and in particular what kind of bounds different proofs get you (e.g. Fourier analysis gives log type bounds, regularity lemma arguments give Ackermann type bounds, etc.)


Also, some proofs are more amenable to generalisation (to longer arithmetic progressions) than others.


My question is 



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> If we are only concerned with brevity and directness (i.e. not with a deeper theoretical understanding or sharp quantitative bounds), what is the shortest proof of Roth's theorem?
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Running through the arguments I know, it seems like the shortest one may be Roth's original one (with a couple of simplifications): show there is a large Fourier coefficient and deduce some sort of density increment argument and iterate it - a good exposition is in Tao and Vu, or Ben Green's notes at <http://people.maths.ox.ac.uk/greenbj/notes.html> . With all details fleshed out, this could probably be done in 8 pages or an hour lecture.


The odd thing is that this proof also gives fairly good quantitative bounds; if $r\_3(N)$ is the size of the largest subset of $\{1,...,N\}$ without three term arithmetic progressions, then even a crude version of this argument gives
$$ r\_3(N)=O\left(\frac{N}{\log\log N^c}\right)$$
whereas all Roth needs is $o(N)$. Hence my second question,



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> Is it inevitable that the most direct and simple proofs would also lead to fairly good quantitative bounds? 
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Finally, an exercise in Tao and Vu mentions that Behrend's example of a lower bound for $r\_3(N)$ shows that simple pigeonhole type arguments couldn't be used to prove Roth's theorem. Hence,



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> What other proof techniques wouldn't work with Roth's theorem?
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What is the shortest route to Roth's theorem?
There's a short-cut in Roth's approach if one only cares to get $o(N)$. Adolf Hildebrand told me so, and here is my shortest writeup.


**Notation:** Let $r(N),\rho(N)$ be the largest cardinality and density of a subset of $[N]$ that is free of 3-term APs, and let $\rho=\lim \rho(N)$, which must exist by [Fekete's subadditive lemma](http://seaneberhard.blogspot.hu/2013/01/feketes-lemma-and-sum-free-sets.html) since $r(N+M)\leq r(N)+r(M)$. Let $A\subset [N]$ witness $r(N)$, and let $S(x) = \sum\_{a\in A} e( a x)$ (where $e(ax)=\exp(2\pi i a x)$, naturally). Let $T(x)=\sum\_{n=1}^N e(nx)$.


**Lemmata:** Now $r(N)=|A|=I:=\int\_0^1 S(x)^2 S(-2x)dx$, since $A$ is 3-free. By Parseval, $|A|=\int\_0^1 |S(x)|^2dx$. By expanding $S(x)^2 T(-2x)$ and exchanging integration and summation, $(|A|/2)^2 \leq I\_0:=\int\_0^1 S(x)^2 T(-2x)$.


**Main Engine:** As long as $0< M\leq N$, $$\sup\_{x\in {\mathbb R}} |S(x)-\rho(M) T(x)| \leq N(\rho(M)-\rho(N)) + 10 M \sqrt{N}.$$ Proof is by circle method; set $E(x):=S(x)-\rho(M)T(x)$. Three steps are 


* $|E(a/q)|\leq N(\rho(M)-\rho(N))+2Mq$, the hard one;
* $|E(a/q+\beta)| \leq N(\rho(M)-\rho(N))+2Mq +2\pi|\beta|NMq$;
* $|E(x)| \leq N(\rho(M)-\rho(N))+10M\sqrt{N}$.


**Main Lemma:** By Main Engine and Lemmata, $\Delta:=|I-\rho(M)I\_0|$ is at most $(N(\rho(M)-\rho(N))+10M\sqrt{N})|A|$. By Lemmata, $\Delta$ is at least $|A|(\rho(M)\rho(N)N/4-1)$. Therefore, $$\rho(N)\rho(M)\leq 4(\rho(M)-\rho(N))+50M/\sqrt{N}.$$


**Conclusion:** Let $N\to\infty$ and then $M\to\infty$ to get $\rho^2\leq 0$.


The step labeled "the hard one" in the Main Engine is tricky, and uses the fact that we can restrict $A$ to arithmetic progressions and still get a 3-free set. The length of the progressions we restrict to is connected to $M$.


I suppose the point is that a write-up can be almost arbitrarily short or extremely long, depending on what the reader can be trusted to fill in.


