  We present a proof of Roth's theorem that follows a slightly different
structure to the usual proofs, in that there is not much iteration. Although
our proof works using a type of density increment argument (which is typical of
most proofs of Roth's theorem), we do not pass to a progression related to the
large Fourier coefficients of our set (as most other proofs of Roth do).
Furthermore, in our proof, the density increment is achieved through an
application of a quantitative version of Varnavides's theorem, which is perhaps
unexpected.
