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We state the integer form of Freiman theorem for integer sets: for every $K\ge 1$ there are finite constants $r(K),C(K)$ such that every finite nonempty $A\subset \mathbb{Z}$ with doubling at most $K$ lies in a proper generalized arithmetic progression of rank at most $r(K)$ and cardinality at most $C(K)|A|$. An arbitrary abelian group requires the coset-progression version, allowing a subgroup factor. We prove the integer statement, with no claim of optimized constants.

We use the allowed Plünnecke inequality and basic Freiman isomorphism facts, and explicitly prove the modelling, Fourier and covering steps. Two geometry-of-numbers consequences are used as permitted: a cyclic rank-$d$ phase-radius-$\rho$ Bohr set contains a proper symmetric progression of rank at most $d+1$ and size at least $c(d,\rho )$ times the cyclic group order; and a rank-$r$ integer progression with parameter-box volume $V$ can be contained in a proper progression of rank at most $r$ and size at most $c_{r}V$. These follow from successive minima and lattice basis reduction. All these constants depend only on the displayed rank and radius, not on the ambient order or step sizes.

First construct a dense cyclic model of a large subset. Put $D=8A-8A$ and $q=8|D|$. Plünnecke gives $|D|\le K^{16}|A|$. Choose an auxiliary prime $p$ so large that every nonzero integer in $D$ stays nonzero modulo $p$, and $p>2q$. Arbitrarily large primes exist: a purported finite list would miss a prime divisor of one plus their product. For a uniform nonzero multiplier $\lambda (\mathrm{mod}p)$, each nonzero $d\in D$ has $\lambda d$ uniformly distributed among the nonzero residues. The forbidden residues are those represented by nonzero multiples of $q$ with absolute value less than $p$; there are at most $2(p-1)/q$ of them. The union bound gives failure probability at mostThus choose $\lambda$ avoiding them for every nonzero $d\in D$.

Represent the dilated elements of $A$ in $[0,p-1]$, divide this interval into nine consecutive subintervals, and retain a most-populated one. It defines $A^{\prime }\subseteq A$ with $|A^{\prime }|\ge |A|/9$. If $u(a)$ is the chosen representative of $\lambda a$, eight-term sums of representatives from this subinterval have a range shorter than $p$: eight times its width is less than $8p/9$. Consequently every original eight-term relation lifts from equality modulo $p$ to equality of the representative sums as integers.

Now reduce these representatives modulo $q$. If their two eight-term sums agree modulo $q$, their integer difference is a multiple of $q$ of absolute value below $p$. A nonzero such difference would give a forbidden residue $\lambda d$ for some nonzero $d\in D$. If $d=0$, its representative difference is already zero. Hence there are no new relations either. The map is therefore a Freiman s-isomorphism of order eight onto $B\subseteq \mathbb{Z}/q\mathbb{Z}$. It is injective, by padding shorter relations to eight terms with a fixed retained point. Its density satisfiesThis proves the cyclic Freiman model of a small-doubling integer set needed below.

Next we prove the cyclic Bogolyubov lemma. Use normalized finite-group Fourier analysis on $\mathbb{Z}/q\mathbb{Z}$. The convolutionhas Fourier expansion $r(x)={\sum }_{\chi }|\widehat{1_B}(\chi ){|}^4\chi (x)$. Let $\Gamma$ consist of those characters whose Fourier coefficient has modulus at least ${\beta }^{3/2}/\sqrt{2}$. Parseval gives $|\Gamma |\le 2{\beta }^{-2}$, and the fourth-power tail outside $\Gamma$ is at most ${\beta }^4/2$. The zero character belongs to $\Gamma$ and contributes ${\beta }^4$.

For the phase-distance Bohr set $\mathcal{B}=\{x:\Vert \arg \chi (x)/(2\pi )\Vert \le 1/10\text{ for all }\chi \in \Gamma \}$, every retained character has real part at least $\cos (\pi /5)>3/4$. ThereforeThe support of this convolution is $2B-2B$, so $\mathcal{B}\subseteq 2B-2B$. Apply the stated progression in a Bohr set from successive minima consequence. It supplies a symmetric proper progression $P_B\subseteq 2B-2B$ of rank bounded by a function of $K$ and size at least $c(K)q$, hence at least $c(K)|A|$ after adjusting the constant.

Lift this progression back to the integers. The inverse Freiman map inducesEquality of two representations uses a four-term relation on each side, so $\psi$ is well defined and injective. If $y,z,y+z$ all lie in this difference set, the relation between their representations involves at most six original terms on either side. The order-eight model therefore gives $\psi (y+z)=\psi (y)+\psi (z)$ and $\psi (0)=0$. Along every coordinate path within the symmetric progression box, all partial sums remain in $P_B$. It follows thatThus $P=\psi (P_B)$ is a symmetric proper integer progression of the same rank and cardinality, and $P\subseteq 2A-2A$. This proves the Freiman lifting of a progression step, including why a sufficiently high-order model was selected.

It remains to cover all of $A$, not merely its large modeled subset. Since $P\subseteq 2A-2A$, Plünnecke givesChoose $X\subseteq A$ maximal such that the translates $x+P$ are pairwise disjoint. Their union lies in $A+P$, so $|X|\le K^5|A|/|P|\le C_1(K)$. For every $a\in A$, maximality makes $a+P$ meet some $x+P$, giving $a\in x+P-P$. ConsequentlyThis proves the needed Ruzsa covering lemma directly.

Write $P=\{{\sum }_{i=1}^r{n}_i{w}_i:|n_i|\le L_i\}$ and $X=\{x_1,\dots ,x_s\}$. The setcontains $X+(P-P)$ and hence $A$. Its rank is at most $r+s$, bounded only by $K$, and its parameter-box volume is at most $2^{r+s}|P|$. Also $|P|\le |2A-2A|\le K^4|A|$, so this volume is at most $C_2(K)|A|$. The allowed integer properification consequence now contains $Q$ in a proper progression of bounded rank and size at most $C(K)|A|$. ThereforeThis proves Freiman's theorem, using geometry of numbers only in the two explicitly stated places.

The proof of Szemerédi theorem for four-term progressions follows a uniformity-versus-density-increment iteration. Fix a positive starting density; the quantitative parameters below depend only on that density.

First apply the counting argument in part (i). A sufficiently long set with no nonconstant four-term arithmetic progression cannot have small balanced Gowers U3 norm. Thus its balanced function has a quantitatively large cube average. The Derivative identity for the Gowers U3 norm expresses this as an average of fourth Fourier moments of the multiplicative derivatives $x\mapsto f(x+h)\overline{f(x)}$. Many shifts therefore have a derivative with a large Fourier coefficient.

The next step is the inverse argument. Choose these frequencies as a function of the shifts. The relations between repeated derivatives force a large part of the resulting frequency graph to have many additive quadruples. A Balog-Szemerédi-Gowers theorem extraction followed by Freiman theorem for integer sets organizes this graph on a bounded-complexity additive model. Integrating the organized derivative frequencies yields correlation of $f$ with a locally quadratic phase on a structured region. On integer intervals this is generally a local quadratic phase on a Bohr-type region, or equivalent two-step structure; it need not be one global quadratic polynomial on the entire interval.

The quantitative density-increment lemma obtained from that inverse argument says that either the interval is bounded in size by a function of its density, or there is a nonempty integer progression of length at least $c(\delta )N^{c(\delta )}$ on which the density increases by at least $c(\delta )>0$. The passage from correlation to this increment partitions the structured region into progressions on which the phase varies very little, controls discarded boundaries, and uses the balanced function's zero mean to find a positive-density gain. Both the gain and the size exponent are bounded away from zero for densities above the fixed starting density.

Reidentify the new progression with an interval and repeat. The property of having no nonconstant four-term progression is preserved under this affine identification. Density can increase only finitely many times before it exceeds one. Starting with $N$ sufficiently large ensures that the successive power-type length losses never reach the exceptional bounded-size case during those finitely many steps. This contradiction proves that every fixed positive density has a threshold beyond which a four-term progression must occur. The counting lemma, inverse structural argument, density increment and final iteration are the main steps; the deep inverse step is only outlined here, as permitted by this part of the question.