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For $k\ge 2$, the following precise cardinality-controlled cyclic Freiman model suffices. If $A\subseteq \mathbb{Z}$ is nonempty, $|A|=n$ and $D=kA-kA$ has $|D|\le Cn$, there is $A^{\prime }\subseteq A$ withHere ${\cong }_k$ means a bijection preserving and reflecting all equalities of $k$-term sums, a Freiman s-isomorphism of order $s=k$. The same construction works for $k=1$ if only injectivity is needed.

Choose an auxiliary prime $P$ larger than $N$ and every nonzero absolute value in $D$, and large enough to embed the finite set $A$ injectively modulo $P$. Multiply its residues by a uniformly chosen nonzero residue $u$. For a fixed $d\in D\setminus \{0\}$, $ud$ is uniform among the $P-1$ nonzero residues. The forbidden residues areof number at most $2(P-1)/N$. The union bound says that the probability some nonzero $d\in D$ enters $\mathcal{F}$ is at most $2(|D|-1)/N<1$. Fix a multiplier avoiding every forbidden event.

Let $\tilde{a}\in \{0,\dots ,P-1\}$ represent $ua(\mathrm{mod}P)$. Partition that range into $k+1$ half-open intervals of length $P/(k+1)$ and retain the most populated interval; it supplies $A^{\prime }$ with $|A^{\prime }|\ge n/(k+1)$. Define the model map by $\theta (a)=\tilde{a}(\mathrm{mod}N)$. For two $k$-term sums of retained representatives, their difference $T$ has $|T|<P$. If the original sums are equal, then $T\equiv 0(\mathrm{mod}P)$, hence $T=0$, so the model preserves the equality.

Conversely, if the model sums are equal, then $T$ is a multiple of $N$. If their original difference $d\in D$ were nonzero, it would satisfy $ud\equiv T(\mathrm{mod}P)$ with $0<|T|<P$, contradicting the forbidden-event choice. If $T=0$ directly, $ud\equiv 0$ implies $d=0$ too by the choice of $P$. Thus the equality is reflected. Padding equal one-term images with $k-1$ copies of a fixed retained element shows that the map is injective. This proves the Freiman s-isomorphism and the claimed bound on the cyclic modulus, independently of the diameter of $A$.

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.

A Freiman 2-isomorphism between subsets of abelian groups is a bijection $\varphi$ such thatBoth directions matter: an injective homomorphism on individual elements may still create new pair-sum relations. More generally a Freiman s-isomorphism preserves equalities of sums of $s$ elements in both directions. For a nonempty finite set, its doubling constant is $\sigma [A]=|A+A|/|A|$. A Freiman 2-isomorphism preserves this constant, since $a+b\mapsto \varphi (a)+\varphi (b)$ is a well-defined bijection of the two sumsets.

First let $A\subseteq \mathbb{Z}$ have $n$ elements and doubling at most $K$. The permitted Plünnecke-Ruzsa inequality gives $|2A-2A|\le K^{4}n$. We will obtain the half-size prime cyclic Freiman model for every prime $p>4K^{4}n$. Choose an auxiliary prime $q>p$ larger than the absolute value of every member of $2A-2A$. Such a prime exists: for any integer bound $M$, every prime divisor of $M!+1$ exceeds $M$.

Choose $\lambda$ uniformly from the nonzero residues modulo $q$. For each nonzero $d\in 2A-2A$, the residue $\lambda d$ is uniform among the nonzero residues. PutThere are at most $2(q-1)/p$ such residues, and therefore the probability that any of the nonzero differences is dilated into $T$ is at mostThe first cardinal estimate is an upper bound even if some listed residues coincide; the slightly looser probability bound is convenient. Thus there exists $\lambda$ with no such bad difference.

For this $\lambda$, let $t_a\in \{0,\dots ,q-1\}$ represent $\lambda a$. Split these representatives into the two intervals $[0,(q-1)/2]$ and $[(q+1)/2,q-1]$. Take $A^{\prime }$ from the more populated interval, so $|A^{\prime }|\ge n/2$. For any four elements of $A^{\prime }$, put $D=t_{a_1}+t_{a_2}-t_{a_3}-t_{a_4}$. Their common half-interval gives $|D|<q$, andIf the original difference is zero, then $D$ is a multiple of $q$ of absolute value less than $q$, so $D=0$. If the original difference is nonzero, $D$ is nonzero, and it cannot be a multiple of $p$: otherwise its residue belongs to $T$, contrary to the choice of $\lambda$. Therefore $a\mapsto t_a(\mathrm{mod}p)$ preserves pair-sum relations in both directions. It is injective as well, by applying the relation test to $a+a_0=b+a_0$ with any fixed $a_0\in A^{\prime }$. HenceThis proves the stated polynomial bound, including the prescribed prime rather than just an unspecified cyclic modulus.

Now let $A\subseteq {\mathbb{F}}_2^{\infty }$. Finiteness of $A$ places it in some finite-dimensional coordinate space, so models exist. Choose a Freiman 2-isomorphism $A\to C\subseteq {\mathbb{F}}_2^m$ with $m$ minimal. The doubling constant of $C$ is at most $K$.

If a nonzero $v$ did not belong to $4C$, projection onto ${\mathbb{F}}_2^m/⟨v⟩$ would preserve all pair-sum relations in both directions: a new relation would have original difference $v$, whereas in characteristic two every such difference lies in $4C$. Projection is also injective on $C$, since $C+C\subseteq 4C$ by adding any element of $C$ twice. This would give a model in dimension $m-1$, contradicting minimality. Consequently every nonzero vector belongs to $4C$; zero belongs as well, by adding one element four times. Thus $4C={\mathbb{F}}_2^m$.

Using the Plünnecke-Ruzsa inequality once more givesThis is the minimal binary Freiman models have full fourfold sumset argument. It also covers a singleton, with $m=0$.

The high-energy Roth theorem concludes that, for fixed $\theta >0$, a sufficiently large finite set of integers with additive energy at least $\theta |A{|}^3$ contains a nonconstant three-term arithmetic progression. The size threshold depends on $\theta$, and there is no assumption about the diameter of $A$.

Here is why additive energy replaces interval subset density. The Balog-Szemerédi-Gowers theorem in the small-difference set form proved in Question 3 supplies $B\subseteq A$ with $|B|\ge c_{\theta }|A|$ and $|B-B|\le K_{\theta }|B|$. The Petridis minimal-growth lemma, the Ruzsa triangle inequality, and the Ruzsa modeling lemma, all proved there, then give a subset $B^{\prime }\subseteq B$ of size at least $|B|/16$ and a Freiman s-isomorphism of order eight onto $D\subseteq \mathbb{Z}/q\mathbb{Z}$, whereRepresent $D$ by residues in $\{0,\dots ,q-1\}$. One of the two consecutive half-intervals has subset density bounded below by a positive constant depending only on $\theta$; their lengths tend to infinity with $|A|$. Apply the Roth theorem on three-term arithmetic progressions proved in part (i) to this interval. The resulting three distinct residues satisfy $d_0+d_2=2d_1$ modulo $q$. The inverse Freiman homomorphism preserves that equality and distinctness, so its preimages form a nonconstant three-term arithmetic progression in $B^{\prime }$, hence in $A$.

Large additive energy therefore gives the same three-term progression conclusion after passing to a dense finite model. A forward reference to the detailed lemmas in Question 3 avoids reproving them here.

We will prove a polynomial progression in a high-energy fourfold difference set. The route is to extract a large subset with a small difference set, construct a dense Freiman s-isomorphism model, and use the Bogolyubov lemma and the pigeonhole principle. To respect the proof requirement, every ingredient beyond the elementary rules for Freiman homomorphisms is established below.

Assume $n\ge 1$. Since three entries of an additive quadruple determine the fourth, $E(A)\le n^3$, so a nonvacuous hypothesis has $0<\theta \le 1$. Let $r(t)=|\{(a,b)\in A^2:a+b=t\}|$. Then $E(A)={\sum }_{t}r(t{)}^2$ and ${\sum }_{t}r(t)=n^2$. Define the popular sum set $S=\{t:r(t)\ge \theta n/2\}$. The contribution to additive energy from its complement is at most $\theta n^3/2$. Since $r(t)\le n$, the bipartite graph on two copies of $A$ whose edges have sums in $S$ has at least $\theta n^2/2$ edges. Moreover $|S|\le 2n/\theta$.

Put $\delta =\theta /2$ and $\tau ={\delta }^2/16$. Here is a dependent random choice argument producing the small difference set. Independently choose two right vertices, allowing repetition, and let $X$ be their common left neighbourhood. The Cauchy-Schwarz inequality gives $\mathbb{E}|X|\ge {\delta }^{2}n$. Call an ordered pair of left vertices bad if its common right neighbourhood has fewer than $\tau n$ elements. The expected number $b(X)$ of bad ordered pairs in $X$ is at most ${\tau }^2{n}^2$, because a fixed bad pair survives both random choices with probability at most ${\tau }^2$. ThusChoose a realization attaining at least this expectation. Then $|X|\ge 3{\delta }^{2}n/4$, and $b(X)\le {\delta }^{2}n|X|/64\le |X{|}^2/48$. Remove from $X$ every vertex having more than $|X|/4$ bad partners in $X$. At most $|X|/12$ vertices are removed. The remaining set $B$ has $|B|\ge {\delta }^{2}n/2$, and any $a,b\in B$ have at least $|X|/2$ vertices $z\in X$ for which both $(a,z)$ and $(z,b)$ are good.

For such a $z$, there are at least $\tau n$ right vertices $y$ adjacent to $a,z$, and at least $\tau n$ right vertices $y^{\prime }$ adjacent to $z,b$. These paths givea representation with four members of $S$. For fixed $a,b$, the tuple of four sums determines $y,z,y^{\prime }$ uniquely, so there are at least $|X|{\tau }^2{n}^2/2$ distinct tuples. Select one ordered pair $a,b$ for each member of $B-B$. Tuples belonging to different differences are disjoint. Counting all possible tuples in $S^4$ provesIn particular, with $m=|B|$, we have $m\ge {\theta }^{2}n/8$ and $|B-B|\le Km$ for $K=2^{25}{\theta }^{-12}$. This is the required small-difference set form of the Balog-Szemerédi-Gowers theorem, proved by counting paths rather than cited.

We next control higher iterated sumsets. Choose a nonempty $X\subseteq B$ minimizing $\rho =|X-B|/|X|$, so $\rho \le K$. We prove the Petridis minimal-growth lemma in the form $|X-B+C|\le \rho |X+C|$ for finite $C$. Order $C=\{c_1,\dots ,c_t\}$, and let $X_i$ consist of those $x\in X$ for which $x+c_i$ is new in the union of the translates $X+c_j$. Then ${\sum }_i|X_i|=|X+C|$. Every member of $(X\setminus X_i)-B+c_i$ already occurs in an earlier translate of $X-B$. By minimality, $|(X\setminus X_i)-B|\ge \rho |X\setminus X_i|$, also when the complement is empty. Thus the new contribution from $X-B+c_i$ is at most $\rho |X_i|$. Sum over $i$ to prove the lemma. Iterating with $C=-(j-1)B$ gives $|X-jB|\le K^j|X|$.

For completeness, the Ruzsa triangle inequality in the needed form isFor each $d\in U-W$, fix $d=u_d-w_d$. The map $(d,v)\mapsto (u_d-v,v-w_d)$ is injective: adding the coordinates recovers $d$, and then $v$. Take $U=kB$, $W=\ell B$, $V=X$ and apply the preceding bounds. We obtain the Plünnecke-Ruzsa inequalityIn particular $T=8B-8B$ has at most $K^{16}m$ elements.

We now prove a slightly wasteful version of the Ruzsa modeling lemma, avoiding any assumption about the diameter of $B$. Set $q=8|T|$. Choose a prime $p$ larger than all $|t|$, $t\in T$, and much larger than $q$. Arbitrarily large primes exist by the elementary argument that a prime divisor of one plus the product of all primes up to a given bound exceeds that bound. For a uniformly chosen nonzero multiplier $r$ modulo $p$, each nonzero $t\in T$ has $rt$ uniform among the nonzero residues. Call a multiplier forbidden if some such $rt$ agrees modulo $p$ with $jq$, where $|jq|<p$. There are at most $2p/q+1\le 3p/q$ such residues. For $p\ge 4$, a union bound therefore makes the probability of being forbidden less thanChoose a nonforbidden $r$. Represent $rB$ by integers in $[0,p-1]$ and partition that interval into sixteen consecutive pieces of diameter less than $p/16$. One piece contains the images of a subset $B^{\prime }\subseteq B$ with $|B^{\prime }|\ge m/16$. Let $\varphi (b)$ be the integer representative of $rb$ in that piece, reduced modulo $q$.

A difference of eight sums of these representatives has absolute value less than $p/2$. If its original difference in $B$ is zero, it is a multiple of $p$, hence zero. Conversely, if its reduction modulo $q$ is zero, it equals $jq$ with $|jq|<p$; our choice of $r$ forces its original difference to be zero. This proves that $\varphi$ is a Freiman s-isomorphism of order eight. Injectivity follows by padding a one-term equality to eight terms with a fixed element. Consequently $D=\varphi (B^{\prime })\subseteq G=\mathbb{Z}/q\mathbb{Z}$ has subset densityNo primality of $q$ is needed.

Here is the cyclic Bogolyubov lemma with all its Fourier analysis on a finite abelian group details. Use $\hat{f}(r)={\mathbb{E}}_{x}f(x)e(-rx/q)$, $f=1_D$, and normalized convolution on a finite group. A finite geometric series proves orthogonality, and expanding the finite sums proves the Parseval identity on a finite group, Fourier inversion on a finite group, and $\widehat{f\ast g}=\hat{f}\hat{g}$. For $\tilde{f}(x)=f(-x)$, the nonnegative functionhas support exactly $2D-2D$ and Fourier coefficients on a finite abelian group $|\hat{f}(r){|}^4$. Put $\Gamma =\{r\ne 0:|\hat{f}(r)|\ge {\alpha }^{3/2}/\sqrt{8}\}$. The Parseval identity on a finite group gives $|\Gamma |\le 8{\alpha }^{-2}$. The total fourth moment of the frequencies outside $\Gamma \cup \{0\}$ is at most ${\alpha }^4/8$.

If $\Vert rx/q\Vert \le 1/16$ for every $r\in \Gamma$, then $\mathrm{Re}e(rx/q)\ge \cos (\pi /8)>3/4$. The zero frequency contributes ${\alpha }^4$ with no loss. Separating the large and small frequencies in Fourier inversion on a finite group yieldsThus $2D-2D$ contains the Bohr set expressed in distance coordinates by $\Vert rx/q\Vert \le 1/16$. This is consistent with the existing character-distance convention for a Bohr set: here it is enough that each character lies in the corresponding short arc around one.

We finally prove the needed nonwrapping progression in a cyclic Bohr set, including for composite $q$. Write $d=|\Gamma |$ and $Q=\lfloor q^{1/(d+1)}\rfloor$. Partition the $d$-dimensional unit cube into $Q^d$ boxes. The $Q^d+1$ points with coordinates $jr/q$ modulo one, $0\le j\le Q^d$, give by the pigeonhole principle an integer $1\le t\le Q^d$ with $\Vert rt/q\Vert \le 1/Q$ for all $r\in \Gamma$. ThenThese points are distinct modulo $q$, since their integer representatives lie between zero and $Q^{d+1}/16\le q/16$. Their number is at least $q^{1/(d+1)}/32$. For $d=0$ the same argument simply chooses $t=1$.

The Freiman s-isomorphism induces a bijection $\Phi :2B^{\prime }-2B^{\prime }\to 2D-2D$: define it on a representation by taking the same two positive and two negative images. Equality of representations uses an order-four relation, and injectivity uses the converse relation. If three consecutive members $y_{j-1},y_j,y_{j+1}$ of $P$ satisfy $y_{j-1}+y_{j+1}=2y_j$, their preimages satisfy the same equality, since its expansion is an equality of eight-term sums in $B^{\prime }$. Hence the inverse image of $P$ is an ordinary arithmetic progression in $2B^{\prime }-2B^{\prime }\subseteq 2A-2A$, with distinct terms.

To remove the multiplicative constant from the length bound, let $R=\lceil 8{\alpha }_0^{-2}\rceil$ and $\beta =1/(R+1)$. The length just proved is at least $c_{\theta }{n}^{\beta }$, where $c_{\theta }={\theta }^{2\beta }/32>0$. Choose $N_0\ge 3$ so that $c_{\theta }{n}^{\beta }\ge n^{\beta /2}$ for $n\ge N_0$, and set $\gamma =\min \{\beta /2,\log 3/\log N_0\}$. For $2\le n<N_0$, two distinct members of $A$ give a nonzero $u\in A-A$, and $\{-u,0,u\}\subseteq 2A-2A$ has length three, at least $n^{\gamma }$. For $n=1$, the singleton $\{0\}$ suffices. In every nonempty case,with $\gamma >0$ depending only on $\theta$. If empty sets are permitted, the required length is zero and there is no substantive assertion.

A bijection $\varphi :A\to B$ is a Freiman s-isomorphism when, for every $a_1,\dots ,a_s,a_1^{\prime },\dots ,a_s^{\prime }\in A$,holds if and only if

Because $A\subseteq {\mathbb{Z}}^2$ is finite, choose $R$ with $A\subseteq [-R,R{]}^2$ and an integer $M>2sR$. DefineAn equality of $s$-term sums in ${\mathbb{Z}}^2$ plainly gives equality after applying this linear map. Conversely, equality of the images giveswhere $|{\Delta }_x|\le 2sR<M$. Therefore ${\Delta }_x=0$, and then ${\Delta }_y=0$. The same estimate with one term on each side shows that $\varphi$ is injective on $A$. Hence $\varphi$ is a Freiman s-isomorphism from $A$ to the subset $\varphi (A)\subseteq \mathbb{Z}$.

Consider the graphThe hypothesis says exactly that $\Gamma$ has at least $\delta N^3$ additive quadruples. By the Balog-Szemerédi-Gowers theorem, it contains ${\Gamma }^{\prime }$ with $|{\Gamma }^{\prime }|{\gg }_{\delta }N$ and $|{\Gamma }^{\prime }+{\Gamma }^{\prime }|{\ll }_{\delta }|{\Gamma }^{\prime }|$.

Part b gives a Freiman s-isomorphism from ${\Gamma }^{\prime }$ to a set $D\subseteq \mathbb{Z}$, where it is enough to take any fixed $s\ge 2$. The set $D$ has bounded doubling, with a bound depending only on $\delta$. Part c therefore provides a proper generalized arithmetic progression $Q$ of rank $O_{\delta }(1)$ such that

Write $Q$ as a proper parameter box. Since $|Q|{\gg }_{\delta }N$, one of its $O_{\delta }(1)$ side lengths tends to infinity with $N$. Averaging over all lines parallel to that side gives a line on which $D\cap Q$ has density bounded below in terms of $\delta$. The Szemerédi theorem quoted in the question then gives, once $N$ is sufficiently large in terms of $k$ and $\delta$, a nonconstant $k$-term arithmetic progression in $D$.

The inverse Freiman isomorphism sends it to a $k$-term arithmetic progression in ${\Gamma }^{\prime }\subseteq \Gamma$, because each relation between three consecutive terms is an additive-quadruple relation. Write this progression asIts first-coordinate difference cannot be zero: the graph of a function has only one point above each $x$. Hence $d\ne 0$. On the nonconstant progressionwe haveTaking $a=e/d$ and $b=y_0-(e/d)x_0$, both in $\mathbb{Q}$, proves the claim.

For every $\theta >0$, some $\gamma (\theta )>0$ has this property: any nonempty finite $A\subseteq \mathbb{Z}$ with $E(A)\ge \theta |A{|}^3$ has an arithmetic progression of length at least $|A{|}^{\gamma (\theta )}$ in $2A-2A$. The small-difference-set form of the Balog-Szemerédi-Gowers theorem, Ruzsa modelling lemma, cyclic Bogolyubov lemma and nonwrapping progression in a cyclic Bohr set prove it. An order-eight Freiman s-isomorphism suffices to lift the progression because its consecutive second-difference equations expand into equalities of eight-term sums.