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The additive energy of a finite set $S\subseteq G$ in an abelian group $G$ counts additive quadruples, with order and repeated members allowed:At a fixed sum $s$, independently choosing two of its $r_{S+S}(s)$ ordered pairs gives $r_{S+S}(s{)}^2$ additive quadruples. Thus additive energy weights heavily represented sums more strongly than the cardinality of the sumset does.

For $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$, the sumset representation functions giveThe arithmetic progression has additional coincidences such as $0+3=1+2$. The set $B$ still has coincidences forced by commutativity, including $0+1=1+0$, so its additive energy is not merely the sixteen identical-pair comparisons.

The same additive energy can be counted using the difference set. Define $r_{S-S}(t)=|\{(a,b)\in S\times S:a-b=t\}|$. The identity $a+b=c+d$ is equivalent to $a-c=d-b$, and the coordinate rearrangement $(a,b,c,d)\mapsto (a,c,d,b)$ is a bijection. ConsequentlyFor $A$, the difference counts from $-3$ through $3$ are again $1,2,3,4,3,2,1$. For $B$, the zero difference has four representations and each of the twelve nonzero differences has one, giving $4^2+12\cdot 1^2=28$. The sumset representation functions for sums and differences can look different while their squared totals agree.

For every nonempty finite set $S\subseteq G$ in an abelian group $G$,The lower bound follows from the Cauchy-Schwarz inequality and the total sumset representation function count:For the upper bound, any three members $a,b,c$ determine at most one fourth member $d=a+b-c$ of an additive quadruple. See Yufei Zhao's MIT lecture 25, pages 1-3, for representation counts and the energy bounds.

In the two integer examples, the lower bounds are $256/7$ and $256/10$, and the upper bound is $64$. For the subgroup $H=\{0,2,4\}$ of $\mathbb{Z}/6\mathbb{Z}$, each element of $H$ has three sum representations, so $E(H)=3\cdot 3^2=27$ and both bounds are equalities.

For nonempty $S$, dividing additive energy by $|S{|}^3$ gives a relative measure between $1/K(S)$ and one. Small doubling constant therefore forces relatively large additive energy. A converse requires care: the Balog-Szemerédi-Gowers theorem extracts a large subset with small doubling constant, rather than asserting that all of $S$ has small doubling constant.

For fixed $c>0$ and prime $p$, if $A\subseteq {\mathbb{F}}_p^N$ has at least $c|A{|}^3$ additive quadruples, then $2A-2A$ contains a vector subspace of size at least $c^{\prime }(c,p)|A|$. Combine the Balog-Szemerédi-Gowers theorem, the Freiman-Ruzsa theorem over a finite field, and the Finite-field Bogolyubov lemma.

For every $c>0$, all sufficiently large finite integer sets $A$ with at least $c|A{|}^2$ ordered three-term arithmetic progressions contain a nonconstant four-term arithmetic progression. Cauchy-Schwarz inequality gives $E(A)\ge c^2|A{|}^3$. The Balog-Szemerédi-Gowers theorem supplies a large small-doubling subset. A bounded-modulus Freiman 2-isomorphism gives a positive-density cyclic model; apply the assumed four-term case of Szemerédi's theorem to representatives, then lift the two consecutive second-difference relations.

If a dense set has too few four-term arithmetic progressions, its balanced function has large $U^3$ norm. A Frequency graph extracted from a large Gowers U3 norm, followed by the Balog-Szemerédi-Gowers theorem and a Freiman theorem, yields correlation with a quadratic phase. Restricting to a suitable level set gives a density increment, and iteration proves the Finite-field Szemerédi theorem for four-term arithmetic progressions.

There are absolute $c,C>0$ such that finite sets $A,B$ of size $n$, a bipartite graph of density at least $\delta$, and a restricted sumset of size at most $Kn$ yield subsets $A^{\prime },B^{\prime }$ with $|A^{\prime }|,|B^{\prime }|\ge c{\delta }^{C}n$ and $|A^{\prime }+B^{\prime }|\le C{\delta }^{-C}{K}^{C}n$. A graph formed by popular representations of sums turns this into the usual large-additive energy form of the Balog-Szemerédi-Gowers theorem.

For fixed $\theta >0$, a sufficiently large finite set of integers with additive energy at least $\theta |A{|}^3$ has a nonconstant three-term arithmetic progression. The Balog-Szemerédi-Gowers theorem and Ruzsa modelling lemma reduce the problem to the Roth theorem on three-term arithmetic progressions in a dense cyclic model.

One quantitative form of the Balog-Szemerédi-Gowers theorem is: if a nonempty finite set $A$ in an abelian group has $|A|=n$ and additive energy $E(A)\ge \eta n^3$, where $0<\eta \le 1$, then there is $B\subseteq A$ with $|B|\ge c_{\eta }n$ and both $|B-B|$ and $|B+B|$ bounded by $K_{\eta }|B|$. The constants depend only on $\eta$. We prove explicit, nonoptimal polynomial bounds, thereby proving the required Balog–Szemerédi statement.

Let $r(s)=|\{(a,a^{\prime })\in A^2:a+a^{\prime }=s\}|$. Then $E(A)={\sum }_{s}r(s{)}^2$. Declare $S=\{s:r(s)\ge \eta n/2\}$ to be the set of popular sums. The remaining sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$. Since $r(s)\le n$, it follows that ${\sum }_{s\in S}r(s)\ge \eta n^2/2$. Also $|S|\le 2n/\eta$. Form a bipartite graph on two copies of $A$, joining $a$ to $a^{\prime }$ when $a+a^{\prime }\in S$. Its density is at least $\rho =\eta /2$.

We prove the needed four-step path lemma for a dense bipartite graph. Call an ordered pair of left vertices bad if its number of common right neighbors is less than $\tau n$, where $\tau ={\rho }^2/32$. For a uniformly chosen right vertex $y$, let $U=N(y)$ and let $b(U)$ count bad ordered pairs in $U^2$. ThenConsequently some $y$ satisfiesFor this $U$, $|U|\ge \rho n/2$ and $b(U)\le \rho n|U|/16\le |U{|}^2/8$. Delete vertices having more than $|U|/4$ bad partners in $U$. At most half are deleted; the retained set $B$ has $|B|\ge |U|/2\ge \rho n/4$. For any $a,b\in B$, at least $|U|/2$ middle vertices $u\in U$ are good partners of both. Each such $u$ supplies at least $(\tau n{)}^2$ walks $a,v,u,w,b$ of four edges. Hence every pair in $B$ has at leastsuch walks. Repetition of vertices is allowed; it does not affect the counting argument.

Each walk has edge labelsall in $S$, and $a-b=s_1-s_2+s_3-s_4$. For a fixed pair $a,b$, these four labels determine $v,u,w$, so distinct walks have distinct label quadruples. For each element of $B-B$, fix one representing pair $a,b$. Label quadruples for distinct differences cannot overlap. ThereforeUsing $\rho =\eta /2$ and $|S|\le 2n/\eta$ yieldsThis proves the small-difference-set form of the Balog-Szemerédi-Gowers theorem. To obtain a small sumset too, write $K=2^{24}{\eta }^{-10}$ and apply the already proved Plünnecke inequality to starting set $-B$ and summand set $B$. Some nonempty $X\subseteq -B$ has $|X+2B|\le K^2|X|$. For any fixed $x\in X$, $x+2B\subseteq X+2B$, whence $|B+B|\le K^2|B|$. Thus the conventional small-sumset conclusion also follows with constants depending only on $\eta$.

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.

Assume $n>0$; the empty case is immediate. Write $r(s)=\#\{(a,b)\in A^2:a-b=s\}$. The additive energy in the hypothesis is ${\sum }_{s}r(s{)}^2$: rearranging $x+y=z+w$ gives $x-z=w-y$. Since $r(s)\le n$, the energy is at most $n^3$, so the substantive range is $0<\alpha \le 1$.

Let $S=\{s:r(s)\ge \alpha n/2\}$. The contribution of the other differences to the energy is at most $(\alpha n/2){\sum }_{s}r(s)=\alpha n^3/2$. ConsequentlyMake a bipartite graph with two copies of $A$, joining a left vertex $a$ to a right vertex $b$ when $a-b\in S$. Put $\rho =\alpha /2$. This graph has at least $\rho n^2$ edges, and $|S|\le n/\rho$.

We prove the needed dense-graph selection rather than quote the Balog-Szemerédi-Gowers theorem. A pair of left vertices is bad if it has fewer than $\tau n$ common right neighbours, where $\tau ={\rho }^2/32$. Choose a right vertex $y$ uniformly, let $U=N(y)$, and let $b(U)$ be the number of ordered bad pairs in $U$. Averaging the degrees gives $\mathbb{E}|U|\ge \rho n$. Every bad pair lies in fewer than $\tau n$ of these neighbourhoods, so $\mathbb{E}b(U)\le \tau n^2$. Therefore some $y$ satisfiesFor this neighbourhood, $|U|\ge \rho n/2$ and $b(U)\le \rho n|U|/16\le |U{|}^2/8$. Delete vertices having more than $|U|/4$ bad partners in $U$. At most half are deleted, so the remaining set $B$ has $|B|\ge |U|/2\ge \rho n/4$.

For any $b,b^{\prime }\in B$, at least $|U|/2$ vertices $v\in U$ are good partners of both. Each such $v$ has at least $\tau n$ common neighbours with $b$ and at least $\tau n$ with $b^{\prime }$. Independently choosing those neighbours gives at leastfour-edge walks $b,u,v,w,b^{\prime }$. Repeated vertices cause no difficulty. This proves the four-step path lemma for a dense bipartite graph in the required form.

The labels of such a walk are $s_1=b-u$, $s_2=v-u$, $s_3=v-w$, $s_4=b^{\prime }-w$, all in $S$, andFor a fixed pair $(b,b^{\prime })$, its label tuple uniquely recovers $u=b-s_1$, $v=u+s_2$, and $w=v-s_3$, so distinct walks have distinct label tuples. Choose one pair representing each element of the difference set $B-B$. Tuples belonging to different differences are disjoint because their alternating sum determines that difference. HenceWe have obtained explicit constants for the small-difference-set form of the Balog-Szemerédi-Gowers theorem:

For a finite nonempty set $A$, its additive energy iswhere $r_{A+A}(s)$ counts ordered representations. Since ${\sum }_s{r}_{A+A}(s)=|A{|}^2$, Cauchy-Schwarz givesThus $|A+A|\le K|A|$ implies $E(A,A)\ge |A{|}^3/K$: a small doubling constant forces many additive quadruples.

A polynomial form of the Balog-Szemerédi-Gowers theorem gives the converse after passing to a subset. There are absolute constants $c,C>0$ such that if $E(A,A)\ge \eta |A{|}^3$, $0<\eta \le 1$, then some $A^{\prime }\subseteq A$ satisfiesThe exponent and constants here need not be optimal.

Precisely, a graph form of the Balog-Szemerédi-Gowers theorem states the following. If $A,B$ are finite sets of size $n$ in an abelian group, $\Gamma \subseteq A\times B$ has at least $\delta n^2$ edges, and its restricted sumsethas size at most $Kn$, then there are $A^{\prime }\subseteq A$, $B^{\prime }\subseteq B$ with $|A^{\prime }|,|B^{\prime }|\ge c{\delta }^{C}n$ andThese are uniform absolute constants, valid for $0<\delta \le 1$ and $K\ge 1$.

To apply this theorem to additive energy, let $n=|A|$ and retain the sums with $r_{A+A}(s)\ge \eta n/2$. The discarded sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$ to the additive energy. Since $r(s)\le n$, the retained pairs number at least $\eta n^2/2$. There are at most $2n/\eta$ retained sums. They therefore define a bipartite graph of density at least $\eta /2$ with restricted sumset size at most $2n/\eta$. The graph theorem supplies large subsets $A^{\prime },B^{\prime }$ with polynomially bounded $|A^{\prime }+B^{\prime }|$.

For completeness, turn this cross-sumset bound into a doubling constant bound. The Ruzsa triangle inequality gives $|A^{\prime }-A^{\prime }|\le |A^{\prime }+B^{\prime }{|}^2/|B^{\prime }|$. If $D=|A^{\prime }-A^{\prime }|/|A^{\prime }|$, the Plünnecke inequality, applied to the pair $A^{\prime },-A^{\prime }$, gives $|A^{\prime }+A^{\prime }|\le D^2|A^{\prime }|$. The minimal-growth proof below in Question 5 applies to arbitrary finite pairs and proves this particular expansion estimate as well. All losses are fixed powers of ${\eta }^{-1}$, proving the displayed converse after increasing $C$.

The requested dense bipartite common-neighbour lemma is a dependent random choice assertion. Call an ordered pair in $V^2$ bad when it has fewer than $ϵn$ common neighbours. Choose $w\in W$ uniformly, and put $S=N(w)\subseteq V$. ThenFor each bad pair, the probability that both endpoints lie in $S$ is its number of common neighbours divided by $n$, which is less than $ϵ$. Thus the expected number $b(S)$ of bad ordered pairs in $S^2$ is at most $ϵm^2$. ConsequentlyChoose a neighbourhood $V^{\prime }=S$ attaining at least this expectation. It has $|V^{\prime }|\ge \delta m/\sqrt{2}\ge \delta m/2$, andHence at least the proportion $1-2ϵ/{\delta }^2$ of its ordered pairs have at least $ϵn$ two-edge connections. As usual these connections are counted by common neighbours, including a return walk for a diagonal ordered pair; assume $\delta >0$, as the formula requires.

The relevance to the graph theorem is that many two-edge connections turn a sparse collection of permitted sums into additive control on a large vertex subset. On a connection $a\to b\to a^{\prime }$, $a-a^{\prime }=(a+b)-(a^{\prime }+b)$ expresses a difference using two elements of the restricted sumset. The abundance of connections, followed by standard cleaning and further path counting, bounds the number of differences without throwing away most vertices. This is the role of dependent random choice in obtaining polynomial losses in the graph form of the Balog-Szemerédi-Gowers theorem.

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.

The Balog-Szemerédi-Gowers theorem states that if a finite set $A$ has at least $\delta |A{|}^3$ additive quadruples, then it has a subset $A^{\prime }\subseteq A$ such thatwhere $C$ is an absolute constant.

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.

The hypothesis says that the additive energy satisfies $E(A)\ge c|A{|}^3$. The Balog-Szemerédi-Gowers theorem supplies $A^{\prime }\subseteq A$ withBy the Freiman-Ruzsa theorem over a finite field, $A^{\prime }$ lies in a subspace $H$ withThus $A^{\prime }$ has density at least $C_2^{-1}$ in $H$. Applying the Finite-field Bogolyubov lemma inside $H$ gives a subspaceof codimension bounded in terms of $c,p$ alone. Thereforewhich is the additive energy produces a large subspace in a fourfold difference set result.

Let $A\subseteq {\mathbb{F}}_5^n$ have density at least $\delta$ and put $f=1_A-\mathbb{E}{1}_A$. If $A$ has too few four-term progressions, part ii forces $\Vert f{\Vert }_{U^3}$ to be large. Part iii then produces a frequency graph with large additive energy. The Balog-Szemerédi-Gowers theorem extracts a large piece with small doubling, and a finite-field Freiman theorem makes the frequency selection approximately affine-linear there. Integrating these approximately linear derivative frequencies produces correlation of $f$ with a quadratic phase, as in the inverse theorem for the Gowers U3 norm over a finite field.

Restricting to a suitable level set of that quadratic phase produces a density increment on a structured affine subspace. Iterating these increments cannot continue indefinitely because density is at most one. Once $n$ is sufficiently large in terms of $\delta$, the iteration must instead terminate with the expected nontrivial four-term progression. This is the density-increment proof of the finite-field four-term progression theorem.

Part (iii) gives at least ${\eta }^2|A{|}^3$ additive quadruples. By the Balog-Szemerédi-Gowers theorem, there is $A^{\prime }\subseteq A$ withThe Freiman-Ruzsa theorem places $A^{\prime }$ in a coset progression $P$ whose rank is bounded in terms of $\eta$ and which satisfiesThus $A^{\prime }$ has density at least $C_1(\eta {)}^{-1}$ inside $P$.

The Szemerédi theorem in a bounded-rank coset progression says that, for fixed rank and density, every sufficiently large such progression has a nontrivial four-term progression in every subset of that density. If $|A|\ge C(\eta )$ for a sufficiently large $C(\eta )$, then $|P|\ge |A^{\prime }|\ge c(\eta )|A|$ crosses this threshold. It gives a nontrivial four-term progression in $A^{\prime }$, which is also contained in $A$. This proves the claim with a constant depending only on $\eta$.

The hypothesis says that the normalized additive energy of $A$ is at least $\eta$. By the Balog-Szemerédi-Gowers theorem, for an absolute $C_0$ there is $A^{\prime }\subseteq A$ such thatThe finite-field Bogolyubov-Ruzsa consequence of the Freiman-Ruzsa theorem over a finite field says that a set of doubling at most $K$ has a vector subspacewith $|V|\ge p^{-K^{C_1}}|A^{\prime }|$ for an absolute $C_1$. Taking $K={\eta }^{-C_0}$ and enlarging the absolute exponent givesThis is an energy form of the Bogolyubov lemma: the usual lemma assumes positive density in an ambient group, whereas the Balog-Szemerédi-Gowers theorem first extracts a dense structured model from the many additive quadruples. The resulting bound depends on the energy parameter rather than on the possibly tiny ambient density of $A$.