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Choose a generator $\theta$ of the multiplicative group of the quadratic finite field extension. Its $p$ shifts $\theta +t$ define $p$ distinct exponents modulo $p^2-1$. Equality of two sums of exponents gives $(\theta +s)(\theta +t)=(\theta +u)(\theta +v)$. Since $1,\theta$ are independent over the prime field, $s+t=u+v$ and $st=uv$, so the unordered pairs coincide. Thus the exponents form a Sidon set in the cyclic group. Taking integer representatives preserves the Sidon property. Primes within $o(\sqrt{N})$ below $\sqrt{N}$ give $\sqrt{N}-o(\sqrt{N})$ representatives in an interval of length $N$.

Write $m=|A|$, choose $u=\lfloor N^{3/4}\rfloor$, and let $A_i$ count the points of $A$ in the length-$u$ window ending at $i$, for $1\le i\le N+u$. Each element occurs in precisely $u$ such windows, so ${\sum }_i{A}_i=um$. The Cauchy-Schwarz inequality consequently givesOn the other hand, a pair $a<b$ of difference $d=b-a<u$ belongs to exactly $u-d$ windows; larger differences contribute none. The Sidon set condition permits at most one pair for each positive difference $d$. ThereforeCombining these two estimates proves the sliding-window upper bound for Sidon sets:If $D=(N+u)/u$, the final inequality for this quadratic polynomial implies $m\le \sqrt{N+u}+D$; otherwise $m(m-D)>N+u$. With the chosen $u$, $D=O(N^{1/4})$ and $\sqrt{N+u}=\sqrt{N}+O(N^{1/4})$. ThusTogether with part (i)'s construction, this proves that the maximum cardinality is $\sqrt{N}+o(\sqrt{N})$. The lower and upper error terms need not have the same size to establish that asymptotic conclusion.

We use the Bose-Chowla Sidon construction, deriving its difference property explicitly. Let $p$ be an odd prime and choose a generator $\theta$ of the multiplicative group of a finite field of the quadratic finite field extension ${\mathbb{F}}_{p^2}$. Such a generator cannot belong to ${\mathbb{F}}_p$, so $1,\theta$ are linearly independent over the prime field. For each $t\in {\mathbb{F}}_p$, the nonzero field element $\theta +t$ has a unique exponent $a_t$ modulo $p^2-1$, with ${\theta }^{a_t}=\theta +t$. The $p$ exponents are distinct.

Suppose $a_s-a_t=a_u-a_v$ in the cyclic group. Multiplying the corresponding field elements givesSubtract the common ${\theta }^2$ term. Linear independence of $1,\theta$ gives $s+v=u+t$ and $sv=ut$. Thus the two unordered pairs are the roots of the same monic quadratic polynomial, so $\{s,v\}=\{u,t\}$. Either $s=u,v=t$, giving the same ordered difference, or $s=t,v=u$, giving two zero differences. This is exactly the Sidon set property.

Choose the integer representatives in $\{1,\dots ,p^2-1\}$. Equality of integer differences implies equality modulo $p^2-1$, so their Sidon property persists in the integers. Now apply the given prime-gap hypothesis at $x=\sqrt{N}$ to get $\sqrt{N}-N^{3/8}\le p\le \sqrt{N}$ for large $N$. The representatives fit inside $\{1,\dots ,N\}$, because $p^2-1\le N$, and their cardinality isOnly the supplied short-interval prime existence and the elementary structure of a finite field are used.

First consider a $K_{t,t}$-free bipartite graph with parts $X,Y$, each of size at most $n$, and $m$ edges. Count pairs $(S,y)$ in which $S$ is a $t$-element subset of the graph neighbourhood of $y\in Y$. Any $t$ vertices of $X$ have at most $t-1$ common neighbours, since $t$ common neighbours would form the complete bipartite graph $K_{t,t}$. ConsequentlyIf the average degree of a vertex is bounded in terms of $t$, then $m=O(n)$ and the required result is immediate. Otherwise, convexity of $x\mapsto \left(\genfrac{}{}{0ex}{}{x}{t}\right)$ and the preceding inequality giveand hence $m=O(n^{2-1/t})$. Every graph has a bipartite spanning subgraph containing at least half its edges, obtained by choosing a random bipartition. Applying the bipartite estimate to that subgraph proves the Kővári–Sós–Turán theorem in the required form:

The stated "nice" subsets of the finite cyclic group ${\mathbb{Z}}_n$ are Sidon sets. Ifthen $a+d=c+b$. Niceness forces $(a,b)=(c,d)$; the alternative pairing would force $a=b$. Thus the $|A|(|A|-1)$ ordered nonzero differences are distinct elements of the $n-1$ nonzero residue classes. Thereforeso

For a Sidon set $A\subseteq [1,N]$, put $m=|A|$ and let $A_i$ count points in a window of $u$ consecutive integers. Each point occurs in $u$ windows, so $\sum A_i=um$. Cauchy-Schwarz gives $\sum \left(\genfrac{}{}{0ex}{}{A_i}{2}\right)\ge [u^2{m}^2/(N+u)-um]/2$. Each positive difference occurs at most once and contributes $u-d$ windows, giving the upper bound $u(u-1)/2$. Thus $m^2\le (N+u)(1+(m-1)/u)$. Taking $u=\lfloor N^{3/4}\rfloor$ gives the displayed estimate, which combines with the Bose-Chowla Sidon construction to determine the asymptotic maximum.