Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 7 / 6

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.

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.