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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.

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.