Given a prime \(p\text{,}\) let \(G_p\) be the graph with vertex set \(\mathbb{Z}_p\times \mathbb{Z}_p\) where \((x,y)\) is adjacent to \((x',y')\) if \(x+x'=yy'\) and \((x,y)\neq (x',y')\text{.}\) For any \((x,y)\) and any choice of \(y'\text{,}\) there is a unique \(x'\) for which the equation is satisfied. Thus, each \((x,y)\) has degree either \(p-1\) or \(p\text{,}\) depending on whether or not the equation is satisfied with \((x',y')=(x,y)\text{.}\) Therefore, the number of edges is at least
\begin{equation*}
\frac{1}{2}p^2(p-1) = \Theta(n^{3/2})
\end{equation*}
where \(n=p^2\) is the number of vertices.
Now, let us show that \(G_p\) does not have any \(4\)-cycles. Let \((x,y)\) be a vertex and let \((x_1,y_1)\) and \((x_2,y_2)\) be two distinct neighbours of it. Then
\begin{equation*}
x+x_1 = yy_1
\end{equation*}
and
\begin{equation*}
x+x_2=yy_2\text{.}
\end{equation*}
Subtracting these equations, we get
\begin{equation*}
x_1-x_2 = y(y_1-y_2)\text{.}
\end{equation*}
Thus, given the choice of \(x_1,x_2,y_1\) and \(y_2\text{,}\) provided that \((x_1,y_1)\neq (x_2,y_2)\text{,}\) the value of \(y\) is uniquely determined from this equation. Taking the equation \(x+x_1=yy_1\) then uniquely determines \(x\text{.}\) Therefore, \((x,y)\) is the unique common neighbour of \((x_1,y_1)\) and \((x_2,y_2)\text{,}\) and so \(G_p\) does not contain a \(4\)-cycle.