Graphs with no 4-cycle

1. Count the pairs $(v,\{x,y\})$ where $x\ne y$ are both neighbours of $v$. For a fixed $v$ there are $\binom{d(v)}2$ pairs of neighbours, so the total is $\sum_v\binom{d(v)}2$.

Fix a pair $\{x,y\}$. If it had two distinct common neighbours $v,w$, then $x\,v\,y\,w$ would be a $4$-cycle. So each pair $\{x,y\}$ occurs for at most one $v$, and the total is at most $\binom n2$.

2. Let $n=5$ and suppose there are $e\ge7$ edges. The function $f(x)=\frac{x(x-1)}2$ is convex, so by Jensen's inequality
$$\sum_{v}f(d(v))\ge5f\left(\frac{2e}5\right)\ge5f\left(\frac{14}5\right)=12.6>10=\binom52,$$
contradicting part 1. So $e\le6$.

Example with $6$ edges. Take two triangles sharing one vertex $c$ (a "bowtie"): triangles $c,a,b$ and $c,d,e$. Every cycle lies inside one of the two triangles, because $c$ is the only vertex joining them. So there is no $4$-cycle.

Remark. The same method gives, for large $n$, that a $C_4$-free graph has at most about $\frac12n^{3/2}$ edges.

Lesson. Double count (vertex, pair of neighbours); forbidden configurations bound the count from above and convexity bounds it from below.