I only have one in russian, but the topic is easy, so I can as well explain it here.

Suppose there is a graph with $$$n$$$ vertices and $$$m$$$ edges.

First I'm gonna prove, that there are $$$O(m \sqrt m)$$$ triangles. For that I'll split vertices into light, when $$$deg_v \le \sqrt m$$$ and heavy, when $$$deg_v \gt \sqrt m$$$. Then there are $$$O(\sqrt m)$$$ heavy vertices. Each triangle has either a light or a heavy edge, that means, the edge connects both vertices of that type.

Fix a light edge, there are $$$O(m)$$$ of these, for each edge there is $$$O(\sqrt m)$$$ ways to pick the third vertice, since it must lie in the intersection of two adjacency lists.

For a heavy edge we'll fix one of the endpoints and the third vertice, there are $$$O(m)$$$ variants to do that since it's an edge of the graph. There are $$$O(\sqrt m)$$$ ways to find a heavy vertice since there are only $$$O(sqrt m)$$$ in total.

This proves an upperbound on the number of triangles. To get the lowerbound just construct a full graph on $$$\sqrt{2m}$$$ vertices. Also this is an algorithm to find the triangles.

But there is another approach which is easier to code, so some people use it instead (maybe without knowing the proof, I've only learned it today)

That is to order vertices by their degree (and number in case of same degrees). Then we want to find all triples $$$(v, u, w)$$$ such that $$$v \lt u \lt w$$$ and for all three pairs there's an edge. So in a way we orient edges. Now the key observation is that outdegree is bounded by $$$\sqrt {2m}$$$. For light vertices the bound is achieved by their degree and for heavy vertices the bound is achieved by the number of the vertices to the right of them. So this code

for v
    for u in g[v]
        mark[u] = true
    for u in g[v]
        for w in g[u]
            if mark[w]:
                 #(v, u, w) triangle
    for u in g[v]
        mark[u] = false

works in $$$O(m \sqrt m)$$$, if all of the edges are to some bigger vertice and there are no repeated edges.