For $n\in N_+$, define f(n) to be that for any n-vertice graph G, if any two triangle in G don't have a common edge, then G has at most f(n) triangles.

Do we have some good estimates for f(n)?

By triangle removal lemma, we can prove for any $\varepsilon>0$, while n is large, we have $f(n)<\varepsilon n^2$.

(Intuitively, for a regular partition of G, if any triangle contributes an edge to a "low density part" (or some ignored part), then the "low density part" can not suffer so much edge. So there's some triangle which every edge contained in a "high density part", and then we'll get $cn^3$ triangles, which can't independent in edge.)

If we can prove that $f(n)<\frac{\varepsilon n^2}{ln\ n}$ for large n, then we can prove that there exists infinitely many triples of primes which forms an arithmetic sequence, in a combinatorial way.

For $n\in N_+$, define $f(n)$ to be the maximum number of triangles in a graph $G$ with $n$ vertices, taken over all $n$-vertex graphs having the property where no two triangles have a common edge.

Do we have some good estimates for $f(n)$?

By triangle removal lemma, we can prove for any $\varepsilon>0$, while $n$ is large, we have $f(n)<\varepsilon n^2$.

(Intuitively, for a regular partition of $G$, if any triangle contributes an edge to a "low density part" (or some ignored part), then the "low density part" can not suffer so much edge. So there's some triangle which every edge contained in a "high density part", and then we'll get $cn^3$ triangles, which can't independent in edge.)

If we can prove that $f(n)<\frac{\varepsilon n^2}{ln\ n}$ for large $n$, then we can prove that there exists infinitely many triples of primes which forms an arithmetic sequence, in a combinatorial way.