For which $n,m$ is $K_{n,m}$ planar? Justify an answer.
I made conclusion that for $K_{n,m}$, where $2 \leq n,m $ then $K_{n,m}$ is not planar not depenting whether $n$ or $m$ is odd or even. Is it true? I want to use Theorem which states that is a triangle- free graph $e(G) \leq 2 n(G)-4$. Since bipartite graphs are triangle free this should hold however $mn \leq 2(m+n) - 4$ does not hold for $n=4$,$m=5$ for example and for numbers bigger. But is this really enough for the proof?
Thanks in advance for help
We know that $K_{3,3}$ is non-planar. So when we have $m \ge 3$ and $n \ge 3$, the graph will have $K_{3,3}$ as a subgraph, so by Kuratowski's Theorem it is non-planar. For the rest, you only should look $K_{2,m}$ for all $m$, which is obviously planar (you can draw it and see). So in general, it is planar for $n \le 2$ and when this condition is satisfied, it is independent from $m$.