1 on Some Problems in Graph Theory , Combinatorial Analysis and Combinatorial Number Theory

1. G(n) is a graph of n vertices and G(n ; e) is a graph of n vertices and e edges. Is it true that if every induced subgraph of a G(10n) of 5n vertices has more than 2n 2 edges then our G(10n) contains a triangle? It is easy to show that if true this result is best possible . To see this let A i , i =1, 2, . . . , 5, be sets of 2n vertices, put A, = A 6 and join every vertex of A, to every vertex of A; + , . This G(10n ; 20n 2) has of course no triangle and every induced subgraph of 5n vertices contains at least 2n2 edges . Equality is of course possible : choose A,, A, and half the vertices of A, Simonovits pointed out to me that a graph of completely different structure also shows that the conjecture, if true, is best possible . Consider the Petersen graph, which is a G(10 ; 15) . Replace each vertex by a set of n vertices and replace every edge of the Petersen graph by the n 2 edges of a K(n, n) . This gives a G(10n ; 15n 2) and it is easy to see that every induced subgraph of 5n vertices has at least 2n2 edges . The fact that two graphs of different structure are extremal perhaps indicates that the conjecture is either false or difficult to prove . I certainly hope that the latter is the case . It is perhaps tempting to conjecture that my graph has the following extremal property . If a G(10n) has no triangle and every induced subgraph of 5n vertices has at least 2n2 edges, then our graph can have at most 20n2 edges. Perhaps the graph of Simonovits has the smallest number of edges among all extremal graphs; perhaps in fact these two graphs are the only extremal graphs . Many generalizations are possible ; the triangle could be replaced by other graphs . Is it true that every G((4h+2)n), every induced subgraph