Question 1086341
Number of diagonals = d = (n/2)(n-3) -> n^2 - 3n - 2d = 0
a) Solve for n, when d = 665
n^2 - 3n - 1330 = 0
This can be factored as (n-38)(n+35) = 0
We take the positive solution, n = 38
For d = 406, we have the equation n^2 - 3n - 812 = 0
This does not give an integer solution for n.  Thus no polygon can have 406 diagonals.