For the case of n and r odd, take one point on each of the
even circuits, the Jr(r— l)+s double points, another point
on each of the r odd circuits and n — 4 more points on an even
circuit.
Since n — 2 is odd, the (n — 2)-ic meets each odd circuit in
three points at least, and the n-ic and (n — 2)-ic meet in at
least

2{%r(r-l)+s} + 2r + 2(D-r+l)+(n-3) = n(n-2) + l
real points, which is impossible.
Ex. 1. Show that not every non-singular n-ic can be projected into
a closed curve, if n is a given number greater than 4.
[If n is odd, the curve can never be closed, for it has an odd circuit.
For n = 6 take

(tf-tf-V, (a; 2 + 8^ + 32^ + 28) (a; 2 + 8y 2 - 32^ + 28) = e.]

Ex. 2. Let (j> = be the equation of six lines parallel to the line of
inflexions of a two-circuited cubic / = 0, four being on one side and
two on the other side of the line of inflexions, and all meeting the odd
circuit of the cubic in three points. Then / 2 = + ?0 are (1) a sextic
with twenty-four inflexions and eleven ovals, (2) a sextic with eighteen
inflexions and nine ovals (i. e. protectable into such a sextic).
[To find all possible arrangements of the circuits of a non-singular
n-ic is a difficult problem, which ha9 been solved only for re = 3, 4, 6.
