A Unique Intersection Pattern for Plane Curves As discussed in the note on Cramer's Paradox, any two algebraic plane curves of degree n intersect in exactly n2 complex points (counting multiplicities and points at infinity), and each such curve is uniquely determined by a set of n(n+3)/2 general points. The apparent paradox (for n greater than 2) is resolved by the fact that of the n2 points of intersection between two curves of degree n there are only n(n+3)/2 - 1 degrees of freedom. Thus the n2 points can be partitioned into n(n+3)/2 - 1 independent points and (n-1)(n-2)/2 dependent points. For the particular case n = 10 the 102 intersection points consist of 82 independent points and 62 dependent points, so we can imagine taking the 64 points of intersection of a pair of octics as the independent intersection points for a family of decatics, such that the 36 dependent points might be the intersection points of two sextics. This possibility arises because with n = 10 the quantities n(n+3)/2 - 1 and (n-1)(n-2)/2 are both non-zero squares. No other value of n satisfies this condition, so the case n = 10 is unique. To prove uniqueness, we can solve each of the conditions for n, which gives n = (-3 ± r)/2 and n = (3 ± s)/2 where r and s are integers such that The Pell equation on the left has two sequences of solutions, r = 5, 23, 133, 775, etc., and r = 7, 37, 215, 1253, etc. Each of these sequences satisfies the recurrence rk - 6rk-1 + rk-2 = 0, so we can give the values of rk as explicit functions of k, leading to the explicit expressions for the corresponding positive values of n shown below. The other Pell equation has just a single sequence of solutions, s = 1, 3, 17, 99, etc., and again these values satisfy the recurrence sj - 6sj-1 + sj-2 = 0, so we can give the corresponding values of n explicitly as (We omit the trivial solutions n = 1 and n = 0 given by the other sign.) It's easy to verify that the only common value of nk (or nk') and nj" for small indices k and j is n1 = n2" = 10. For larger indices equality is ruled out by placing bounds on the difference between nk (or nk') and nj". We have which vanishes if and only if The smaller root of the characteristic equation (3 minus the square root of 8) is much less than 1, so as the indices k and j increase, these terms become negligible, so we need to satisfy the approximate equality Clearly we must have j > k, so this can be written as The minimum magnitude of the quantity in square brackets for integer values of j-k is 2, which occurs with j-k = 1. Thus in order to achieve equality the leading factor cannot exceed approximately 6. The argument of this factor is 5.828..., so the only possible solution occurs with k = 1. For any higher values of k, equality is ruled out. This proves that the only solution is the one with k = 1 and j = 2, which gives n = 10. By similar reasoning we can rule out equality between nk' and nj" for any indices k and j. Return to MathPages Main Menu