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 n^{2} 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 n^{2} points of intersection between two curves of degree n there are only n(n+3)/2 - 1 degrees of freedom. Thus the n^{2} 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 10^{2} intersection points consist of 8^{2} independent points and 6^{2} 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 r_{k} - 6r_{k-1} + r_{k-2} = 0, so we can give the values of r_{k} 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 s_{j} - 6s_{j-1} + s_{j-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 n_{k} (or n_{k}') and n_{j}" for small indices k and j is n_{1} = n_{2}" = 10. For larger indices equality is ruled out by placing bounds on the difference between nk (or n_{k}') and n_{j}". 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 n_{k}' and n_{j}" for any indices k and j. |