There is exactly one square with its vertices on the curve y = x3 + ax2 + bx + c, where a, b, c are real. Show that the square must have side 721/4.
Solution
By a translation along the x-axis we can take a = 0. By a translation along the y-axis we can take c = 0, so the curve is y = x3 + bx. We call the curve C. The center of the square O must be at the origin, otherwise we could get a second square by rotating through 180o. Suppose one of the vertices has coordinates (r, s). Then the opposite vertex must be (-r, -s). Another vertex is obtained by rotating (r, s) through 90o about O, so it must be (-s, r), and the fourth vertex must be (s, -r).
The vertices also lie on the cubic -x = y3 + by, which we call C'. Now suppose P (u, v) is any point which lies on both cubics. Then R (-u, -v) must also lie on both conics. Since C' is obtained from C by a rotation of 90o, the point Q (v, -u) obtained from P by a rotation of 90o must lie on on C. But Q is also obtained from R by a rotation of -90o and R lies on C, so Q must also lie on C'. Hence S (-v, u) must also lie on both cubics. Thus we have a square PQRS with vertices on C. But it is clear from the graph that (apart from the intersection at the origin), the two cubics generally intersect in 0 or 8 points. 0 intersections gives 0 squares on C, and 8 intersections gives 2 squares on C.
The only way we can get just 4 intersections, and a single square, is if the cubics touch (at each of four points).
Note that (r, s) satisfy s = r3 + br, -r = s3 + bs. Hence s + (s3 + bs)3 + b(s3 + bs) = 0, or s(s8 + 3bs6 + 3b2s4 + (b3 + b)s2 + b2 + 1) = 0. Thus the 8th degree polynomial x8 + 3bx6 + 3b2x4 + (b3 + b)x2 + b2 + 1 must be the square of the polynomial with roots ±r, ±s. In other words, it must be (x2 - r2)2(x2 - s2)2 = x8 - 2(r2 + s2)x6 + (r4 + 4r2s2 + s4) - 2r2s2(r2 + s2)x2 + r4s4.
Comparing coefficients:
3b = -2(r2 + s2) (1)
b3 + b = -2r2s2(r2 + s2) (2)
3b2 = r4 + 4r2s2 + s4 (3)
b2 + 1 = r4s4 (4)
Substituting (1) in (2), we get b(b2 + 1) = 3br2s2. Squaring, (b2 + 1)2 = 9r4s4 = 9(b2 + 1), from (4). Hence b2 = 8. But if d is the side of the square, then d2 = 2(r2 + s2). Squaring, d4 = 9b2 (from (1)). Hence d4 = 72.
© John Scholes
jscholes@kalva.demon.co.uk
2 Jan 2003
Last corrected/updated 2Jan 03