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Notes on Explorations 401

An equivalent diophantine equation is X3 + Y3 = U3 + V3. One polyno-
mial solution is given by

(X, Y, U, V) = (ax4 + bxy3, oly4 + cx3y, ax4 + cx$, ay4 + bx3y),

where a, b, c are constants chosen to satisfy b(3a2 - b2) = c(3a2 - c2). For
example, (a, b,c) = (7,11,2) will work. (Cf. Victor Thebault, El. Math. 8
(1953), 47.)
For a discussion of other solutions to this equation as well as to X4+Y4 =
U4 + V4 in polynomials of two variables, see G.H. Hardy & E.M. Wright,
An Introduction to the Theory of Numbers, (Oxford, 4th ed., 1960), pages
199-201.
(b) A simple solution is (x - 1, x, x + 1, x + 2). The solutions
(6, 23, 32, 39), (39, 70, 91, lOS), (108, 157, 194, 225), etc. are instances of

A = 2x3 - 5x = 2y3 - 3y2 - (7/2)y + (9/4)

I3 = 2x3 + 2x2 - x + 1 = 2y3 - y2 - (3/2)y + (7/4)

C = 2x3 + 4x2 + x - 2 = 2~” + y” - (3/2)y - (7/4)

D = 2x3 + 6x2 + x - 3 = 2y3 + 3y2 - (7/2)y - (9/4).

The variables are related by y = x + l/2. Note that D(x) = A(x + 1) =
-A(-x - 1) and C(x) = -B(-x - 1).
(c) This equation is satisfied by

(X,Y,Z)=(2xy+rx2,y2-x2,x2+rxy+y2).

A complex number method of solving the related equation

is to let 2 = I(y+ xcose) + ixsinel. Then

z2 = lb2 - x2) + 2x(y + x cos e)(cos e + i sin e)12

yields the solution

(X, y, z> = (2x(y + xcose),y2 -x2, x2 + ~XYCOS~ + y2).

These equations were the subject of lively correspondence in the Reader
Reflections column of the Mathematics Teacher; see 78 (1985), 238, 663;
79 (1986) 158, 522; 80 (1987) 343.
(d) Observing that (12)3 = l2 + 22, (23)~ = 22 + 32, etc., we find that,
for numbers of base 2k + 1,

k(2k + 1) + (k + 1) = k2 + (k + 1)“.