May the Product of Planes Be a Sphere?

Show that it is impossible to find (real or complex) numbers a, b, c, A, B, and C such that the equation

holds identically for independent variables x, y, z.

Solution

[Stanford Problem Book, #54.3, Mathematical Discovery, #2.66]

Let's expand the right-hand side of the hypothetical identity:

Now equating corresponding (i.e., by the same powers of the variables) coefficients on both sides of the identity, we obtain

We derive from (2) that

and multiplying these three equations, we derive further that

or

Yet we derive from (1) that

Consequently, the hypothetical identity is impossible.

Solution 2

If the identity at hand holds for any values of the three variables x, y, z, it is bound to hold for x, y, z that satisfy the system

for any choice of r and R. The two equations determine a plane each, thus, unless the planes are parallel, the triples of the solutions to the system lie on a straight line. While the left-hand side of the identity satisfies

Since this is true for any selection of r and R, we may as well choose both positive, making the above a sphere with the radius √rR. We arrive at a contradiction by observing that no sphere contains a straight line.

In case where the two planes are parallel, the vectors (a, b, c) and (A, B, C) are proportional:

for some α. Choosing again positive r and R makes α positive and the left-hand side of the hypothetical identity represent a sphere. Now, since a sphere does not contain a plane, the result is contradictory in this case also.

Reference

1. G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009
2. G. Polya, Mathematical Discovery, John Wiley & Sons, 1981