The Universe of Disco

Symmetric functions
I used to teach math at the John Hopkins CTY program, which is a well-regarded summer math camp. Kids would show up and finish a year (or more) of high-school math in three weeks. We'd certify them by giving them standardized tests, which might carry some weight with their school. But before they were allowed to take the standardized test, they had to pass a much more difficult and comprehensive exam that we'd made up ourselves.

The most difficult question on the Algebra III exam presented the examinee with some intractable third degree polynomial—say x³ + 4x² - 2x + 6—and asked for the sum of the cubes of its roots.

You might like to match your wits against the Algebra III students before reading the solution below.

In the three summers I taught, only about two students were able to solve this problem, which is rather tricky. Usually they would start by trying to find the roots. This is doomed, because the Algebra III course only covers how to find the roots when they are rational, and the roots here are totally bizarre.

Even clever students didn't solve the problem, which required several inspired tactics. First you must decide to let the roots be p, q, and r, and, using Descartes' theorem, say that

x³ + bx² + cx + d = (x - p)(x - q)(x - r)

But expanding the right-hand side of the equation above yields:

x³ + bx² + cx + d = x³ - (p + q + r)x² + (pq + pr + qr)x - pqr

-b3

=

p3 + q3 + r3 + 3p2q + 3p2r + 3q2r + 3pq2 + 3pr2 + 3qr2 + 6pqr

And you now need to figure out how to get rid of the unwanted terms. The 6pqr term is not hard to eliminate, since it is just -6d, and if you notice this, it will probably inspire you to try combinations of the others. In fact, the answer is:

p3 + q3 + r3

=

-b3 + 3bc - 3d

So for the original polynomial, x³ + 4x² - 2x + 6, we know that the sum of the cubes of the roots is -4³ + 3·4·(-2) - 3·6 = -64 - 24 - 18 = -106, and we calculated it without any idea what the roots actually were.

Or, to take an example that we can actually check, consider x³ - 6x² + 11x - 6, whose roots are 1, 2, and 3. The sum of the cubes is 1 + 8 + 27 = 36, and indeed -b³ + 3bc - 3d = 6³ + 3·(-6)·11 + 18 = 216 - 198 + 18 = 36.

This was a lot of algebra III, but once you have seen this example, it's not hard to solve a lot of similar problems. For instance, what is the sum of the squares of the roots of x² + bx + c? Well, proceeding as before, we let the roots be p and q, so x² + bx + c = (x - p)(x - q) = x² - (p + q)x + pq, so that b = -(p + q) and c = pq. Then b² = p² + 2pq+ q², and b² - 2c = p² + q².

In general, if F is any symmetric function of the roots of a polynomial, then F can be calculated from the coefficients of the polynomial without too much difficulty.

Anyway, I was tinkering around with this at breakfast a couple of days ago, and I got to thinking about b² - 2c = p² + q². If roots p and q are both integers, then b² - 2c is the sum of two squares. (The sum-of-two-squares theorem is one of my favorites.) And the roots are integers only when the discriminant of the original polynomial is itself a square. But the discriminant in this case is b² - 4c. So we have the somewhat odd-seeming statement that when b² - 4c is a square, then b² - 2c is a sum of two squares.

I found this surprising because it seemed so underconstrained: it says that you can add some random even number to a fairly large class of squares and the result must be a sum of two squares, even if the even number you added wasn't a square itself. But after I tried a few examples to convince myself I hadn't made a mistake, I was sure there had to be a very simple, direct way to get to the same place.

It took some fiddling, but eventually I did find it. Say that b² - 4c = a². Then b and a must have the same parity, so p = (b + a)/2 is an integer, and we can write b = p + q and a = p - q where p and q are both integers.

Then c = (b² - a²)/4 is just pq, and b² - 2c = p² + q².

So that's where that comes from.

It seems like there ought to be an interesting relationship between the symmetric functions of roots of a polynomial and their expression in terms of the coefficients of the polynomial. The symmetric functions of degree N are all linear combinations of a finite set of symmetric functions. For example, any second-degree symmetric function of two variables has the form a(p² + q²) + 2bpq. We can denote these basic symmetric functions of two variables as F_i,j(p, q) = Σpⁱq^j. Then we have identities like (F_1,0)² = F_2,0 + F_1,1 and (F_1,0)³ = F_3,0 + 3F_2,1.

Maybe I'll do an article about this in a week or two.

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