##
USAMO 1973 #4 *August 19, 2009*

*Posted by lumixedia in algebra, Problem-solving.*

Tags: algebra, contest math, olympiad math, USAMO, USAMO 1973

trackback

Tags: algebra, contest math, olympiad math, USAMO, USAMO 1973

trackback

A fairly straightforward algebra problem. Could appear on a modern AMC-12, though the decoy answers would have to be carefully written.

**USAMO 1973 #4.** Determine all the roots, real or complex, of the system of simultaneous equations

**Solution.** First note that

Let , so that are the roots of the polynomial . Then we have

Summing these three equalities together gives

or

from which . So are the roots of the polynomial , and the only solution is .

An interesting note is that the scan of the original test shows the last inequality as . While the official solution makes clear that was intended, the problem is only slightly trickier as originally printed. Here’s how to do the typoed-up version. Let . The calculation of stands. We also have, just like before,

or

so we can write . By multiplying one of the equalities we used before by , we get

as well as the two analogous equations in and , and summing these up gives

Similarly, since we know

as well as the two analogous equations, we can sum them up to get

which gives as before, so we conclude that as before.

You can see from the intermediate step that giving for the last equation would also have worked and given the same answer. Using makes things a little more complicated. The analogous solution gives the quadratic , so in addition to as before, we can also take , making the polynomial or , so we have a second solution , , where is an imaginary cube root of unity. For still higher powers things probably continue to get worse.

I think a general system of equations in variables, say , for elements of some algebraically closed field can be solved up to permutation. Indeed, the polynomials generate the ring of symmetric functions in the . So by writing the elementary symmetric functions in terms of the (which we can do in any polynomial ring), yields what is. Then note that the polynomial . We know what the polynomial to the right by the above, so we can find the roots in to get the up to a permutation.

The subtlety is in solving a system like , since then one has to use Newton’s sums and so forth to determine the various relations that emerge. I’d like to add that in order to conclude that the power symmetric functions generate the symmetric functions one has to specify that the underlying ring is a field of characteristic zero. (On the other hand, the elementary symmetric functions generate the symmetric functions over any commutative ring.) I’ll be writing a post about symmetric functions soon if anyone wants details.

Ok, I must have forgotten that subtlety. I’ll read your post to clarify my understanding.

Maybe “subtlety” was too strong a word. All I meant is that even if you accept that your proposal works for it doesn’t follow that you can always solve for the up to permutation for arbitrary increasing sequences as in the extension; this is clearly false for , for example, since you can at best solve up to sign. This is actually an interesting question to think about; is relative primality sufficient?

Actually, my reference to a “subtlelty” (didn’t realize you had used that word for something else) was the fact that the power sums only generate the symmetric polynomials over a field of characteristic zero.

I did not expect the methods to generalize to arbitrary increasing sequences .

By the way, is it enough to conclude that the power sums generate the symmetric polynomials, if the ring contains ?

Yep. In practice, the choice of coefficient ring is irrelevant since the interesting relations among the common classes of symmetric functions can be defined over or even and then one can take tensor products to recover the general case.

Well, not irrelevant. The positive characteristic case is very different from the characteristic zero case.