USAMO 1973 #4 August 19, 2009
Posted by lumixedia in algebra, Problem-solving.
Tags: algebra, contest math, olympiad math, USAMO, USAMO 1973
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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 ![{x=1-\sqrt[3]{20}}](http://s0.wp.com/latex.php?latex=%7Bx%3D1-%5Csqrt%5B3%5D%7B20%7D%7D&bg=ffffff&fg=000000&s=0 "{x=1-\sqrt[3]{20}}")
, ![{y=1-\omega\sqrt[3]{20}}](http://s0.wp.com/latex.php?latex=%7By%3D1-%5Comega%5Csqrt%5B3%5D%7B20%7D%7D&bg=ffffff&fg=000000&s=0 "{y=1-\omega\sqrt[3]{20}}")
, ![{z=1-\omega^2\sqrt[3]{20}}](http://s0.wp.com/latex.php?latex=%7Bz%3D1-%5Comega%5E2%5Csqrt%5B3%5D%7B20%7D%7D&bg=ffffff&fg=000000&s=0 "{z=1-\omega^2\sqrt[3]{20}}")
where 
is an imaginary cube root of unity. For still higher powers things probably continue to get worse.
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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.