[ Background required: knowledge of basic algebra and polynomial operations. ]
After a spate of posts on non-IMO related topics, we’re back on track.
Here, we shall look at polynomials in n variables, e.g. P(x, y, z) when n = 3. Such polynomials are said to be symmetric if it remains the same after we swap any two variables. E.g. P(x, y) = xy + 3 is symmetric since P(x, y) = P(y, x). Likewise, is symmetric while is not symmetric.
Let’s consider the case n=2. Two obvious symmetric polynomials in x,y are the sum and product:
Key result. Any symmetric polynomial in x and y with integer coefficients can be expressed as a polynomial in P and Q with integer coefficients.
In particular, if the sum and product of x and y are integers, then is also an integer (for any natural number n) although x and y themselves may not be integral: e.g. , y = 3 – \sqrt 2$.
Example. Consider the polynomial . This can be written as and then we use the substitution . So .
Better Method. Instead of tediously calculating in this manner, we denote for integer n and proceed to derive a recurrence relation in . First, suppose we fix x and y; let be the quadratic equation with roots x and y. Then and . Now since x and y are both roots of that quadratic equation, we have:
Multiplying the two equations by respectively, we obtain:
Adding them, we get: . Since this holds for any x and y, we see that as polynomials in x and y.
Example. Now we can express as a polynomial in P and Q with just a bit of work. Starting with and , we have:
Now we can use this technique to solve the following problems.
Problem 1. Solve the simultaneous equations .
Solution. We consider the quadratic equation whose roots are x and y. In other words, as polynomials. From the above computations, we have and thus
Solving for q gives . The first case gives complex roots while the second case gives or . ♦
Note. An alternate solution exists, via the substitution . Try it!
Problem 2 (AIME 1990 Q15) Let a, b, x, y be real numbers such that:
Compute . [ Note: it may not be possible to arrive at the answer via guesswork; in fact, a, b, x, y may well be irrational, but the above sums turn out to be rational. ]
Solution. We consider the quadratic equation whose roots are x and y. In other words, as polynomials. As above, we get and . Multiplying the two equations by and respectively, we get:
Now we denote . Upon adding the above two equations, we get a recurrence:
We are given: , , , . Substituting into the recurrence relation above, we obtain:
which is wholly linear! So we easily solve that to obtain and the final answer is:
Exercise 1. Suppose x and y are roots of . Find the quadratic equation whose roots are and .
[Hint : if you do it “right”, the calculation should not be too tedious.]
Exercise 2. (USSR) Suppose α and β are roots of the equation ; γ and δ are roots of the equation . Prove that
Exercise 3. Prove that if P(x, y) is a symmetric polynomial which is divisible by (x – y), then it is in fact divisible by .
(To be continued … )