[ 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 … )

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