When we move on to *n*=3 variables, we now have, as basic building blocks,

These are just the coefficients of in the expansion of . Once again, any symmetric polynomial in *x*, *y*, *z* with integer coefficients can be expressed as a polynomial in *P*, *Q* and *R* with integer coefficients.

**Example**. If we let , then using the technique described in the previous post, we obtain the recurrence relation:

Starting from , , , we obtain:

- ;
- …

**Example 1**. Factor the polynomial .

**Solution**. Since that’s a symmetric polynomial, we can express it as a polynomial in *P*, *Q* and *R*. From the above, we have , so:

.

So . ♦

**Note**. To be fair, this method worked only because the individual factors were also symmetric. Sometimes one gets something like the following.

**Example 2**. Let and . Express the polynomial as a polynomial in *P*, *Q*, *R*.

**Solution**. No, we’re not going to expand the monster. Instead, we note that it is homogeneous of degree 6. Since *P*, *Q* and *R* are homogeneous of degrees 1, 2, and 3 respectively, we must have:

for some constants . That’s too much to solve linearly, so we prune down the cases further by singling out one variable: *x*. Indeed, the degree of *x* in *A* is 4, with leading coefficient . So:

- and can’t appear since they contain higher powers of
*x*, i.e. ; - also, only appears in and , with respective leading terms and . So .
- with only three unknowns left, we can substitute explicit values into
*x*,*y*,*z*to solve for them, or we can substitute*x*=*y*to first weed out from the coefficient of .

It turns out so we get

**Exercise 1**. Factor the polynomial *A*. [*Hint (highlight to read): deduce from the above that the factors aren’t symmetric, so given any factor, we can obtain new ones by permuting the variables. Such a factor has degree at most 2, let z=0 and conclude the degree is in fact exactly 2. Guess the factor and prove it works. *]

**Example 3**. Solve the system of simultaneous equations:

where *x*, *y* and *z* are distinct complex numbers.

**Solution**. Let *s* = *x*+*y*+*z* and rewrite the above equations as:

Hence, taking *s* as a constant, we see that *x*, *y* and *z* are all roots of the cubic equation , where *T* is unknown. Since *x*, *y* and *z* are distinct, these are precisely all the roots. Expanding the polynomial gives and the sum of the roots is thus given by 3*s*. Yet the sum of the roots is also *x*+*y*+*z* = *s*. Thus *s*=0 and *x*, *y* and *z* are roots of . So and other permutations. ♦

**Exercise 2**. Solve the simultaneous equations:

**Exercise 3**. Suppose *x*, *y* and *z* are non-zero real numbers satisfying:

Prove that one of *x*, *y*, *z* is equal to 1.

**Exercise 4**. Find a recurrence relation for in terms of *P*, *Q* and *R*.

**Exercise 5**.(*) Find positive integers *x*, *y*, *z* > 10 satisfying the Diophantine equation .

(To be concluded…)