Symmetric Polynomials (II)

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

$P(x,y,z) = x+y+z,\quad Q(x,y,z) = xy+yz+zx, \quad R(x,y,z) = xyz.$

These are just the coefficients of $T^i$ in the expansion of $(T-x)(T-y)(T-z)$ . 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 $S_n = x^n + y^n + z^n$ , then using the technique described in the previous post, we obtain the recurrence relation:

$S_{n+3} = P\cdot S_{n+2} - Q\cdot S_{n+1} + R\cdot S_n.$

Starting from $S_0 = 3$ , $S_1 = P$ , $S_2 = (x+y+z)^2 - 2(xy+yz+zx) = P^2 - 2Q$ , we obtain:

$S_3 = P\cdot S_2 - Q\cdot S_1 + R\cdot S_0 = P^3 - 3PQ + 3R$ ;
$S_4 = P\cdot S_3 - Q\cdot S_2 + R\cdot S_1 = P^4 - 4P^2 Q + 2Q^2 + 4PR$ …

Let’s solve some problems now.

Example 1. Factor the polynomial $x^3 + y^3 + z^3 - 3xyz$ .

Solution. Since that’s a symmetric polynomial, we can express it as a polynomial in P, Q and R. From the above, we have $x^3 + y^3 + z^3 = P^3 - 3PQ + 3R$ , so:

$x^3 + y^3 + z^3 -3xyz= P^3 - 3PQ = P(P^2 - 3Q)$ .

So $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$ . ♦

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 $U = xy^2 + yz^2 + zx^2$ and $V = x^2 y + y^2 z + z^2 x$ . Express the polynomial $A = U^2 + UV + V^2$ 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:

$A = a_1 P^6 + a_2 P^4 Q + a_3 P^3 R + a_4 P^2 Q^2 + a_5 PQR + a_6 Q^3 + a_7 R^2,$

for some constants $a_1, a_2, \ldots$ . 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 $x^4(y^2 + yz + z^2)$ . So:

$P^6$ and $P^4 Q$ can’t appear since they contain higher powers of x, i.e. $a_1 = a_2 = 0$ ;
also, $x^4$ only appears in $P^3 R$ and $P^2 Q^2$ , with respective leading terms $x^4 yz$ and $x^4 (y+z)^2$ . So $a_3 = -1, a_4 = 1$ .
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 $a_6$ from the coefficient of $x^6$ .

It turns out $a_5 = a_7 = 0$ so we get $A = -P^3 R + P^2 Q^2 - Q^3.$

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:

$(x+y)^3 = z, \quad (y+z)^3 = x, \quad (z+x)^3 = y,$

where x, y and z are distinct complex numbers.

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

$(s-z)^3 = z, \quad (s-x)^3 = x, \quad (s-y)^3 = y.$

Hence, taking s as a constant, we see that x, y and z are all roots of the cubic equation $(s-T)^3 = T$ , where T is unknown. Since x, y and z are distinct, these are precisely all the roots. Expanding the polynomial gives $T^3 - 3s\cdot T^2 + (3s^2 - 1)T - s^3 = 0$ and the sum of the roots is thus given by 3s. Yet the sum of the roots is also x+y+z = s. Thus s=0 and x, y and z are roots of $T^3 + T = 0$ . So $(x,y,z) = (-i, 0, +i)$ and other permutations. ♦