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260 Answers to Exercises and Solutions to Problems

have degree 3. No polynomial has a term in x3, y3 or z3, for such terms

would result in terms in the sum in x6, y6 or z6, respectively, with positive

coefficients. No j(x, y, z) can have terms in yz2, for such a term would

produce in the sum terms in y2z4 with positive coefficients. Since y2z4 can

come only from squaring yz2 (and not from a product of y2z and z3), there

would be no cancellation of such terms. Similarly, there are no terms in zx2

and xy”. Hence, each f(x, y,z) has the form ut2y + by2z + cz2x + dxyz.

But the sum of any number of squares of such f(x, y, z) would produce a

term in x2y”z2 with a positive coefficient (certainly, not -3x2y2z2).

9.12. See the hints for a possible strategy. Three polynomials with the

required property are

x2 + (1 - fi)y2

UY2 +3:

-uxy + y

where u is the positive square root of 2(fi - 1).

9.13. Let n 2 3. Then

F,, = (x+y)” -xc”- y”

= (x + y)2[(x + yy2 - (x”-2 + f-2)]

+ xy[xy"-3+xn-3y+2x"-2+2yn--2]

= (x2 + xy + y2)F,,-2 + xy[(x + Y)“-~ + x*-~

+ y-2 + xy-3 + xn-3yJ

= QF,,-2 + PG,-3.

Similarly, G, = QG,,-2 + PF,,-3. These equations can be used as a basis
of an induction argument that F, is a polynomial in P and Q when n is
odd and G, is a polynomial in P and Q when n is even, once it has been
checked that this is so when n 5 3.
9.14. We have PO = 1, PI = xy + xz + YZ, P2 = (XY + xz + ~z)~ +
(x + Y>(X + %)(Y + z>. A ssume as an induction hypothesis that P, is sym-
metric in (2, y, 2) for m < 12.
It is clear that P,+l g symmetric in x and y. It suffices to show that
Pn+l(X, Y, z) = Pn+l( z, y, x). First, observe that, for m^5 n,

Pm+l(x, Y, %) - Pm++, Y, x> = (x + 4Qm(x, Y, z>

where

Q&x, Y, %> = (Y+%)~,(x,Y, %+I)--(x+~)Pm(x+L Y, z)+(x--z)Pm(x, Y, 2).

Since the left side vanishes for m 5 n - 1, so also does Qm(x, y, z). We
show that Qn(x, y, z) = 0.