Advanced book on Mathematics Olympiad

Algebra 391

whereQ(x)=b(x 1 −x)(x 2 −x)···(xn−x). This completes the proof.
(Romanian Mathematical Olympiad, 1979, proposed by M. ̧Tena)

154.Denote the zeros ofP(x)byx 1 ,x 2 ,x 3 ,x 4 , such thatx 1 +x 2 =4. The first Viète
relation givesx 1 +x 2 +x 3 +x 4 =6; hencex 3 +x 4 =2. The second Viète relation can
be written as

x 1 x 2 +x 3 x 4 +(x 1 +x 2 )(x 3 +x 4 )= 18 ,

from which we deduce thatx 1 x 2 +x 3 x 4 = 18 − 2 · 4 =10. This, combined with the
fourth Viète relationx 1 x 2 x 3 x 4 =25, shows that the productsx 1 x 2 andx 3 x 4 are roots of
the quadratic equationu^2 − 10 u+ 25 =0. Hencex 1 x 2 =x 3 x 4 =5, and thereforex 1 and
x 2 satisfy the quadratic equationx^2 − 4 x+ 5 =0, whilex 3 andx 4 satisfy the quadratic
equationx^2 − 2 x+ 5 =0. We conclude that the zeros ofP(x)are 2+i, 2 −i, 1 + 2 i, 1 − 2 i.

155.Ifa≥0,b≥0,c≥0, then obviouslya+b+c>0,ab+bc+ca≥0, and
abc≥0. For the converse, letu=a+b+c,v=ab+bc+ca, andw=abc, which
are assumed to be positive. Thena, b, care the three zeros of the polynomial

P(x)=x^3 −ux^2 +vx−w.

Note that ift<0, that is, ift=−swiths>0, thenP(t)=s^3 +us^2 +vs+w>0;
hencetis not a zero ofP(x). It follows that the three zeros ofP(x)are nonnegative,
and we are done.

156.Taking the conjugate of the first equation, we obtain

x+y+z= 1 ,

and hence

1

x

+

1

y

+

1

z

= 1.

Combining this withxyz=1, we obtain

xy+yz+xz= 1.

Therefore,x, y, zare the roots of the polynomial equation

t^3 −t^2 +t− 1 = 0 ,

which are 1,i,−i. Any permutation of these three complex numbers is a solution to the
original system of equations.

157.Dividing by the nonzeroxyzyieldsxz+yx+zy=yz+zx+xy=r. Leta=xy,b=yz,

c=zx. Thenabc=1,^1 a+^1 b+^1 c=r,a+b+c=r. Hence