Advanced book on Mathematics Olympiad

392 Algebra

a+b+c=r,

ab+bc+ca=r,

abc= 1.

We deduce thata, b, care the solutions of the polynomial equationt^3 −rt^2 +rt− 1 =0.
This equation can be written as

(t− 1 )[t^2 −(r− 1 )t+ 1 ]= 0.

Since it has three real solutions, the discriminant of the quadratic must be positive. This
means that(r− 1 )^2 − 4 ≥0, leading tor∈(−∞,− 1 ]∪[ 3 ,∞). Conversely, all such
rwork.

158.Consider the polynomialP(t)=t^5 +qt^4 +rt^3 +st^2 +ut+vwith rootsa, b, c, d, e.
The condition from the statement implies thatqis divisible byn. Moreover, since

∑

ab=

1

2

(∑

a

) 2

−

1

2

(∑

a^2

)

,

it follows thatris also divisible byn. Adding the equalitiesP(a)=0,P(b)=0,
P(c)=0,P(d)=0,P(e)=0, we deduce that

a^5 +b^5 +c^5 +d^5 +e^5 +s(a^2 +b^2 +c^2 +d^2 +e^2 )+u(a+b+c+d+e)+ 5 v

is divisible byn. But sincev=−abcde, it follows that

a^5 +b^5 +c^5 +d^5 +e^5 − 5 abcde

is divisible byn, and we are done.
(Kvant(Quantum))

159.LetP(x)=anxn+an− 1 xn−^1 +···+a 0. Denote its zeros byx 1 ,x 2 ,...,xn. The
first two of Viète’s relations give

x 1 +x 2 +···+xn=−

an− 1

an

,

x 1 x 2 +x 1 x 3 +···+xn− 1 xn=

an− 2

an

.

Combining them, we obtain

x^21 +x 22 +···+xn^2 =

(

an− 1

an

) 2

− 2

(

an− 2

an

)

.

The only possibility isx 12 +x^22 +···+xn^2 =3. Given thatx 12 x^22 ···xn^2 =1, the AM–GM
inequality yields