Advanced book on Mathematics Olympiad

48 2 Algebra

can be factored over the complex numbers as

P(x)=an(x−x 1 )(x−x 2 )···(x−xn).

Equating the coefficients ofxin the two expressions, we obtain

x 1 +x 2 +···+xn=−

an− 1

an

,

x 1 x 2 +x 1 x 3 +···+xn− 1 xn=

an− 2

an

,

···

x 1 x 2 ···xn=(− 1 )n

a 0

an

.

These relations carry the name of the French mathematician F. Viète. They combine

two ways of looking at a polynomial: as a sum of monomials and as a product of linear

factors. As a first application of these relations, we have selected a problem from a 1957

Chinese mathematical competition.

Example.Ifx+y+z=0, prove that

x^2 +y^2 +z^2

2

·

x^5 +y^5 +z^5

5

=

x^7 +y^7 +z^7

7

.

Solution.Consider the polynomialP(X)=X^3 +pX+q, whose zeros arex, y, z. Then

x^2 +y^2 +z^2 =(x+y+z)^2 − 2 (xy+xz+yz)=− 2 p.

Adding the relationsx^3 =−px−q,y^3 =−py−q, andz^3 =−pz−q, which hold
becausex, y, zare zeros ofP(X), we obtain

x^3 +y^3 +z^3 =− 3 q.

Similarly,

x^4 +y^4 +z^4 =−p(x^2 +y^2 +z^2 )−q(x+y+z)= 2 p^2 ,

and therefore

x^5 +y^5 +z^5 =−p(x^3 +y^3 +z^3 )−q(x^2 +y^2 +z^2 )= 5 pq,

x^7 +y^7 +z^7 =−p(x^5 +y^5 +z^5 )−q(x^4 +y^4 +z^4 )=− 5 p^2 q− 2 p^2 q=− 7 p^2 q.

The relation from the statement reduces to the obvious

− 2 p

2

·

5 pq

5

=

− 7 p^2 q

7

.