Advanced book on Mathematics Olympiad

Algebra 399

in the ring of polynomials with integer coefficients. Since the two polynomials are monic
and have integer coefficients, it suffices to prove that the zeros of the second are also
zeros of the first, with at least the same multiplicity.
Note that ifζis a primitiverth root of unity, thenζis a zero ofxj−1 precisely when
jis divisible byr. So the multiplicity ofζas a zero of the polynomial(xm− 1 )(xm−^1 −
1 )···(x− 1 )ismr, while its multiplicity as a zero of(xk+m− 1 )(xk+m−^1 − 1 )···(xk+^1 − 1 )
ism+rk−kr. The claim now follows from the inequality

⌊

m+k

r

⌋

−

⌊

k

r

⌋

≥

⌊m

r

⌋

.

This completes the solution.
(communicated by T.T. Le)

171.The equationQ(x)=0 is equivalent to

n

P(x)P′′(x)−(P′(x))^2

P(x)^2

+

[

P′(x)

P(x)

] 2

= 0.

We recognize the first term on the left to be the derivative ofP

′(x)
P(x). Denoting the roots of
P(x)byx 1 ,x 2 ,...,xn, the equation can be rewritten as

−n

∑n

k= 1

1

(x−xk)^2

+

(n

∑

k= 1

1

x−xk

) 2

= 0 ,

or

n

∑n

k= 1

1

(x−xk)^2

=

( n

∑

k= 1

1

x−xk

) 2

.

If this were true for some real numberx, then we would have the equality case in the
Cauchy–Schwarz inequality applied to the numbersak=1,bk=x−^1 xk,k= 1 , 2 ,...,n.
This would then further imply that all thexi’s are equal, which contradicts the hypothesis
that the zeros ofP(x)are distinct. So the equality cannot hold for a real number, meaning
that none of the zeros ofQ(x)is real.
(D.M. Batine ̧tu, I.V. Maftei, I.M. Stancu-Minasian, ̆ Exerci ̧tii ̧si Probleme de Analiza ̆
Matematic ̆a(Exercises and Problems in Mathematical Analysis), Editura Didactic ̆a ̧si
Pedagogica, Bucharest, 1981) ̆

172.We start with the identity

P′(x)

P(x)

=

1

x−x 1

+

1

x−x 2

+···+

1

x−xn

,forx   =xj,j= 1 , 2 ,...,n.