Advanced book on Mathematics Olympiad

398 Algebra

168.Assume without loss of generality that deg(P (z))=n≥deg(Q(z)). Consider the
polynomialR(z)=(P (z)−Q(z))P′(z). Clearly, deg(R(z))≤ 2 n−1. Ifωis a zero of
P(z)of multiplicityk, thenωis a zero ofP′(z)of multiplicityk−1. Henceωis also a
zero ofR(z), and its multiplicity is at leastk. So thenzeros ofP(z)produce at leastn
zeros ofR(z), when multiplicities are counted.
Analogously, letωbe a zero ofP(z)−1 of multiplicityk. Thenωis a zero ofQ(z)−1,
and hence ofP(z)−Q(z). It is also a zero of(P (z)− 1 )′=P′(z)of multiplicityk−1.
It follows thatωis a zero ofR(z)of multiplicity at leastk. This gives rise to at leastn
more zeros forR(z).
It follows thatR(z), which is a polynomial of degree less than or equal to 2n−1, has
at least 2nzeros. This can happen only ifR(z)is identically zero, hence ifP(z)≡Q(z).
(Soviet Union University Student Mathematical Olympiad, 1976)

169.LetQ(x)=xP (x). The conditions from the statement imply that the zeros ofQ(x)
are all real and distinct. From Rolle’s theorem, it follows that the zeros ofQ′(x)are real
and distinct.
LetH(x)=xQ′(x). Reasoning similarly we deduce that the polynomialH′(x)has
all zeros real and distinct. Note that the equationH′(x)=0 is equivalent to the equation

x^2 P′′(x)+ 3 xP′(x)+P(x)= 0 ;

the problem is solved.
(D. Andrica, published in T. Andreescu, D. Andrica, 360Problems for Mathematical
Contests, GIL, 2003)

170.Differentiating the product, we obtain

P′(x)=

∑n

k= 1

kxk−^1 (xn− 1 )···(xk+^1 − 1 )(xk−^1 − 1 )···(x− 1 ).

We will prove that each of the terms is divisible byPn/ 2 (x). This is clearly true if
k>n 2 .
Ifk≤n 2 , the corresponding term contains the factor

(xn− 1 )···(xn/^2 +^2 − 1 )(xn/^2 +^1 − 1 ).

That this is divisible byPn/ 2 (x)follows from a more general fact, namely that for any
positive integerskandm, the polynomial

(xk+m− 1 )(xk+m−^1 − 1 )···(xk+^1 − 1 )

is divisible by

(xm− 1 )(xm−^1 − 1 )···(x− 1 )