Advanced book on Mathematics Olympiad

Algebra 389

of course is impossible. Hence only linear and constant polynomials satisfy the given
relation.
(Romanian Team Selection Test for the International Mathematical Olympiad, 1979,
proposed by D. Bu ̧sneag)

149.Letx=

√

2 +^3

√

3. Then^3

√

3 =x−

√

2, which raised to the third power yields
3 =x^3 − 3

√

2 x^2 + 6 x− 2

√

2, or

x^3 + 6 x− 3 =( 3 x^2 + 2 )

√

2.

By squaring this equality we deduce thatxsatisfies the polynomial equation

x^6 −^6 x^4 −^6 x^3 +^12 x^2 −^36 x+^1 =^0.

(Belgian Mathematical Olympiad, 1978, from a note by P. Radovici-M ̆arculescu)

150.Note thatrandsare zeros of bothP(x)andQ(x). So on the one hand,Q(x)=
(x−r)(x−s), and on the other,randsare roots ofP(x)−Q(x). The assumption that
this polynomial is nonnegative implies that the two roots are double; hence

P(x)−Q(x)=(x−r)^2 (x−s)^2 =Q(x)^2.

We find thatP(x)=Q(x)(Q(x)+ 1 ). Because the signs ofP(x)andQ(x)agree,
the quadratic polynomialQ(x)+1 is nonnegative. This cannot happen because its
discriminant is(r−s)^2 − 4 >0. The contradiction proves that our assumption was false;
hence for somex 0 ,P(x 0 ) < Q(x 0 ).
(Russian Mathematical Olympiad, 2001)

151.BecauseP( 0 )=0, there exists a polynomialQ(x)such thatP(x)=xQ(x). Then

Q(k)=

1

k+ 1

,k= 1 , 2 ,...,n.

LetH(x)=(x+ 1 )Q(x)−1. The degree ofH(x)isnandH(k)=0 fork= 1 , 2 ,...,n.
Hence

H(x)=(x+ 1 )Q(x)− 1 =a 0 (x− 1 )(x− 2 )···(x−n).

In this equalityH(− 1 )=−1 yieldsa 0 =(−^1 )

n+ 1

(n+ 1 )!. Forx=m,m>n, which gives

Q(m)=

(− 1 )n+^1 (m− 1 )(m− 2 )···(m−n)+ 1

(n+ 1 )!(m+ 1 )

+

1

m+ 1

,

and so