390 Algebra
P (m)=(− 1 )m+^1 m(m− 1 )···(m−n)
(n+ 1 )!(m+ 1 )+
m
m+ 1.
(D. Andrica, published in T. Andreescu, D. Andrica, 360Problems for Mathematical
Contests, GIL, 2003)
152.Adding and subtracting the conditions from the statement, we find thata 1 +a 2 +
··· +ananda 1 −a 2 + ··· +(− 1 )nanare both real numbers, meaning thatP( 1 )and
P(− 1 )are real numbers. It follows thatP( 1 )=P( 1 )andP(− 1 )=P(− 1 ). Writing
P(x)=(x−x 1 )(x−x 2 )···(x−xn), we deduce
( 1 −x 1 )( 1 −x 2 )···( 1 −xn)=( 1 −x 1 )( 1 −x 2 )···( 1 −xn),
( 1 +x 1 )( 1 +x 2 )···( 1 +xn)=( 1 +x 1 )( 1 +x 2 )···( 1 +xn).Multiplying, we obtain
( 1 −x 12 )( 1 −x 22 )···( 1 −xn^2 )=( 1 −x^21 )( 1 −x^22 )···( 1 −x^2 n).This means thatQ( 1 )=Q( 1 ), and henceb 1 +b 2 +···+bnis a real number, as desired.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu)
153.If such aQ(x)exists, it is clear thatP(x)is even. Conversely, assume thatP(x)is
an even function. WritingP(x)=P(−x)and identifying coefficients, we conclude that
no odd powers appear inP(x). Hence
P(x)=a 2 nx^2 n+a 2 n− 2 x^2 n−^2 +···+a 2 x^2 +a 0 =P 1 (x^2 ).Factoring
P 1 (y)=a(y−y 1 )(y−y 2 )···(y−yn),we have
P(x)=a(x^2 −y 1 )(x^2 −y 2 )···(x^2 −yn).Now choose complex numbersb,x 1 ,x 2 ,...,xnsuch thatb^2 =(− 1 )naandx^2 j =yj,
j= 1 , 2 ,...,n. We have the factorization
P(x)=b^2 (x^21 −x^2 )(x 22 −x^2 )···(x^2 n−x^2 )
=b^2 (x 1 −x)(x 1 +x)(x 2 −x)(x 2 +x)···(xn−x)(xn+x)
=[b(x 1 −x)(x 2 −x)···(xn−x)][b(x 1 +x)(x 2 +x)···(xn+x)]
=Q(x)Q(−x),