Advanced book on Mathematics Olympiad

388 Algebra

xP 1 (x)=(x− 9 )P 1 (x+ 1 ).

Arguing as before, we find thatP 1 (x)=(x− 1 )(x− 9 )P 2 (x). Repeating the argument,
we eventually find thatP(x)=x(x− 1 )(x− 2 )···(x− 10 )Q(x), whereQ(x)satisfies
Q(x)=Q(x+ 1 ). It follows thatQ(x)is constant, and the solution to the problem is

P(x)=ax(x− 1 )(x− 2 )···(x− 10 ),

whereais an arbitrary constant.

147.Having odd degree,P(x)is surjective. Hence for every rootriofP(x)=0 there

exists a solutionaito the equationP(ai)=ri, and triviallyai =ajifri =rj. Then

P(P(ai))=0, and the conclusion follows.

(Russian Mathematical Olympiad, 2002)

148.First solution: Letmbe the degree ofP(x), and write

P(x)=amxm+am− 1 xm−^1 +···+a 0.

Using the binomial formula for(x±n^1 )mand(x±^1 n)m−^1 we transform the identity from

the statement into

2 amxm+ 2 am− 1 xm−^1 + 2 am− 2 xm−^2 +am

m(m− 1 )

n^2

xm−^2 +Q(x)

= 2 amxm+ 2 am− 1 xm−^1 + 2 am− 2 xm−^2 +R(x),

whereQandRare polynomials of degree at mostm−3. If we identify the coefficients

of the corresponding powers ofx, we find thatamm(mn 2 −^1 )=0. Butam    =0, being the

leading coefficient of the polynomial; hencem(m− 1 )=0. So eitherm=0orm=1.

One can check in an instant that all polynomials of degree 0 or 1 satisfy the required

condition.

Second solution: Fix a pointx 0. The graph ofP(x)has infinitely many points in common

with the line that has slope

m=n

(

P

(

x 0 +

1

n

)

−P(x 0 )

)

and passes through the point(x 0 ,P(x 0 )). Therefore, the graph ofP(x)is a line, so the

polynomial has degree 0 or 1.

Third solution: If there is such a polynomial of degreem≥2, differentiating the given

relationm−2 times we find that there is a quadratic polynomial that satisfies the given

relation. But then any point on its graph would bethevertex of the parabola, which