Advanced book on Mathematics Olympiad

Real Analysis 521

424.We first show thatP(x)has rational coefficients. Letkbe the degree ofP(x),
and for eachn, letxnbe the rational root ofP(x)=n. The system of equations in the
coefficients

P(xn)=n, n= 0 , 1 , 2 ,...,k,

has a unique solution since its determinant is Vandermonde. Cramer’s rule yields rational
solutions for this system, hence rational coefficients forP(x). Multiplying by the product
of the denominators, we may thus assume thatP(x)has integer coefficients, sayP(x)=
akxk+···+a 1 x+a 0 , thatak>0, and thatP(x)=Nnhas a rational solutionxnfor
alln≥1, whereNis some positive integer (the least common multiple of the previous
coefficients).
Becausexnis a rational number, its representation as a fraction in reduced form has
the numerator a divisor ofa 0 −nand the denominator a divisor ofak.Ifm =n, then
xm =xn,so

|xm−xn|≥

1

ak

.

Let us now show that under this hypothesis the derivative of the polynomial is con-
stant. Assume the contrary. Then lim|x|→∞|P′(x)|=∞. Also, limn→∞P(xn) =
limn→∞n=∞. Hence|xn|→∞, and so|P′(xn)|→∞, forn→∞.
For somen, among the numbersxn,xn+ 1 ,xn+ 2 two have the same sign, call themx
andy. Then, by the mean value theorem, there exists acnbetweenxandysuch that

P′(cn)=

P(y)−P(x)

y−x

.

Taking the absolute value, we obtain

|P′(cn)|≤

(n+ 2 )−n

|y−x|

≤ 2 ak,

where we use the fact thatxandyare at least 1/akapart. Butcntends to infinity, and so
|P′(cn)|must also tend to infinity, a contradiction. This shows that our assumption was
false, soP′(x)is constant. We conclude thatP(x)is linear.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by M. Dad ̆arlat) ̆

425.Arrange thexi’s in increasing orderx 1 ≤x 2 ≤ ··· ≤xn. The function

f(a)=|a−x 1 |+|a−x 2 |+···+|a−xn|

is convex, being the sum of convex functions. It is piecewise linear. The derivative at
a pointa, in a neighborhood of whichfis linear, is equal to the difference between the