Problem 5 Polynomials

Problem 5: Polynomials

USAMO 1975 #3

Problem: A polynomial p(x) of degree n satisfies p(0) = 0, p(1) = n p(n) = n+1 . Find P (n + 1).

1 2,

p(2) =

2 3,

... ,

Solution: The polynomial Q(x) = (x + 1)P (x) − x is of the degree n + 1 and has n + 1 zeroes: 0, 1, . . . , n. Therefore Q(x) = Cx(x − 1) . . . (x − n) where C is a constant. That gives Cx(x − 1) . . . (x − n) + x x+1 Since the fraction has to be a polynomial, the numerator must be divisible by the denominator, and hence the numerator must vanish for x = −1: P (x) =

C(−1)n+1 (n + 1)! − 1 = 0 ⇐⇒ C =

(−1)n+1 (n + 1)!

So now we have a formula for P (n): P (n + 1) =

(−1)n+1 (n+1)! (n

+ 1)! + n + 1

n+2

n + 1 + (−1)n+1 = = n+2

( 1 n n+2

n odd n even

Practice Problem: If P (x) denotes a fifth degree polynomial such that P (k) = k = 0, 1, 2, 3, 4, 5, determine P (6). (AoPS Intermediate Algebra PostTest)

k k+1

for

Answer: 1

Solution was written by Farenhajt and compiled from Art of Problem Solving Forums.