2005 AIME II Problem 13

Question

Let

P(x)

be a polynomial with integer coefficients that satisfies

P(17) = 10

and

P(24) = 17.

Given that the equation

P(n) = n + 3

has two distinct integer solutions

n_1

and

n_2,

find the product

n_1 \cdot n_2.

Answer

Answer: 418

Concepts:polynomial divisibility factor

Difficulty rating: 2760

Small hint:

Consider

S(x) = P(x) - x - 3,

which equals

-10

at both

x = 17

and

x = 24

Big hint:

If

S(n) = 0,

then

(n-17)(n-24)

divides

10;

those two factors are integers differing by

7

Solution:

Let

S(x) = P(x) - x - 3,

so

S(17) = S(24) = -10.

Since

S(x) + 10

has integer coefficients and vanishes at

17

and

24,

\begin{aligned} S(x) &= -10 \\ &\quad {}+ (x - 17)(x - 24)Q(x) \end{aligned}

for some polynomial

Q

with integer coefficients.

If

P(n) = n + 3

for an integer

n,

then

(n-17)(n-24)Q(n) = 10,

so

(n-17)(n-24)

divides

10.

The factors

n - 17

and

n - 24

are integers differing by

7

whose product divides

10,

so they are

\{2, -5\}

or

\{5, -2\},

giving

n = 19

and

n = 22.

Both occur, for example, for

P(x) = x - 7

- (x-17)(x-24).

Hence

n_1 \cdot n_2 = 19 \cdot 22 = 418.