2016 AIME I Problem 11

Attempt Problem 11 of the 2016 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AIME I solutions, or check the answer key.

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11.

Let

P(x)

be a nonzero polynomial such that

(x - 1)P(x + 1) = (x + 2)P(x)

for every real

x,

and

\left(P(2)\right)^2 = P(3).

Then

P\left(\tfrac{7}{2}\right) = \tfrac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 109

Concepts:polynomial functional equation substitution

Difficulty rating: 2990

Small hint:

Substitute

x = 1,

x = 0,

and

x = -2

to find three roots of

P

Big hint:

Writing

P(x) = x(x - 1)(x + 1)L(x)

forces

L(x + 1) = L(x),

so

L

is a constant; find it from

\left(P(2)\right)^2 = P(3)

Solution:

Setting

x = 1

in the identity gives

0 = 3P(1),

so

P(1) = 0.

Setting

x = 0

gives

-P(1) = 2P(0),

so

P(0) = 0,

and setting

x = -2

gives

-3P(-1) = 0,

so

P(-1) = 0.

Hence

P(x) = x(x - 1)(x + 1)L(x)

for some polynomial

L.

Substituting back,

(x - 1)\,(x + 1)x(x + 2)L(x + 1)

= (x + 2)\,x(x - 1)(x + 1)L(x),

so

L(x + 1) = L(x)

for all real

x,

which forces

L

to be a constant

c.

The normalization

\left(P(2)\right)^2 = P(3)

reads

(6c)^2 = 24c,

so

c = \frac{2}{3}.

Then

P\left(\tfrac{7}{2}\right) = \frac{2}{3} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{9}{2} = \frac{105}{4},

and

m + n = 105 + 4 = 109.