Maths Olympiad Prep

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Algebra Difficulty 5.5 AIME, harder Prove it United States

Problem:
Suppose P(x)P(x) is a polynomial such that P(1)=1P(1) = 1 and

P(2x)P(x+1)=856x+7 \frac{P(2x)}{P(x+1)} = 8 - \frac{56}{x+7}

for all real xx for which both sides are defined. Find P(1)P(-1).

Solution

Solution:
Cross-multiplying gives (x+7)P(2x)=8(x+7)P(x+1)56P(x+1)=8(x+7)P(x+1)56P(x+1)(x+7) P(2x) = 8(x+7) P(x+1) - 56 P(x+1) = 8(x+7) P(x+1) - 56 P(x+1). But more simply, rearrange:

P(2x)P(x+1)=856x+7\frac{P(2x)}{P(x+1)} = 8 - \frac{56}{x+7}

P(2x)P(x+1)=8(x+7)56x+7=8x+5656x+7=8xx+7\Rightarrow \frac{P(2x)}{P(x+1)} = \frac{8(x+7) - 56}{x+7} = \frac{8x + 56 - 56}{x+7} = \frac{8x}{x+7}

So:

P(2x)P(x+1)=8xx+7\frac{P(2x)}{P(x+1)} = \frac{8x}{x+7}

Cross-multiplied:

(x+7)P(2x)=8xP(x+1)(x+7) P(2x) = 8x P(x+1)

Suppose P(x)P(x) has degree nn and leading coefficient cc. Then P(2x)P(2x) has leading term c(2x)n=c2nxnc (2x)^n = c 2^n x^n, and P(x+1)P(x+1) has leading term cxnc x^n. So the left side's leading term is xc2nxn=c2nxn+1x \cdot c 2^n x^n = c 2^n x^{n+1}, and the right side's leading term is 8xcxn=8cxn+18x \cdot c x^n = 8c x^{n+1}. Equate:

c2n=8c    2n=8    n=3c 2^n = 8c \implies 2^n = 8 \implies n = 3.

So P(x)P(x) is cubic.

Now, if x=0x=0, the right side is 00, so the left side is 7P(0)=0    P(0)=07 P(0) = 0 \implies P(0) = 0.

So P(x)=xQ(x)P(x) = x Q(x) for some quadratic Q(x)Q(x).

Plug into the original equation:

P(2x)=2xQ(2x)P(2x) = 2x Q(2x), P(x+1)=(x+1)Q(x+1)P(x+1) = (x+1) Q(x+1)

So:

2xQ(2x)(x+1)Q(x+1)=8xx+7\frac{2x Q(2x)}{(x+1) Q(x+1)} = \frac{8x}{x+7}

Cross-multiplied:

(x+7)2xQ(2x)=8x(x+1)Q(x+1)(x+7) 2x Q(2x) = 8x (x+1) Q(x+1)

If x=0x=0, both sides are 00.

Divide both sides by xx (for x0x \neq 0):

2(x+7)Q(2x)=8(x+1)Q(x+1)2(x+7) Q(2x) = 8(x+1) Q(x+1)

Or:

(x+7)Q(2x)=4(x+1)Q(x+1)(x+7) Q(2x) = 4(x+1) Q(x+1)

Now, try x=1x = -1:

(1+7)Q(2)=4(0)Q(0)    6Q(2)=0    Q(2)=0(-1+7) Q(-2) = 4(0) Q(0) \implies 6 Q(-2) = 0 \implies Q(-2) = 0

So Q(x)Q(x) has root at x=2x = -2, so Q(x)=(x+2)R(x)Q(x) = (x+2) R(x) for some linear R(x)R(x).

Now, Q(2x)=(2x+2)R(2x)Q(2x) = (2x+2) R(2x), Q(x+1)=(x+3)R(x+1)Q(x+1) = (x+3) R(x+1)

Plug into the previous equation:

(x+7)(2x+2)R(2x)=4(x+1)(x+3)R(x+1)(x+7) (2x+2) R(2x) = 4(x+1)(x+3) R(x+1)

Divide both sides by 22:

(x+7)(x+1)R(2x)=2(x+1)(x+3)R(x+1)(x+7)(x+1) R(2x) = 2(x+1)(x+3) R(x+1)

If x=3x = -3:

(3+7)(3+1)R(6)=2(3+1)(3+3)R(2)(-3+7)(-3+1) R(-6) = 2(-3+1)(-3+3) R(-2)

(4)(2)R(6)=2(2)(0)R(2)=0(4)(-2) R(-6) = 2(-2)(0) R(-2) = 0

So R(6)=0R(-6) = 0

So R(x)R(x) has root at x=6x = -6, so R(x)=(x+6)S(x)R(x) = (x+6) S(x), S(x)S(x) constant.

Therefore,

P(x)=x(x+2)(x+6)SP(x) = x(x+2)(x+6) S

Since P(x)P(x) is cubic, SS is a constant.

Given P(1)=1P(1) = 1:

P(1)=137S=21S=1    S=121P(1) = 1 \cdot 3 \cdot 7 \cdot S = 21 S = 1 \implies S = \frac{1}{21}

Therefore,

P(x)=121x(x+2)(x+6)P(x) = \frac{1}{21} x(x+2)(x+6)

So,

P(1)=121(1)15=521P(-1) = \frac{1}{21} \cdot (-1) \cdot 1 \cdot 5 = \frac{-5}{21}

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