Harvard-MIT Mathematics Tournament

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

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

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

So:

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

Cross-multiplied:

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

Suppose $P(x)$ has degree $n$ and leading coefficient $c$. Then $P(2x)$ has leading term $c (2x)^n = c 2^n x^n$, and $P(x+1)$ has leading term $c x^n$. So the left side's leading term is $x \cdot c 2^n x^n = c 2^n x^{n+1}$, and the right side's leading term is $8x \cdot c x^n = 8c x^{n+1}$. Equate:

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

So $P(x)$ is cubic.

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

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

Plug into the original equation:

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

So:

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

Cross-multiplied:

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

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

Divide both sides by $x$ (for $x \neq 0$):

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

Or:

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

Now, try $x = -1$:

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

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

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

Plug into the previous equation:

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

Divide both sides by $2$:

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

If $x = -3$:

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

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

So $R(-6) = 0$

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

Therefore,

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

Since $P(x)$ is cubic, $S$ is a constant.

Given $P(1) = 1$:

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

Therefore,

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

So,

$P(-1) = \frac{1}{21} \cdot (-1) \cdot 1 \cdot 5 = \frac{-5}{21}$