2016 AIME I Problems/Problem 11

Problem

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 $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution 1

Plug in $x=1$ to get $(1-1)P(1+1) = 0 = (1+2)P(1) \Rightarrow P(1) = 0$ . Plug in $x=0$ to get $(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0$ . Plug in $x=-1$ to get $(-1-1)P(-1+1) = (-1+2)P(-1)\Rightarrow$ (-2)P(0)=P(-1)\Rightarrow P(-1) = 0 $. So$ P(x) = x(x-1)(x+1)Q(x) $for some polynomial$ Q(x) $. Using the initial equation, once again, <cmath>(x-1)P(x+1) = (x+2)P(x)</cmath> <cmath>(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))</cmath> <cmath>(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)</cmath> <cmath>Q(x+1) = Q(x)</cmath> From here, we know that$ Q(x) = C $for a constant$ C $, so$ P(x) = Cx(x-1)(x+1) $. We know that$ \left(P(2)\right)^2 = P(3) $. Plugging those into our definition of$ P(x) $:$ (C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = 0 $or$ \frac{2}{3} $. So we know that$ P(x) = \frac{2}{3}x(x-1)(x+1) $. So$ P(\frac{7}{2}) = \frac{2}{3} \cdot \frac{7}{2} \cdot (\frac{7}{2} - 1) \cdot (\frac{7}{2} + 1) = \frac{105}{4} $. Thus, the answer is$ 105 + 4 = \boxed{109}$.

Solution 2

So from the equation we see that $x-1$ divides $P(x)$ and $(x+2)$ divides $P(x+1)$ so we can conclude that $x-1$ and $x+1$ divide $P(x)$ . This means that $1$ and $-1$ are roots of $P(x)$ . Plug in $x = 0$ and we see that $P(0) = 0$ so $0$ is also a root.

Suppose we had another root that is not those $3$ . Notice that the equation above indicates that if $r$ is a root then $r+1$ and $r-1$ is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.

That means $P(x) = cx(x-1)(x+1)$ . We can use $P(2)^2 = P(3)$ to get $c = \frac{2}{3}$ . Plugging in $\frac{7}{2}$ is now trivial and we see that it is $\frac{105}{4}$ so our answer is $\boxed{109}$

Solution 3

Although this may not be the most mathematically rigorous answer, we see that $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$ . Using a bit of logic, we can make a guess that $P(x+1)$ has a factor of $x+2$ , telling us $P(x)$ has a factor of $x+1$ . Similarly, we guess that $P(x)$ has a factor of $x-1$ , which means $P(x+1)$ has a factor of $x$ . Now, since $P(x)$ and $P(x+1)$ have so many factors that are off by one, we may surmise that when you plug $x+1$ into $P(x)$ , the factors "shift over," i.e. $P(x)=(A)(A+1)(A+2)...(A+n)$ , which goes to $P(x+1)=(A+1)(A+2)(A+3)...(A+n+1)$ . This is useful because these, when divided, result in $\frac{P(x+1)}{P(x)}=\frac{A+n+1}{A}$ . If $\frac{A+n+1}{A}=\frac{x+2}{x-1}$ , then we get $A=x-1$ and $A+n+1=x+2$ , $n=2$ . This gives us $P(x)=(x-1)x(x+1)$ and $P(x+1)=x(x+1)(x+2)$ , and at this point we realize that there has to be some constant $a$ multiplied in front of the factors, which won't affect our fraction $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$ but will give us the correct values of $P(2)$ and $P(3)$ . Thus $P(x)=a(x-1)x(x+1)$ , and we utilize $P(2)^2=P(3)$ to find $a=\frac{2}{3}$ . Evaluating $P \left ( \frac{7}{2} \right )$ is then easy, and we see it equals $\frac{105}{4}$ , so the answer is $\boxed{109}$

Solution 4

As above, we find that $P(2)=4$ . Now for integers $n\ge 2$ , we know that $P(n+1)=\frac{n+2}{n-1}P(n).$ Applying this repeatedly, we find that $P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).$ Therefore, as $P(2)=4$ , we find $P(n+1)=\frac{2}{3}(n+2)(n+1)n$ for all positive integers $n\ge2$ . This cubic polynomial matches the values $P(n+1)$ for infinitely many numbers, hence the two polynomials are identically equal. In particular, $P\left(\frac72\right)=\frac23\cdot\frac92\cdot\frac72\cdot\frac52=\frac{105}{4}$ , and the answer is $\boxed{109}$ .

2016 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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