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

We substitute $x=2$ into $(x-1)P(x+1)=(x+2)P(x)$ to get $P(3)=4P(2)$ . Since we also have that $\left(P(2)\right)^2 = P(3)$ , we have that $P(2)=4$ and $P(3)=16$ . We can also substitute $x=1$ , $x=0$ , and $x=3$ into $(x-1)P(x+1)=(x+2)P(x)$ to get that $0=P(1)$ , $-1P(1)=2P(0)$ , and $2P(4)=5P(3)$ . This leads us to the conclusion that $P(0)=P(1)=0$ and $P(4)=40$ .

We next use finite differences to find that $P$ is a cubic polynomial. Thus, $P$ must be of the form of $ax^3+bx^2+cx+d$ . It follows that $d=0$ ; we now have a system of $3$ equations to solve. We plug in $x=1$ , $x=2$ , and $x=3$ to get

$a+b+c=0$ $8a+4b+2c=4$ $27a+9b+3c=16$

We solve this system to get that $a=\frac{3}{2}$ , $b=0$ , and $c=-\frac{3}{2}$ . Thus, $P(x)=\frac{3}{2}x^3-\frac{3}{2}x$ . Plugging in $x=\frac{7}{2}$ , we see that $P\left(\frac{7}{2}\right)=\frac{105}{4}$ . Thus, $m=105$ , $n=4$ , and our answer is $m+n=\boxed{109}$ .

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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