2016 AIME I Problems/Problem 11

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 $m+n=\boxed{109}$ .

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2016 AIME I Problems/Problem 11