2016 AIME I #11

This morning, I came across this problem on the AoPS community. I thought I'd give it a go, and it turned out having a functional equation (pretty rare on a computational contest). The solution came out to be pretty nice with an inductive root theorem proof. Check out the problem and solution below!

Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$, and $(P(2))^2=P(3)$. Then $P\Big(\frac{7}{2}\Big)=\frac{m}{n}$, where $m$ and $n$ are positive integers. Find $m+n$.

Solution: First, we plug in $x=-2$ to the given functional equation to get that $$(-3)P(-1)=0\Rightarrow P(-1)=0.$$ Similarly, we can plug in $x=1$ to the functional equation to find that $P(1)=0$.
Now, we investigate the other roots of the polynomial. Suppose that $P(k)=0$, where $k$ is a real number other than $-1$ or $1$. Plugging in $x=k$ to the above functional equation gives us that $$(k-1)P(k+1)=(k+2)P(k)\Rightarrow (k-1)P(k+1)=0 \Rightarrow P(k+1)=0.$$ The last arrow in the above sequence follows from the fact that $k\not=1\Leftrightarrow k-1\not=0$. Similarly, we can plug in $x=k-1$ to the above functional equation to get $$(k-2)P(k)=(k+1)P(k-1)\Rightarrow 0=(k+1)P(k-1)\Rightarrow P(k-1)=0.$$ Here again, our final conclusion was due to that fact that $k\not=-1\Leftrightarrow k+1\not=0$.
Our above analysis leads us to the conclusion that if $k$ is a root of $P(x)$ other that $-1 \text{ or } 1$, then so are $k-1$ and $k+1$. We can use inspection that see that $x=0$ could be a root of the polynomial. However, for any other real number $k$, this condition would imply that $P(x)$ is a polynomial of infinite degree, contradicting the problem statement. Finally, we are given that $(P(2))^2=P(3)$, suggesting the use of $x=2$ in the above functional equality. $$P(3)=4P(2)\Rightarrow P(2)=4.$$ Hence, we can rearrange the statement as $\frac{P(3)}{P(2)}=P(2)=4$. Now, we need to consider two cases based on whether or not $x=0$ is a root of the polynomial and see which case meets the above condition.
a. Letting P(x)=a(x-1)(x+1), we get that $$\frac{P(3)}{P(2)}=\frac{a(2)(4)}{a(1)(3)}=\frac{8}{3}\not=4.$$ Therefore, $x=0$ is a root of $P(x)$.
b. Letting $P(x)=ax(x-1)(x+1)$, we find that $$\frac{P(3)}{P(2)}=\frac{a(2)(3)(4)}{a(1)(2)(3)}=4$$ as desired. We have verified that the degree-three polynomial with roots of $x=-1,0,1$ does indeed fit the conditions for $P(x)$. Now, we solve for the constant $a$ using the known value of $P(2)$: $$P(2)=4=a(1)(2)(3)\Leftrightarrow a=\frac{2}{3}.$$ We finish by getting the desired value: $$P\Big(\frac{7}{2}\Big)=\Big(\frac{2}{3}\Big)\Big(\frac{5}{2}\Big)\Big(\frac{7}{2}\Big)\Big(\frac{9}{2}\Big)=\frac{105}{4}\Rightarrow m=105, n=4\Rightarrow m+n=105+4=\boxed{109}.$$