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:
$$P\Big(\frac{7}{2}\Big)=\Big(\frac{2}{3}\Big)\Big(\frac{5}{2}\Big)\Big(\frac{7}{2}\Big)\Big(\frac{9}{2}\Big)=\frac{105}{4}\Rightarrow m=105, n=4\Rightarrow m+n=105+4=\boxed{109}.$$
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