AIME 2016 I · 第 11 题

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,

$$(x-1)P(x+1) = (x+2)P(x)$$ $$(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))$$ $$(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)$$ $$Q(x+1) = Q(x)$$ From here, we know that $Q(x) = C$ for a constant $C$ ($Q(x)$ cannot be periodic since it is a polynomial), 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

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 $P(x)$ (if we shift the function right by 1, we get $(x-2)P(x) = (x+1)P(x-1)$, and from here we can see that $x+1$ divides $P(x)$). 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 one of 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

Substituting $x=2$ into the given equation, we find that $P(3)=4P(2)=P(2)^2$. Therefore, either $P(2)=0$ or $P(2)=4$. 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

$$P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).$$ If $P(2)=0$, this shows that $P(x)$ has infinitely many roots, meaning that $P(x)$ is identically equal to zero. But this contradicts the problem statement. Therefore, $P(2)=4$, and 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 $\boxed{109}$.

Solution 5

We can find zeroes of the polynomial by making the first given equation $0 = 0$. Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$, respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \cdot P(2) = 0$, which means $P(2)$ is not necessarily $0$. If $P(2) = 0$, then plugging in $2$ to the equation yields $P(3) = 0$, plugging in $3$ to the equation yields $P(4) = 0$, and so on, a contradiction of "nonzero polynomial". So $2$ is not a zero. Note that plugging in $x = 0$ to the equation does not yield any additional zeros. Thus, the only zeroes of $P(x)$ are $-1, 0,$ and $1$, so $P(x) = a(x + 1)x(x - 1)$ for some nonzero constant $a$. We can plug in $2$ and $3$ into the polynomial and use the second given equation to find an equation for $a$. $P(2) = 6a$ and $P(3) = 24a$, so:

$$(6a)^2 = 24a \implies 36a^2 = 24a \implies a = \frac23$$ Plugging in $\frac72$ into the polynomial $\frac23(x + 1)x(x - 1)$ yields $\frac{105}{4}$. $105 + 4 = \boxed{109}$.

Solution 6

Plug in $x=2$ yields $P(3)=4P(2)$. Since also $(P(2))^2=P(3)$, we have $P(2)=4$ and $P(3)=16$. Plug in $x=3$ yields $2P(4)=5P(3)$ so $P(4)=40$.

Repeat the action gives $P(2)=4$, $P(3)=16$, $P(4)=40$, $P(5)=80$, and $P(6)=140$.

Since $P(x)$ is a polynomial, the $k$th difference is constant, where $k=\deg(P(x))$. Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant.

$$4,16,40,80,140$$ $$12,24,40,60$$ $$12,16,20$$ $$4,4,4$$ Since the 3rd difference of $P(x)$ is constant, we can conclude that $\deg(P(x))=3$.

Let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$. Plug in the values for $x$ and solve the system of 4 equations gives $(a_3,a_2,a_1,a_0)=(\frac{2}{3},0,-\frac{2}{3},0)$

Thus $P(x)=\frac{2}{3}x^3-\frac{2}{3}x$ and $P(\frac{7}{2})=\frac{105}{4}\Longrightarrow m+n=\boxed{109}$

~ Nafer

-Note that the coefficient of $x^3$ is $\frac{2}{3}$ because $\frac{2}{3}\cdot\deg(x^3)!=\frac{2}{3}\cdot3!=4$, the 3rd difference of P(x). ~inaccessibles

Note(Key Idea)

Their are a couple of solution methods such as constant $n$th differences and using intuition on the roots. But the main idea in this problem was the fact that, $P(\frac{7}{2})$ is hard to get by itself, cause we can only find the output for integral input values from the information we have. This is crucial as it helps solves this mischievous problem.

Thus, we can solve for $P(x)$, and this is best approached by finding its roots.

~Bigbrain_2009

== See also