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Difference between revisions of "2016 AIME I Problems/Problem 11"

(Created page with "We substitute <math>x=2</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get <math>P(3)=4P(2)</math>. Since we also have that <math>\left(P(2)\right)^2 = P(3)</math>, we have...")
 
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Let <math>P(x)</math> be a nonzero polynomial such that <math>(x-1)P(x+1)=(x+2)P(x)</math> for every real <math>x</math>, and <math>\left(P(2)\right)^2 = P(3)</math>. Then <math>P(\tfrac72)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
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We substitute <math>x=2</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get <math>P(3)=4P(2)</math>. Since we also have that <math>\left(P(2)\right)^2 = P(3)</math>, we have that <math>P(2)=4</math> and <math>P(3)=16</math>. We can also substitute <math>x=1</math>, <math>x=0</math>, and <math>x=3</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get that <math>0=P(1)</math>, <math>-1P(1)=2P(0)</math>, and <math>2P(4)=5P(3)</math>. This leads us to the conclusion that <math>P(0)=P(1)=0</math> and <math>P(4)=40</math>.
 
We substitute <math>x=2</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get <math>P(3)=4P(2)</math>. Since we also have that <math>\left(P(2)\right)^2 = P(3)</math>, we have that <math>P(2)=4</math> and <math>P(3)=16</math>. We can also substitute <math>x=1</math>, <math>x=0</math>, and <math>x=3</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get that <math>0=P(1)</math>, <math>-1P(1)=2P(0)</math>, and <math>2P(4)=5P(3)</math>. This leads us to the conclusion that <math>P(0)=P(1)=0</math> and <math>P(4)=40</math>.
  

Revision as of 13:45, 4 March 2016

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 $m + n$.

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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