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 be a nonzero polynomial such that
for every real
, and
. Then
, where
and
are relatively prime positive integers. Find
.
We substitute into
to get
. Since we also have that
, we have that
and
. We can also substitute
,
, and
into
to get that
,
, and
. This leads us to the conclusion that
and
.
We next use finite differences to find that is a cubic polynomial. Thus,
must be of the form of
. It follows that
; we now have a system of
equations to solve. We plug in
,
, and
to get
We solve this system to get that ,
, and
. Thus,
. Plugging in
, we see that
. Thus,
,
, and our answer is
.