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2016 AIME Problems
11
Polynomial FE
Polynomial FE
Source: 2016 AIME I #11
March 4, 2016
2016 AIME I
AIME
polynomial
Problem Statement
Let
P
(
x
)
P(x)
P
(
x
)
be a nonzero polynomial such that
(
x
−
1
)
P
(
x
+
1
)
=
(
x
+
2
)
P
(
x
)
(x-1)P(x+1)=(x+2)P(x)
(
x
−
1
)
P
(
x
+
1
)
=
(
x
+
2
)
P
(
x
)
for every real
x
x
x
, and
(
P
(
2
)
)
2
=
P
(
3
)
\left(P(2)\right)^2 = P(3)
(
P
(
2
)
)
2
=
P
(
3
)
. Then
P
(
7
2
)
=
m
n
P(\tfrac72)=\tfrac{m}{n}
P
(
2
7
)
=
n
m
, where
m
m
m
and
n
n
n
are relatively prime positive integers. Find
m
+
n
m + n
m
+
n
.
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