2016 AIME I Problems/Problem 11
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
.