2023 AIME I Problems/Problem 9
Contents
[hide]Problem (Unofficial, please update when official one comes out):
is a polynomial with integer coefficients in the range
, inclusive. There is exactly one integer
such that
. How many possible values are there for the ordered triple
?
Solution
Solution 1
Plugging into
, we get
. We can rewrite into
, where
can be any value in the range. Since
must be
. The problem also asks for unique integers, meaning
can only be one value for each polynomial, so the discriminant must be
.
, and
. Rewrite to be
.
must be even for
to be an integer.
because
. There are 9 pairs of
and 41 integers for
, giving
~chem1kall
Solution 2
a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328?
Solution
Define .
Hence, for
, beyond having a root 2, it has a unique integer root that is not equal to 2.
We have
Thus, the polynomial has a unique integer root and it is not equal to 2.
Following from Vieta' formula, the sum of two roots of this polynomial is .
Because
is an integer, if a root is an integer, the other root is also an integer.
Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0.
Thus,
\[
\left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1)
\]
In addition, because two identical roots are not 2, we have \[ 2 + a \neq - 4 . \]
Equation (1) can be reorganized as \[ 4 b = \left( a - 2 \right)^2 - 16 . \hspace{1cm} (2) \]
Thus, . Denote
.
Thus, (2) can be written as
\[
b = d^2 - 4 . \hspace{1cm} (3)
\]
Because ,
, and
, we have
.
Therefore, we have the following feasible solutions for :
,
,
,
,
.
Thus, the total number of
is 8.
Because can take any value from
, the number of feasible
is 41.
Therefore, the number of is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)