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)