2023 AIME I Problems/Problem 9
Find the number of cubic polynomials , where , , and are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer with .
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[hide]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 . However, plugging in result in . There are 8 pairs of and 41 integers for , giving
~chem1kall
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,
In addition, because two identical roots are not 2, we have
Equation (1) can be reorganized as
Thus, . Denote . Thus, (2) can be written as
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)
Solution
It can be easily noticed that is independent of the condition , and can thus safely take all possible values between and .
There are two possible ways for to be the only integer satisfying : has a double root at or a double root at .
Case 1: has a double root at :
In this case, , or . Thus ranges from to . One of these values, corresponds to a triple root at , which means . Thus there are possible values of . (It can be verified that is an integer).
Case 2: has a double root at :
See the above solution. This yields possible combinations of and .
Thus, in total we have combinations of .
-Alex_Z