Difference between revisions of "2023 AIME I Problems/Problem 9"
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Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>. Hence, for <math>q \left( x \right)</math>, beyond having a root 2, it has a unique integer root that is not equal to 2. | Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>. Hence, for <math>q \left( x \right)</math>, beyond having a root 2, it has a unique integer root that is not equal to 2. | ||
− | We have | + | We have |
Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> has a unique integer root and it is not equal to 2. | Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> has a unique integer root and it is not equal to 2. | ||
Revision as of 16:16, 8 February 2023
Find the number of cubic polynomials , where , , and are integers in , such that there is a unique integer with
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 2
Define . Hence, for , beyond having a root 2, it has a unique integer root that is not equal to 2.
We have
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)