2023 AIME I Problems/Problem 9
Problem
Find the number of cubic polynomials where and are integers in such that there is a unique integer with
Solution 1 (bash)
Plugging and into and equating them, we get . Rearranging, we have Note that the value of won't matter as it can be anything in the provided range, giving a total of possible choices for So what we just need to do is to just find the number of ordered pairs that work, and multiply it by We can start by first dividing both sides by (Note that this is valid since We can rearrange this so it is a quadratic in : Remember that has to be unique and not equal to We can split this into two cases: case being that has exactly one solution, and it isn't equal to ; case being that has two solutions, one being equal to but the other is a unique solution not equal to
There is exactly one solution for and that solution is not This means that the discriminant of the quadratic equation is using that, we have rearranging in a neat way, we have Using the fact that must be a perfect square, we can easily see that the values for can be and Also since it's a "" there will usually be solutions for for each value of The two exceptions for this would be if and For because it would be a which only gives one solution, instead of two. And for because then and the solution for would equal to and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
So we have solutions for each of which give values for except for which only give one. So in total, there are ordered pairs of in this case.
has two solutions, but exactly one of them isn't equal to This ensures that of the solutions is equal to
Let be the other value of that isn't By Vieta: From the first equation, we subtract both sides by and double both sides to get which also equals to from the second equation. Equating both, we have We can easily count that there would be ordered pairs that satisfy that.
However, there's an outlier case in which happens to also equal to and we don't want that. We can reverse engineer and find out that when which we overcounted. So we subtract by one and we conclude that there are ordered pairs of that satisfy this case.
This all shows that there are a total of amount of ordered pairs Multiplying this by (the amount of values for ) we get as our final answer.
~s214425
Solution 2 (factor the difference)
is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either or , with .
In the first case , with (which entails ), so can be and can be any integer from to , giving polynomials (since the coefficients are given by linear functions of and thus are distinct).
In the second case , and can be and can be any integer from to , giving polynomials.
The total is .
~EVIN-
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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