Problem:
Find the number of cubic polynomials , where , and are integers in , such that there is a unique integer with .
Solution:
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 .
Case 1:
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.
Case 2:
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.
The problems on this page are the property of the MAA's American Mathematics Competitions