Talk:2020 AMC 12A Problems/Problem 25
So... guys, I dont seem to understand the first solution. What is k and b and c?
Remarks of Solution 2 and Video Solution 3
For all positive integers the first nonzero solutions to are of the form where
Equivalently, for the intersections of the graphs of and occur in the consecutive branches of namely at
Proof by Graph
Clearly, the equation has no negative solutions, and its positive solutions all satisfy Moreover, none of its solutions is an integer.
Note that the upper bounds of the branches of are along the line (excluded). To prove the claim, we wish to show that for each branch of there is exactly one solution for (from the branch to the branch containing the larger solution of ). In 8:07-11:31 of Video Solution 3 (Art of Problem-Solving), Mr. Rusczyk questions whether two solutions of can be in the same branch of and he concludes that it is impossible in 16:25-16:43.
We analyze the upper bound of Let be one solution of It is clear that We substitute this point to find
We substitute this result back to find By the way, using the precondition that is a root of we can factor its left side easily by the Factor Theorem. Note that for all as quadratic functions always outgrow linear functions.
Now, we perform casework:
- (Trivial Case)
It follows that the graphs of and only intersect at the point which is not on the graph of So, the equation has no solutions in this case, as the inequality has no solutions.
It follows that for the smaller solution is and holds for all
By the Intermediate Value Theorem, for each branch of (where ), we have in between its left output and its right "output", namely Therefore, for the equation there is exactly one solution for each branch of where Now, the proof of the bolded sentence of paragraph 2 is complete.
This case uses the same argument as Case 2. The smaller solution is and for each branch of where the equation has exactly one solution.