1979 AHSME Problems/Problem 26
Problem 26
The function satisfies the functional equation for every pair of real numbers. If , then the number of integers for which is
Solution
We are given that and , so we can let . Thus we have:
Rearranging gives a recursive formula for :
We notice that this is the recursive form for a quadratic, so f(x) must be of the form . To solve for and , we can first work backwards to solve for the values of f(0) and f(-1):
Since : Since : Similarly, since :
Thus we have the system of equations:
Which can be solved to yield , . Therefore, .
Since we are searching for values for which , we have the equation . Subtracting yields , which we can simplify by dividing both sides by : . This factors into , so therefore there are two solutions to : and . Since the problem asks only for solutions that do not equal , the answer is .
See Also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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