1988 AIME Problems/Problem 13
Find if and are integers such that is a factor of .
Let's work backwards! Let and let be the polynomial such that .
Let's find now. Notice that all we care about in finding is that . Therefore, . Undergoing a similar process, , , , and we see a nice pattern. The coefficients of are just the Fibonacci sequence with alternating signs! Therefore, , where denotes the 16th Fibonnaci number and .
Let represent the th number in the Fibonacci sequence. Therefore,
The above uses the similarity between the Fibonacci recursive definition, , and the polynomial .
We can long divide and search for a pattern; then the remainder would be set to zero to solve for . Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is . Since the coefficient of must be zero, this gives us two equations, and . Solving these two as above, we get that .
There are various similar solutions which yield the same pattern, such as repeated substitution of into the larger polynomial.
The roots of are (the golden ratio) and . These two must also be roots of . Thus, we have two equations: and . Subtract these two and divide by to get . But the formula for the nth fibonacci number is (You may want to research this). Thus, we have , so since and are relatively prime, and the answer must be a positive integer less than , we can guess that it equals .
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