Mock AIME II 2012 Problems/Problem 2
Let be a recursion defined such that , and where , and is an integer. If for being a positive integer greater than and being a positive integer greater than 2, find the smallest possible value of .
The sequence goes as follows: Note that this sequence repeats, because once we get to , we subtract from the number under the square root sign and take the square root of that, then when you take the difference of the squares of these two numbers, they have a difference of , so you get back to .
The first perfect square less than is . Therefore .
We present three ways to finding :
Way 1: Now notice that, for , we have . Also, the terms under the radical counts down one for every such that . The number of before a number can be seen to be . Therefore, we can see that . Setting this equal to , we see that . We can see that both and satisfy this, however, since , we disregard this, and we find that , so .
Way 2: Look at the sequence by grouping terms as follows: . Note that the number of these groups of three that we have is equal to the number of terms in the sequence . This has terms in it. Before the last group, we have terms, and is therefore the th term, and .
Both ways lead to and , so the answer is .
To show that no smaller sums can be created, note that the next perfect square will be and will take more than as the value of , and hence all other perfect squares will take more than , and we have as our sum which is greater than since .
We must ignore the cases where , as required by the problem statement. Write for some integers . We want to minimize . Note if is even ( is even) then , and if is odd ( is odd), then . Rewrite and note that is at most . In the case that is odd, we have . This is a quadratic in and is strictly decreasing for , so to minimize we should choose , which gives . Do the case for even as well, you find that is the lowest.