2001 Pan African MO Problems/Problem 4
Let be a positive integer, and let be a real number. Consider the equation: How many solutions () does this equation have, such that:
First, rearrange the equation by expanding, simplifying, and using properties of summation. Next, we can try smaller values of to find patterns. If , then the equation becomes , so or . If , we can graph the equation on the Cartesian plane (with axes as and ). The equation is a circle, and from the bounds, the only solutions are . It seems as if the solution to the set involves only and . To prove this, we need to prove that solutions involving only and would work and that no other solutions are possible.
For the first part, we can use induction. The base case is already covered since the solutions for are or . For the inductive step, assume that , where or for integer values of . If , then , and if , then . Thus, . The inductive step holds, so solutions are all ordered pairs where or for integer values of .
For the second part, note that we can complete the square for to get . We know that for integer values of , , so . Thus, .
If or , then . If , then , so . WLOG, let , making . Thus, , so there are no solutions where .
Thus, for each in the solution to the original equation, where is an integer, can only equal or . Therefore, there are ordered pairs that are solutions to the equation.
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