2003 Pan African MO Problems/Problem 1
Let . Find all functions: such that:
(1) , all ;
(3) , all .
Because , we must have , so . Thus, .
By trying smaller values, we suspect that for all values in the set . To prove this, we would use induction to show that for for all integer values of .
The base case works because . For the inductive step, assume that for for an integer value of . To prove this case, we need to prove that all values from to also work.
Note that . Therefore, we must have . Additionally, given that is an integer that satisfies , we must have . Since is within the range of , we must have . Furthermore, we must also have . Thus, we must have . Since can be any integer that satisfies , the value can be any even integer. Thus all numbers from to are represented in the proof, so the inductive step holds.
Therefore, for for any integer value of , so the only function that satisfies the three conditions is .
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