2003 Pan African MO Problems/Problem 6
Find all functions such that: for .
By letting , we have , and by letting , we have . Substitution and simplification results in Therefore, if , where is a real number, then . Thus, the function has rotational symmetry about .
Additionally, multiplying both sides by results in , and rearranging results in . The equation seems to resemble a rearranged slope formula, and the rotational symmetry seems to hint that is a linear function.
To prove that must be a linear function, we need to prove that the slope is the same from to all the other points of the function. By letting , we have . Additionally, by letting , we have . By rearranging the prior equation, we have The slope from points and is . By substitution, The slope from point to is the same as the slope from point to any other point on the function, so must be a linear function.
Let . Using the function on the original equation results in Thus, can be any linear function, so , where are real numbers.
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