2001 IMO Shortlist Problems/A4
Find all functions , satisfying
for all .
Assume . Take . We get , so . This is a solution, so we can take it out of the way: assume .
. We either have or , so for every , . In particular, .
Assume . We get . This means that ( is defined because ). Assume now that and . We get , and after replacing everything we get , so . Assume now . From we get , and after applying again to we get . We can now see that combine to .
Let . and simply say that is a subgroup of .
Conversely, let be a subgroup of the multiplicative group . Take . It's easy to check the condition .