Difference between revisions of "2001 IMO Shortlist Problems/A4"
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Conversely, let <math>G</math> be a subgroup of the multiplicative group <math>\mathbb R^*</math>. Take <math>f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\ | Conversely, let <math>G</math> be a subgroup of the multiplicative group <math>\mathbb R^*</math>. Take <math>f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\ | ||
− | 0,\ x\not \in G\end{array}</math>. It's easy to check the condition <math>f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)</math>. | + | 0,\ x\not \in G\end{array}\right\}</math>. It's easy to check the condition <math>f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)</math>. |
== Resources == | == Resources == |
Latest revision as of 12:39, 1 November 2020
Problem
Find all functions , satisfying
![$f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$](http://latex.artofproblemsolving.com/9/a/9/9a905d7941ed2e7b78a9b6e0fb68a3dcffb00c90.png)
for all .
Solution
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
.