IMO 2010 #1
by Wolstenholme, Nov 1, 2014, 1:07 AM
Find all function 
such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](//latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where
is greatest integer not greater than 
Solution:
Letting
we have that 
so either 
or 
for all 
We now proceed with casework.
Case 1: for all
This immediately yields the solution
Case 2:
Letting
we have that 
so either 
or 
Case 2.1:
Letting
immediately yields the solution 
Case 2.2:
Letting
yields that 
for all 
Knowing this and letting 
and 
yields $ f(1) = f(2)\left\lfloor{f\left(\frac{1}{2}\right)\right\rfloor = f(2)f(0) = 0 $ which contradicts the fact that
Therefore we have found all possible solutions and they are all easily shown to work so we are done.
such that for all 
the following equality holds ![\[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \]](http://latex.artofproblemsolving.com/f/3/a/f3ac860e90d671f9fb592a6bd55c102a3000daec.png)
where
is greatest integer not greater than 
Solution:
Letting
we have that 
so either 
or 
for all 
We now proceed with casework.Case 1:
for all This immediately yields the solution
Case 2:
Letting
we have that 
so either 
or 
Case 2.1:
Letting
immediately yields the solution 
Case 2.2:
Letting
yields that 
for all 
Knowing this and letting 
and 
yields $ f(1) = f(2)\left\lfloor{f\left(\frac{1}{2}\right)\right\rfloor = f(2)f(0) = 0 $ which contradicts the fact that Therefore we have found all possible solutions and they are all easily shown to work so we are done.
This post has been edited 2 times. Last edited by Wolstenholme, Nov 1, 2014, 1:08 AM
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