Problem : (here)
Find all functions
for all

Solution :
Let be the assertion
is obviously surjective.

Let there exist a such that .

and


But similarly

Hence can only be zero.
Due to surjectivity of ,




Thus is ,
That is


Let there exist some such that

Let
Using the above result,


Hence we get if

From and the result of sign (positive or negative) of depending on ),

Observe that
for

For
Hence we get
Setting above, we get
This implies

Thus

But again because of the fact (after some case making I guess),
we get is additive that is

Also obviously by the above condition and the fact that , the function is monotonous...

Hence by Cauchy equation, the function is linear.
Putting value is the only function satisfying the condition and hence we are done.



Problem (@socrates)(here):[\b]
Determine all functions such that for all

Solution :
Let be the assertion

First I will show that is injective
Let
We can always chose an such that

That is

Hence we have is surjective.


Setting
for all (since we can find such
But this also implies

Thus we have
Using injectivity and setting

That is is linear.

Putting , we get

Hence is the only solution.