[Peter]

Find all functions such that for all :

• ,
• ,
• .

[jastrzab]

There is only one such function: .
Of course so .
Now we prove by induction that . Indeed and
Now as the function is increasing, and we have we get for every

[TTsphn]

Other solution.
We prove by induction that
then it is true.
Suppose
We prove that
Case 1 is a prime then is a composite .
Suppose where
So
But so
Case 2 where
Then easy to check
So
Problem 2



To solve this problem we use result:
If where then
So induction as above solution we get result :

[azo]

jastrzab wrote:

There is only one such function: .
Of course

so .
Now we prove by induction that . Indeed and

Now as the function is increasing, and we have

we get for every

I don't understand - how is the conclusion that f(n) = n reached. Could someone clarify it to me please?

[t0rajir0u]

The values are strictly increasing and between and , so they must take on each value between exactly once and in increasing order.

[Allnames]

Peter wrote:

Find all functions such that for all :

• ,
• ,
• .

Ok,how about
Find all functions such that for all :

• ,
• .

Result

,where

Let try

[joh]

A proof by induction

We will prove that for all by using induction.

First note that which implies that . Assume that is true for . We will show that . We consider two cases.

First case. Assume that is odd and let for some integer , so

and we are done.

Second case. Assume that is even and let for some integer , so

Thus we have

which gives , as desired.

Allnames' problem can be solved analogously by the hypothesis that ..

[ZetaX]

Allnames wrote:

Find all functions such that for all :

• ,
• .

Result

,where

Let try

In fact, it suffices to have for coprime only. It was some China TST, I think.