Answer yes. f(0)=0 let's determine for natural number.Let f(5n)=5f(n) and all natural number divide in infinite multitude with 2003 number n1,n2,...,n2003 where CGD(ni,5)=1 and let for each multitude f(n1)=5(n2),f(n2)=n3,f(n3)=n2,...,f(n2003)=n1. So it's easy to see that this function satisfy all codition.

shobber wrote: Let $\mathbb{N}_0=\{0,1,2 \cdots \}$. Does there exist a function $f: \mathbb{N}_0 \to \mathbb{N}_0$ such that: \[ f^{2003}(n)=5n, \forall n \in \mathbb{N}_0 \]where we define: $f^1(n)=f(n)$ and $f^{k+1}(n)=f(f^k(n))$, $\forall k \in \mathbb{N}_0$?

Yes, there exist a function. Define a sequence a0=n,a1=f(n)...a2003= f2003(n). Using characterstic equation we get: ak=C(5)^(k/2003) where C=n. a1=n(5)^(1/2003)=f(n). Therefore function satisfying the equation are f(n)=n(5)^(1/2003) and f(n)=0.

OGGY_666 wrote: Therefore function satisfying the equation are f(n)=n(5)^(1/2003) and f(n)=0. Could you kindly explain the notation f(n)=n(5)^(1/2003) ? (what is the meaning of (5) ?) And give as example what is the exact value of nonnegative integer $f(2)$ in your example ? Thanks in advance.

Quote: Could you kindly explain the notation f(n)=n(5)^(1/2003) ? (what is the meaning of (5) ?) This means that f(n)=n times 2003rd root of 5. Quote: And give as example what is the exact value of nonnegative integer f(2) in your example ? f(2)=2*5(1/2003) which on calculating gives ~2.001607673. f(f(2))=2*5(2/2003) which on calculating gives ~2.003216639. f(f(f(3)))=2*5(3/2003) which on calculating gives ~2.004826898. ..................... ..................... ..................... f2003(n)=2*5(2003/2003) which gives the value 10.

OGGY_666 wrote: Quote: Could you kindly explain the notation f(n)=n(5)^(1/2003) ? (what is the meaning of (5) ?) This means that f(n)=n times 2003rd root of 5. Quote: And give as example what is the exact value of nonnegative integer f(2) in your example ? f(2)=2*5(1/2003) which on calculating gives ~2.001607673. Which unfortunately does not seem to be a nonnegative integer, as required by the problem statement

Quote: Which unfortunately does not seem to be a nonnegative integer, as required by the problem statement

OGGY_666 wrote: Quote: Which unfortunately does not seem to be a nonnegative integer, as required by the problem statement Isn't 2 a non-negative integer? Are you just joking ? $2$ is indeed a nonnegative integer. But $f(2)=~2.001607673$, as you suggested, is not (or at least was not $45$ years ago, when I was a student)

akhan98 wrote: Answer yes. f(0)=0 let's determine for natural number.Let f(5n)=5f(n) and all natural number divide in infinite multitude with 2003 number n1,n2,...,n2003 where CGD(ni,5)=1 and let for each multitude f(n1)=5(n2),f(n2)=n3,f(n3)=n2,...,f(n2003)=n1. So it's easy to see that this function satisfy all codition. I don't quite understand this solution. Could someone help me out?

Consider the function defined as follows: Group the naturals indivisible by 5 into groups of size $k$ (example: if $k=3$, the groups are $(1,2,3)$, $(4,6,7)$, $(8,9,11)$, so on) and put the first term of each group in set $S_{1}$, the second term of each group in set $S_{2}$, so on until the last term of each group in set $S_{k}$. Now for notation let $S(n)$ denote the nth term of set S. Clearly $S_{1} \cup S_{2} \cup ... \cup S_{k}$ is the set of naturals indivisible by 5, and the sets are pairwise disjoint, so for any natural $n$ indivisible by $5$ there exists unique $t$ and $m$ such that $n = S_{t}(m)$. Then define $f(n)$ as follows: If $t<k$ then $f(n) = S_{t+1}(m)$ If $t=k$ then $f(n) = 5S_{1}(m)$ If $5|n$ then $f(n) = 5^{v_{5}(n)}f(\frac{n}{5^{v_{5}(n)}})$ $f(0)=0$ From this definition it becomes clear that $f^{k}(n)=5n$ for all non-negative integers $n$, so therefore it's possible for all positive integers $k$. Setting $k=2003$ creates a function such that $f^{2003}(n)=5n$ for all non-negative integers $n$, so the answer is that it's possible.

OGGY_666 wrote: Yes, there exist a function. Define a sequence a0=n,a1=f(n)...a2003= f2003(n). Using characterstic equation we get: ak=C(5)^(k/2003) where C=n. a1=n(5)^(1/2003)=f(n). Therefore function satisfying the equation are f(n)=n(5)^(1/2003) and f(n)=0. How do you use a characteristic equation on this sequence?