K 12

Peter

K 12

**Posted:** Fri May 25, 2007 2:25 am

jastrzab

There is only one such function: $f(n)=n$ .
Of course $f(k \cdot 1)=f(k) \cdot f(1)$ so $f(1)=1$ .
Now we prove by induction that $f(2^{k})=2^{k}$ . Indeed $f(2)=2$ and $f(2^{k+1})=f(2^{k}) \cdot f(2)= 2^{k+1}$
Now as the function is increasing, and we have $2^{k+1}-2^{k}=f(2^{k+1})-f(2^{k})$ we get $f(n)=n$ for every $n \in N$

**Posted:** Fri May 25, 2007 2:25 am

TTsphn

#3

Other solution.
We prove by induction that $f(n)=n$
$n=1,2$ then it is true.
Suppose $f(k)=k,\forall k=1,..,k$
We prove that $f(k+1)=k+1$
Case 1 $k+1$ is a prime then $k+2$ is a composite .
Suppose $k+2=ab$ where $a,b>1$
So $f(k+2)=f(a)f(b)=ab=k+2$
But $f(k)<f(k+1)<f(k+2)$ so $f(k+1)=k+1$
Case 2 $k+1=ab$ where $a,b>1$
Then easy to check $f(k+1)=k+1$
So $f(n)=n,\forall n\in N$
Problem 2 $f: N\to N$
$f(mn)=f(m)f(n),\forall \gcd(m,n)=1$
$f(2)=2,f(3)=3$
$f(n+1)>(n)$
To solve this problem we use result:
If $f(n)=n,f(m)=m$ where $m>n$ then $f(k)=k,\forall k\in[n,m]$
So induction as above solution we get result : $f(n)=n,\forall n\in N$

**Posted:** Fri Oct 26, 2007 7:55 am

azo · New Member Offline Joined: 02 Apr 2008 Posts: 6

#4

**Posted:** Mon Jul 07, 2008 11:12 pm

t0rajir0u

#5

The values $f(2^k + a), a = 1, 2, ... 2^k - 1$ are strictly increasing and between $2^k$ and $2^{k+1}$ , so they must take on each value between exactly once and in increasing order.