Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all $m\in \mathbb{Z}$: \[f(f(m))=m+1.\]
Problem
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Tags: function, Functional Equations
jastrzab
25.05.2007 03:25
There are no such functions. Indeed, plugging $m=f(l)$ in our equation we get \[f(f(f(l)))=f(l+1)\] but on the other hand \[f(f(f(l)))=f(l)+1\] and thus \[f(l+1)=f(l)+1\] which means that our function has form \[f(l)=l+k\] where $k\in N$. Plugging this into the equation we obtain $2\cdot k=1$, which is contradiction.
panamath
18.09.2012 03:56
So what you say is, $k=f(0)$, am I right?
Wolowizard
04.02.2015 23:18
My solution: You have f^2(m)=m+1; ( here f^2 denotes f(f()) ), so if you instead of m put f^2(m) you get f(f^2(m))=f^2(m)+1 And then if you put for f^2(m) m+1 you get f(m+1)=m+2 => f(x)=x+1 for every x e Z. End of proof.
Wolowizard
04.02.2015 23:31
Actually in 4 th line you have f(f^2(m))=f(m)+1 => f(x+1)=f(x)+1 and than if we put f(0)=z where z is fro Z we get f(x)=f(x-1) +1 we get z+f(m)=m+1=> 2z=1 => z=1/2 => contradiction.
WalkerTesla
10.05.2016 17:26
WLOG, let $f(0) = a$. It follows by the identity that $f(a)=1$, $f(1) = a+1$, and so on. In general, we have that $f(n) = a+n$, and $f(a+n) = n+1$ for nonnegative integers $n$. We have two cases. $a$ cannot equal $0$ as if $a=0$, we would have $f(f(0)) = f(0) = 0$ which contradicts the identity. If $a<0$, we take the value $n=-a$. This gives us $f(-a) = 0$, and $f(0) = 1-a$. However, we have that $f(0) = a$, and this gives us $1-a = a$ or $a=\dfrac{1}{2}$, which is not an integer. The other case is when $a>0$. In this case, we take $n = a$. This gives us $f(a) = 2a$, but we also have that $f(a)=f(f(0)) = 1$ by the identity. This gives us $2a = 1$ or $a=\dfrac{1}{2}$ which is also not an integer. Because all integer values which $f(0)$ could take on lead to contradictions, it follows that no such functions exist.