Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ satisfying the condition $f(n)-f(n+f(m))=m$ for all $m,n\in \mathbb{Z}$
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Tags: function, induction, algebra proposed, algebra
11.12.2010 12:51
efoski1687 wrote: Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ satisfying the condition $f(n)-f(n+f(m))=m$ for all $m,n\in \mathbb{Z}$ If $f(a)=f(b)$ then $f(f(a))=f(f(b))$ then $a=f(0)-f(f(a))=f(0)-f(f(b))=b$ So $f(a)=f(b)\implies a=b$ or in other words, $f$ is an injective function. $m=0\implies f(n)=f(n+f(0))\implies n=n+f(0)\implies f(0)=0$ (because $f$ is an injective function) $n=0\implies m=f(0)-f(f(m))=-f(f(m))$, for all $m\in\mathbb{Z}$ Substitute this into the original equation: $f(n)+f(f(m))=f(n+f(m))$, for all $m, n \in\mathbb{Z}$ Replace $n$ with $f(n)$: $f(f(m))+f(f(n))=f(f(m)+f(n))$, for all $m, n \in\mathbb{Z}$ Now $f(f(m))=-m$ and $f(f(n))=-n$, so $-(m+n)=f(f(m)+f(n))$, for all $m, n \in\mathbb{Z}$ But $f(f(m+n))=-(m+n)\implies f(f(m+n))=f(f(m)+f(n))$, for all $m, n \in\mathbb{Z}$ Now $f$ is an injective function, so $f(m+n)=f(m)+f(n)$, for all $m, n \in\mathbb{Z}$ Induction can be used to show that $f(m)=mf(1)$, for all $m\in\mathbb{Z}$ So $f(f(m))=f(m)f(1)=mf(1)^2$, for all $m\in\mathbb{Z}$ So $f(f(1))=f(1)^2$ But $f(f(m))=-m$, for all $m\in\mathbb{Z}$ so $f(f(1))=-1$ So $f(1)^2=-1\implies f(1)\not\in\mathbb{Z}$ Therefore no function $f$ satisfies those conditions.
11.12.2010 12:58
Let $P(n,m)$ be assertion to the equation $ f(n)-f(n+f(m))=m $. $P(0,0) \Rightarrow f(0) = f(f(0))$, let $k=f(0)$, so $f(k)=k$ $P(n,0) \Rightarrow f(n) = f(n+f(0)) = f(n+k)\quad (*)$ $P(n,k) \Rightarrow f(n)-f(n+k) = k$, but from $(*)$ we have $k=f(n)-f(n+k)=0$, so $f(0) = 0$. $P(0,n) \Rightarrow f(f(n)) = -n$, so $f$ is a bijective function (every integer is in image of function and $f(x) = f(y) \Rightarrow f(f(x)) = f(f(y)) \Rightarrow -x=-y \Rightarrow x=y$) Let $l$ be such that $f(l) = -1$. $P(n,l) \Rightarrow f(n) - f(n-1) = l$, so $f$ is a linear function $f(n) = ln$ (because $f(0) = 0$). $f(n)-f(n+f(m))=m$ $ln-l(n+lm)=m$ $-l^2m=m$ $l^2=-1$, contraction. So such function doesn't exist.
11.12.2010 16:04
efoski1687 wrote: Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ satisfying the condition $f(n)-f(n+f(m))=m$ for all $m,n\in \mathbb{Z}$ What about with $f:R \rightarrow R $ ?
30.07.2014 21:19
efoski1687 wrote: Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ satisfying the condition $f(n)-f(n+f(m))=m$ for all $m,n\in \mathbb{Z}$ Let $P(n,m)$ be assertion of $f(n)-f(n+f(m))=m$ $(1) $ First of all we see that $f$ is injective. $P(n,0): f(n)=f(n+f(0))$ and since $f$ is injective $n=n+f(0)$ so $f(0)=0$ $P(0,n): f(f(n))=-n$ $(2)$ $P(n,f(n)): f(n)-f(m)=f(n-m)$ $(3)$ So because of $(1)$ and $(2)$: $f(f(-m))=m=f(n)-f(n+f(m))=f(-f(m))$ and since $f$ is injective $f(-m)=-f(m)$ and now inserting $-m$ instead of $m$ in $(3)$ we see $f(n)+f(m)=f(n+m)$ and since $f$ is defined on integers we have $f(n)=cn$. Inserting this in $(1)$ we see $c^2=-1$ , impossible. So no such function exist.
15.09.2015 19:56
Take $n=0$ to get $f(0)-f(f(m)) = m$, from this we easily get $f$ is bijective. Now take $n=-f(m)$ to get $f(-f(m)) +f(0)= m = f(0)-f(f(m))$ thus $f(f(m)) = -f(-f(m))$ for all $m$. Since $f$ is surjective this means $f(t)=-f(-t)$ for all integers $t$. Let $z$ be an integer with $f(z)=0$ and let $m=z$ in the original equation. This gives $f(n)-f(n) = 0= z$. Thus $f(0)=0$. Therefore $f(0)-f(f(m)) = m$ means that $f(f(m))=-m$. Therefore $f(f(-r)) = r$ for any integer $r$. If we let $m=f(-r)$ in the original equation we get: $$f(n) - f(n+r) = f(-r)$$$$f(n)-f(-r) = f(n+r)$$for all integers $n,r$. Furthermore we had already proved that $f(t)=-f(-t)$ for all integers $t$. Thus the above equation is $f(n)+f(r)=f(n+r)$ for all integers $r$. This means that $f$ is additive so the solutions must be of the form $f(n)=cn$ for some integer constant $c$. Though we know that $-n=f(f(n)) = f(cn) = c^2n$ for all integer $n$ which means $c^2=-1$, impossible for any integer $c$. Therefore there are no such functions.
06.08.2021 23:48
Injectivity is trivial. $P(0,0)\Rightarrow f(0)=0$ $P(0,x)\Rightarrow f(f(x))=-x\Rightarrow f(-x)=f(f(f(x)))=-f(x)$ $P(n,f(-m))\Rightarrow f(n+m)=f(n)+f(m)$, so $f$ is linear, no solutions exist.
07.08.2021 00:09
Let $P(n, m)$ be assertion $P(0, m)$: $$f(f(m))=f(0)-m$$so $f$ is bijective, pick $u\in \mathbb{Z}$ s.t. $f(u)=0$ $P(u, u)$: $u=0\Rightarrow f(0)=0$, so now we have $$f(f(m))=-m \Rightarrow f(m)=-f(-m)$$$P(n, f(-m))$: $$f(n+m)=f(n)-f(-m)=f(n)+f(m)$$and this is Cauchy over integers, after substituting $f(x)=cx$ into original we get $$c^2+1=0$$but since image of $f$ does not contain complex numbers - no such $f$ exists.