find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(-f(x)-f(y)) = 1-x-y$ $\quad \forall x,y \in \mathbb{Z}$
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Tags: functional equation, function, algebra
Math-wiz
15.08.2019 14:30
dominicleejun wrote: find all functions $f:\mathbb{Z}\to\mathbb{Z}$ such that $f(-f(x)-f(y)) = 1-x-y$ $\text{\LaTeX}ed$
NikoIsLife
15.08.2019 14:52
The only solution is $\boxed{f(x)=x-1}$ for all integers $x$.
Proof: Denote $P(x,y)$ as the assertion that $f(-f(x)-f(y))=1-x-y$.
From $P(-f(x)-f(0),-f(3)-f(0))$, we have:
$$f(-f(-f(x)-f(0))-f(-f(3)-f(0))=1-(-f(x)-f(0))-(-f(3)-f(0))$$$$f(-(1-x-0)-(1-3-0))=1+f(x)+f(0)+f(3)+f(0)$$$$f(x+1)=f(x)+2f(0)+f(3)+1$$Since $2f(0)+f(3)+1$ is constant, let $2f(0)+f(3)+1=c_1$ for some integer $c_1$. Finally, we get:
$$f(x+1)=f(x)+c_1$$This implies that $f$ is a linear function. Therefore, $f(x)=c_1x+c_2$ for some integers $c_1$ and $c_2$.
By substituting this back to our equation, we find that the only possible solution is $f(x)=x-1$.
DVDthe1st
16.08.2019 00:37
Posted before.
lazizbek42
23.12.2021 21:36
dutch imo tst
jasperE3
30.03.2022 17:19
Let $g(x)=-f(x)$. The equation is: $$P(x,y):g(g(x)+g(y))=x+y-1.$$$P(x,0)\Rightarrow g(g(x)+g(0))=x-1\Rightarrow g$ is bijective. $P(x+y,0)\Rightarrow g(g(x+y)+g(0))=x+y-1=g(g(x)+g(y))\Rightarrow g(x+y)+g(0)=g(x)+g(y)$ Let $h(x)=g(x)-g(0)$, then $h$ is additive so linear. This means that $g$ is linear, so $f$ is linear. Testing, the only solution is $\boxed{f(x)=x-1}$.