How many bijections f: Z rightarrow Z are the | vedclass

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(B) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$. This is Cauchy's functional equation on the set of integers $Z$. The ge

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Two

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(B) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$. This is Cauchy's functional equation on the set of integers $Z$. The general solution is $f(x) = kx$ for some constant $k \in Z$. For $f$ to be a bijection,it must be both injective and surjective. If $f(x) = kx$,then $f$ is injective if $kx_1 = kx_2 \implies x_1 = x_2$,which holds if $k 
eq 0$. For $f$ to be surjective,for any $y \in Z$,there must exist $x \in Z$ such that $f(x) = y$,i.e.,$kx = y$. This implies $x = y/k$. For $x$ to be an integer for every $y \in Z$,$k$ must be a divisor of $1$. Thus,$k = 1$ or $k = -1$. Therefore,the possible functions are $f(x) = x$ and $f(x) = -x$. There are exactly $2$ such bijections.

How many bijections $f: Z \rightarrow Z$ are there such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$?

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