The number of bijective functions f: Z righta | vedclass

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(A) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$. Setting $x=0, y=0$, we get $f(0) = f(0) + f(0)$, which implies $f(0) =

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(A) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$. Setting $x=0, y=0$, we get $f(0) = f(0) + f(0)$, which implies $f(0) = 0$. For any $n \in Z^+$, by induction, $f(nx) = nf(x)$. Let $f(1) = k$, where $k \in Z$. Then $f(n) = nk$ for all $n \in Z$. Since $f(x+y) = f(x) + f(y)$, it follows that $f(x) = kx$ for all $x \in Z$. For $f$ to be a bijection from $Z$ to $Z$, it must be both injective and surjective. If $f(x) = kx$, then $f$ is injective if $k 
eq 0$. For $f$ to be surjective, the range of $f$ must be $Z$. The range of $f(x) = kx$ is the set of all multiples of $k$, i.e.,$\{..., -2k, -k, 0, k, 2k, ...\}$. For this set to be equal to $Z$, we must have $k = 1$ or $k = -1$. If $k = 1$, $f(x) = x$, which is the identity function (bijective). If $k = -1$, $f(x) = -x$, which is also bijective. Thus, there are exactly two such functions.

The number of bijective functions $f: Z \rightarrow Z$ such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$ is:

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