Decide whether there exists a function $f : Z \rightarrow Z$ such that for each $k =0,1, ...,1996$ and for any integer $m$ the equation $f (x)+kx = m$ has at least one integral solution $x$.
Problem
Source: Czech and Slovak Match 1996 P4
Tags: number theory, Function equations, integer equation, algebra
madmathlover
02.10.2017 14:52
Almost ttrivial. If $x = 1997m + k$, where $k$ is any of $0, 1, ..., 1996$, then take $f(x) = m - kx$. Since every integer can be written in the form $1997m + k$, the function is defined for all integers and we are done.
nmd27082001
02.10.2017 16:02
madmathlover wrote: Almost ttrivial. If $x = 1997m + k$, where $k$ is any of $0, 1, ..., 1996$, then take $f(x) = m - kx$. Since every integer can be written in the form $1997m + k$, the function is defined for all integers and we are done. You use the same k so this make no sense it should be:For each $t=1,1996$,$f(1997s+t)=s-t(1997s+t)$
Valjeanpi
14.01.2023 18:48
Hello, I don't understand a step of your solution : Why do you have to specify that every number can be written like 1997m+k ?
Valjeanpi
21.01.2023 17:24
anyone ?
Russell1234
06.05.2024 18:39
Hello, I don't understand why f can depend to k and m ?