Find all $f(x),g(x),h(x)$ be three nonconstant polynomials such:for any postive real number $x$,such $$\{f(x)\}+\{g(x)\}=\{h(x)\}\tag{1}$$
where $\{x\}=x-[x]$.
It seem may use Kronecker theorem: the sequence $(\{na\})$ is dense in $[0,1]$ if $a$ is irrational numbers?But I can't How to solve the problem (1),Thanks
Since $\{f(x)\}=f(x)-k(x), k(x) \in \mathbb Z$ we have $f(x)+g(x)-h(x)=m(x), m(x) \in \mathbb Z$ so by continuity $m(x)=m$ constant and we clearly can replace $h(x)$ with $h(x)+m$ keeping the original identity so $f(x)+g(x)=h(x), [f(x)]+[g(x)]=[h(x)]$ for all $x$ positive, while obviously then $f(x)+g(x)=h(x)$ holds for all $x$ real too.
But now $f',g', h' \ne 0$ so they all have a constant sign from some large $x_0>0$ on, hence $f,g,h$ are strictly monotonic from $x_0$ on.
Since $h(x)$ integer for $x>0$ implies $f(x), g(x)$ integers too as $\{f(x)\} \ge 0$ it follows that at a point $x+\epsilon>x_0$ where $h(x)$ hence $f(x),g(x)$ are integral and $\epsilon >0$ small enough so there are no other integral values of $f,g,h$ in $(x-\epsilon,x +\epsilon)$, $[h(x+\epsilon)]$ jumps by $1$ but also $[f(x+\epsilon)]$ and $[g(x+\epsilon)]$ jump by $1$ (up or down depending if the polynomials increase or decrease there of course). But then the parity of $[f(x+\epsilon)]+[g(x+\epsilon)]$ stays the same as the parity of $[f(x-\epsilon)]+[g(x-\epsilon)]$ while the parity of $[h(x+\epsilon)]$ changes vs the parity of $[h(x-\epsilon)]$ and that contradicts $[f(x\pm \epsilon)]+[g(x \pm \epsilon)]=[h(x \pm \epsilon)]$, so there are no such polynomials.