Let $a,b,c>0$ satisfy for all integers $n$, we have $$\lfloor an\rfloor+\lfloor bn\rfloor=\lfloor cn\rfloor$$Prove that at least one of $a,b,c$ is an integer.
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Tags: algebra, floor function, function
EmilXM
23.10.2020 16:02
$$\lfloor an\rfloor+\lfloor bn\rfloor=\lfloor cn\rfloor \Longrightarrow an-\{an\} + bn-\{bn\} = cn-\{cn\} \Longrightarrow n(a+b-c) = \{an\}+\{bn\}-
\{cn\}$$If $a+b-c > 0$ we can take sufficiently large $n$ such that $\{an\}+\{bn\}-\{cn\} = n(a+b-c) > 2 > \{an\}+\{bn\}-\{cn\}$. Similarly if $a+b-c<0$ we can take $n$ such that $\{an\}+\{bn\}-\{cn\} = n(a+b-c) < -1 < \{an\}+\{bn\}-\{cn\}$. So $a+b-c=0$. Plug in this gives us $$\{an\}+\{bn\}-\{an+bn\}=0 \iff \{an\}+\{bn\}=\{an+bn\}$$Now let $\{a\}=a_0 \neq 0$ and $\{b\} = b_0 \neq 0$.
$$n=1 \Longrightarrow a_0+b_0 = {a_0+b_0} \Longrightarrow a_0+b_0 < 1$$$$n=-1 \Longrightarrow 1-a_0+1-b_0 = 1-a_0-b_0 \Longrightarrow 0=1$$a contradiction. So one of $a_0$ or $b_0$ must be $0$ which means one of $a$ or $b$ must be an integer.
Isarogu777
28.12.2020 03:23
Let $[an]+[bn]=[cn]$ $(*)$. By contradiction, we assume that $a$, $b$, $c$ $\notin Z$ From $(*)$ and $n=-1$ we get $$[-a]+[-b]=[-c] (**)$$. From $(*)$ and $n=1$ we get $$[a]+[b]=[c] (***)$$Adding $(**)$ and $(***)$ we conclude $$([a]+[-a])+([b]+[-b])=[c]+[-c] (****)$$Now, since that $a,b,c \notin Z$ it's easy to proof that $[x]+[-x]=-1$. So we have in $(****)$ that $-2=-1$ we get a contradiction.
Maths_1729
28.12.2020 12:24
$[an]+[bn]=[cn] \implies [a]n +[b]n+[\{a\}n]+[\{b\}n]=[c]n+[\{c\}n]$ now if $n=1$ then $[a]+[b]=[c]$ if $n=-1$ we have $[a]+[b]+1=[c]$ which is clearly not possible. At least one of $\{a\}, \{b\}$ is $0$.