Let $x>1$ be a real number which is not an integer. For each $n\in\mathbb{N}$, let $a_n=\lfloor x^{n+1}\rfloor - x\lfloor x^n\rfloor$. Prove that the sequence $(a_n)$ is not periodic.
Problem
Source: Japan Final MO
Tags: number theory, algebra, number theory unsolved, Sequence, algebra unsolved
Vlad-3-14
14.06.2020 13:30
Suppose it is indeed periodic. Then there exist $a,b$ such that $\lfloor x^{a+bi+1}\rfloor - x\lfloor x^{a+bi}\rfloor=\lfloor x^{a+bi+1}\rfloor x\lfloor x^{a+bi}\rfloor$ for any pair of $\{i,j\}$ Then we have that $\lfloor x^{a+bi+1}\rfloor-\lfloor x^{a+bi+1}\rfloor=x\lfloor x^{a+bi}\rfloor-x\lfloor x^{a+bi}\rfloor=x(\lfloor x^{a+bi}\rfloor-\lfloor x^{a+bi}\rfloor)$. So $x=\frac{\lfloor x^{a+bi+1}\rfloor-\lfloor x^{a+bi+1}\rfloor}{\lfloor x^{a+bi}\rfloor-\lfloor x^{a+bi}\rfloor}$ for any pair $\{i,j\}$. Let us have $f(k)=\lfloor x^{a+bk}\rfloor$ and $g(k)=\lfloor x^{a+bk+1}\rfloor$ Hence for any $\{i,j\}$, $\frac{g(i)-g(j)}{f(i)-f(j)}$ is a constant let us set $j$ and call $f(j)=f$ and $g(j)=g$. So for any $a$ and $b$ we have $\frac{f(a)-f}{g(a)-g}=\frac{f(b)-f}{g(b)-g}$. From here averything is just downhill
historypasser-by
15.06.2020 20:07
Vlad-3-14 wrote: Suppose it is indeed periodic. Then there exist $a,b$ such that $\lfloor x^{a+bi+1}\rfloor - x\lfloor x^{a+bi}\rfloor=\lfloor x^{a+bi+1}\rfloor x\lfloor x^{a+bi}\rfloor$ for any pair of $\{i,j\}$ Then we have that $\lfloor x^{a+bi+1}\rfloor-\lfloor x^{a+bi+1}\rfloor=x\lfloor x^{a+bi}\rfloor-x\lfloor x^{a+bi}\rfloor=x(\lfloor x^{a+bi}\rfloor-\lfloor x^{a+bi}\rfloor)$. So $x=\frac{\lfloor x^{a+bi+1}\rfloor-\lfloor x^{a+bi+1}\rfloor}{\lfloor x^{a+bi}\rfloor-\lfloor x^{a+bi}\rfloor}$ for any pair $\{i,j\}$. Let us have $f(k)=\lfloor x^{a+bk}\rfloor$ and $g(k)=\lfloor x^{a+bk+1}\rfloor$ Hence for any $\{i,j\}$, $\frac{g(i)-g(j)}{f(i)-f(j)}$ is a constant let us set $j$ and call $f(j)=f$ and $g(j)=g$. So for any $a$ and $b$ we have $\frac{f(a)-f}{g(a)-g}=\frac{f(b)-f}{g(b)-g}$. From here averything is just downhill could you explain the first equality? Besides, I think there may be some typos since I can find no "j"
historypasser-by
16.06.2020 19:28
Is my following solution wrong? For I think it is too short for a mo problem (though I didn't actually found the problem in japan mo list) if T is an integer s.t. a(n+T)=a(n) for any integer n, namely [x^(n+1)]-x[x^n]=[x^(n+1+T)]-x[x^(n+T)], then [x^(n+1)]-[x^(n+1+T)]=x([x^n]-[x^(n+T)]), since LHS is an integer, RHS must be 0, so for any integer n, [x^n]=[x^(n+T)], so for any integer k, [x^n]=[x^(n+kT)], which is absurd since RHS is unbounded