Let $a\in\mathbb{R}\setminus\mathbb{Z}$. Prove or disprove that there do not exist three distinct $k_1, k_2, k_3\in\mathbb{N}$ such that $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}\neq 0$, where $\{x\}=x-\lfloor x \rfloor$.

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    $\begingroup$ $\{\sqrt{2}^2\} = \{\sqrt{2}^4\} = \{\sqrt{2}^6\} = 0$ ?? $\endgroup$ – achille hui May 5 '13 at 15:12
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    $\begingroup$ Perhaps a more useful reformulation: If $a\in\mathbb R$ is a solution of $x^{k_2}-x^{k_1}=m$ and $x^{k_3}-x^{k_2}=n$ for positive integers $k_1<k_2<k_3$ and $m,n$, then $a$ is integer. I.e. a nontrivial algebraic integer cannot be a root of two different polynomials of the form $x^{k}-x^l-m$. $\endgroup$ – Bart Michels Oct 25 '14 at 21:04
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    $\begingroup$ Nice problem! If it's a conjecture, you should write "prove or disprove that". $\endgroup$ – Capublanca Oct 28 '14 at 23:49
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    $\begingroup$ And of course if you allow $k \in \mathbb{Z}$ instead, you have $\{\phi^{-1}\}=\{\phi\}=\{\phi^2\}\neq 0$ and a related triplet for $P$. $\endgroup$ – mjqxxxx Oct 29 '14 at 6:23
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    $\begingroup$ special case : math.stackexchange.com/questions/363800/… $\endgroup$ – mercio Oct 31 '14 at 15:21

This is not a complete answer, but I thought the approach might help someone find a solution. We can ask when the following condition holds for positive integers $a>b>c$ and $a>d\ge c$: $$ x^a - x^d - m = P(x)(x^b - x^c - n), $$ where $P(x)$ is some polynomial over $x$, and $m,n\in\mathbb{Z}$. If it does, then $\{x^a\}=\{x^d\}$ and $\{x^b\}=\{x^c\}$ whenever $x$ is a root of $x^b-x^c-n$. If, moreover, $d=b$ and the right-hand trinomial has a root in $\mathbb{R}$ for which $x^a\not\in\mathbb{Z}$, then OP's question has a positive answer.

Clearly $P(x)$ must start with $x^{a-b}$; so $$ x^a-x^d-m=(x^b - x^c - n)(x^{a-b}+Q(x))=x^a-x^{a-b+c}-nx^{a-b}+Q(x)(x^b-x^c-n), $$ or $$ Q(x)(x^b-x^c-n)=x^{a-b+c}+nx^{a-b}-x^d-m. $$ The constant term on the left and right must be $-m$, so $Q(x)=m/n+xR(x)$, giving $$ (xR(x)+m/n)(x^b-x^c-n)=xR(x)(x^{b}-x^{c}-n)+(m/n)x^b-(m/n)x^c-m \\ =x^{a-b+c}+nx^{a-b}-x^d-m, $$ or $$ R(x)(x^b-x^c-n)=x^{a-b+c-1}+nx^{a-b-1}-x^{d-1}-(m/n)x^{b-1}+(m/n)x^{c-1}. $$ We must have $c=1$ at this point, so $$ R(x)(x^b-x-n)=x^{a-b}+nx^{a-b-1}-x^{d-1}-(m/n)x^{b-1}+(m/n). $$ Now, for the highest-order terms to match, we need either $a-b \ge b$ or $d=b+1$. The latter case does not help with OP's question, but setting $d=b+1$ gives $R(x)=-1$ and $$ -x-n=-x^{a-b}-nx^{a-b-1}+(m/n)x^{b-1}-(m/n), $$ so $m=n=1$ and $(a,b,c,d)=(5,3,1,4)$. This gives the known but nontrivial result that $x^5-x^4-1=P(x)(x^3-x-1)$; and $x^3-x-1$ has as a real root the plastic constant $P=1.3247…$. We conclude that $\{P^5\}=\{P^4\}$ and $\{P^3\}=\{P\}$, which is interesting but not exactly what we want.

If we instead set $a=2b+k$ for $k\ge 0$, then $$ R(x)(x^b-x-n)=x^{b+k}+nx^{b+k-1}-x^{d-1}-(m/n)x^{b-1}+(m/n). $$ Restricting ourselves to $d=b$, $$ R(x)(x^b-x-n)=x^{b+k}+nx^{b+k-1}-\left(\frac{m}{n}+1\right)x^{b-1}+\left(\frac{m}{n}\right). $$ It's then possible to try various values for $k$. Interestingly, if $k=1$, then the solution $(a,b,c,d)=(5,2,1,2)$, $m=18$, $n=-3$ arises. This corresponds to $$ x^5-x^2-18 = P(x)(x^2-x+3), $$ which is very close to a positive result. Unfortunately $x^2-x+3$ has only complex roots: $r=1/2 + i\sqrt{11}/2$ and its complex conjugate. But indeed $$ r^5 = 31/2 + i\sqrt{11}/2, \\ r^2 = -5/2 + i\sqrt{11}/2, \\ r = 1/2 + i\sqrt{11}/2, $$ so this can be seen as a complex solution to $\{x^5\}=\{x^2\}=\{x\}\neq 0$.

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    $\begingroup$ Just noticed that this answer was the 1,000,000th post on MSE. :-) $\endgroup$ – MarnixKlooster ReinstateMonica Dec 29 '14 at 8:39
  • $\begingroup$ I also just noticed this @Marnix. Very good post for the milestone! $\endgroup$ – user142198 Mar 24 '15 at 1:41

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