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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 $\{x\}=x-\lfloor x \rfloor$.

J. M. ain't a mathematician
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user72870
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    ${\sqrt{2}^2} = {\sqrt{2}^4} = {\sqrt{2}^6} = 0$ ?? – achille hui May 05 '13 at 15:12
  • Where did you get this problem from? – Karl Kroningfeld May 05 '13 at 15:20
  • It's a my conjecture. I should have said that. – user72870 May 05 '13 at 15:23
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    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 $x^{k}-x^l-m$. – Bart Michels Oct 25 '14 at 21:04
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    Nice problem! If it's a conjecture, you should write "prove or disprove that". – Capublanca Oct 28 '14 at 23:49
  • Certainly there are values of $a$ where there are no $k$'s. For example, $a=\sqrt 2$. If we had ${2^{k_1/2}}={2^{k_2/2}} \neq 0$ we would have a rational expression for $\sqrt 2$. I think you want to ask if there is any $a$ where this is true. – Ross Millikan Oct 29 '14 at 00:26
  • Not an answer, but a closely related example: the plastic constant $P$ satisfies ${P}={P^3}\neq 0$ and ${P^4}={P^5}\neq 0$. – mjqxxxx Oct 29 '14 at 06:21
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    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 $P$. – mjqxxxx Oct 29 '14 at 06:23
  • @barto: The statement "a nontrivial algebraic integer cannot be a root of two different polynomials of the form $x^k-x^l-m$" is certainly false, as $P$ is a counterexample. – mjqxxxx Oct 29 '14 at 06:31
  • Indeed, we need one polynomial's $l$ to be equal to the $k$ of the other. – Bart Michels Oct 29 '14 at 13:58
  • When polynomials $x^{k_2} −x^{k_1}−m$ and $x^{k_3} −x^{k_2}−n$ are not coprime or, at least, reducible? – Alex Ravsky Oct 30 '14 at 17:40
  • I was looking at the polynomial formulation and trying to use Euclid's algorithm, but the presence of the 'mod 1' constants makes it (un)surprisingly tricky to actually get anywhere with that tactic. – Steven Stadnicki Oct 30 '14 at 23:52
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    special case : http://math.stackexchange.com/questions/363800/proving-that-x-is-an-integer-if-the-differences-between-any-two-of-x1919 – mercio Oct 31 '14 at 15:21
  • Related question: https://math.stackexchange.com/questions/4737448/real-number-x-such-that-xn-is-constant-for-all-n-in-s – Gerry Myerson May 12 '24 at 01:05

1 Answers1

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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 $\{x^5\}=\{x^2\}=\{x\}\neq 0$.

mjqxxxx
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