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Let $n \ge 2$ be an integer. Let $a = n+\sqrt{n^2 - n}$. Prove that for any positive integer m, we have $\lfloor a^m\rfloor \equiv -1\mod n$.

Suppose $(x-a)(x-c) = x^2 + ux + v \in \mathbb{Z}[x]$ (the ring of polynomials with integer coefficients) with $0<c<1$. To ensure that such a c exists in the first place, I think we can just pick $c$ to be $1-\{a\}$ where $\{a\} := a-\lfloor a\rfloor$. Note that this choice of c is actually unique, since there is a unique value x between 0 and 1 inclusive so that $a+x$ is an integer. Note that $n-1 < \sqrt{n^2 - n} < n$, so the fractional part of a is just $\sqrt{n^2 - n} - n+1$. Then $ac = (n+\sqrt{n^2 - n}) (n-\sqrt{n^2-n}) = n,$ which is indeed an integer. Then $\lfloor a^m \rfloor = a^m +c^m - 1$. To see why, observe that this holds for $m=1$ by the definition of c. We know from the beginning of this paragraph that both $a+c$ and $ac$ are integers, so $a^2 + c^2 = (a+c)^2 - 2ac$ is an integer. Then assuming $a^m+c^m$ and $a^{m-1}+c^{m-1}$ are both integers, we see that $a^{m+1} + c^{m+1} = (a+c)(a^m + c^m) - ac(a^{m-1} + c^{m-1})$ is an integer. Since $0<c<1$ and $a^m$ is irrational for all m, we see that $\lfloor a^m + c^m\rfloor = a^m+c^m-1$. Now it may be useful to come up with some recurrence relations and to use the fact that $a$ is irrational.

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    $\begingroup$ In the title you ask a question which is not pursued in the question body. Also that question is trivial, take any irrational b in $(0,1),$ $\endgroup$
    – coffeemath
    Commented Oct 22, 2023 at 21:36
  • $\begingroup$ Do you really mean the floor function since, if so, it is trivial as coffeemath says. $\endgroup$
    – badjohn
    Commented Oct 23, 2023 at 2:48
  • $\begingroup$ Is this question still open? $\endgroup$
    – Krave37
    Commented Feb 11 at 9:14

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