This question was inspired by the following:


Is there a real nonintegral $x>1$ s.t. $\lfloor x^n \rfloor$ is square integer for all positive integers $n$? I am asking because the question is interesting in and of itself, but also because the proof techniques should also be interesting.

  • 9
    $\begingroup$ One can get "close" to having $\lfloor x^{n} \rfloor$ being a square. For example, if $x = \frac{7+3\sqrt{5}}{2}$, then $\lfloor x^{n} \rfloor + 3$ is a square for every positive integer $n$. $\endgroup$ Nov 9, 2020 at 2:14
  • $\begingroup$ @Jeremy Rouse please put this as an answer I'd like to give you credit/upvote $\endgroup$
    – Mike
    Nov 9, 2020 at 2:44

2 Answers 2


There is no such number. Suppose $\alpha>1$ is a real number such that $\lfloor \alpha^n \rfloor$ is a square for all $n\in {\Bbb N}$. Put $\beta=\sqrt{\alpha}$.

Now for each $n$ we have $$ m^2 + 1 > \alpha^n \ge m^2 $$ for some integer $m$, so that taking square-roots $$ m + \frac{1}{2m} > \beta^n \ge m. $$ In other words $\beta^n$ has exponentially small fractional part, indeed at most $1/(2m) \approx 1/(2\beta^n)$.

A theorem of Pisot now states that $\beta$ must be a Pisot--Vijayaraghavan (or PV) number (see for example the Wikipedia page on PV numbers). That is $\beta$ is an algebraic integer $>1$ such that all its Galois conjugates are $<1$ in absolute value. Suppose that $\beta$ has degree $k$ and that $\beta_1$, $\ldots$, $\beta_{k-1}$ are its Galois conjugates. Now for all $n$, we must have $$ \beta^n + \beta_1^n + \ldots +\beta_{k-1}^{n} \in {\Bbb Z}, $$ and from our assumption it follows that for large $n$ $$ |\beta_1^n + \ldots +\beta_{k-1}^n |\le \frac{3}{4\beta^n}. $$

Write each $\beta_j$ in polar coordinates as $r_j e^{2\pi i\theta_j}$. Note that $$ \beta r_1 \cdots r_{k-1} \ge 1, $$ since this is the absolute value of the norm of $\beta$. Moreover by Dirichlet's theorem we may find arbitrarily large $n$ with $\Vert n\theta_j \Vert \le 1/10^6$ for all $1\le j\le k-1$. Then, for such $n$, \begin{align*} Re(\beta_1^n + \ldots + \beta_{k-1}^n) &\ge 0.99 (r_1^n +\ldots +r_{k-1}^n) \ge 0.99 (k-1) (r_1\cdots r_{k-1})^{n/(k-1)} \\ &\ge 0.99 (k-1) \beta^{-n/(k-1)}, \end{align*} by AM-GM. That gives a contradiction.

The argument also shows that if $\alpha^n$ is within a bounded distance of a square, then $\beta$ again is a PV number, and moreover its degree $k$ must be $2$ and its norm must be $1$ in size. In other words, we must have $\beta = (r+ \sqrt{r^2 \pm 4})/2$ for some natural number $r$. This is in keeping with Jeremy Rouse's comment above.


At the request of the OP, I am turning my comment into an answer. It is possible to have $\lfloor x^{n} \rfloor$ close to a square for all positive integers $n$. For example, if $x = \frac{7 + 3 \sqrt{5}}{2} = \phi^{4}$, where $\phi = \frac{1 + \sqrt{5}}{2}$, then $\lfloor x^{n} \rfloor + 3$ is a square for every positive integer $n$. For all integers $k$, $\phi^{k} + \left(-\frac{1}{\phi}\right)^{k} = L_{k}$, the $k$th Lucas number. Squaring this identity shows that $\phi^{2k} + 2 \cdot (-1)^{k} + \left(-\frac{1}{\phi}\right)^{2k} = L_{k}^{2}$ and so $L_{k}^{2} = L_{2k} + 2 \cdot (-1)^{k}$. Thus $$ \lfloor x^{n} \rfloor + 3 = x^{n} + x^{-n} + 2 = L_{4n} + 2 = L_{2n}^{2} $$ is the square of an integer.


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