This sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers 
Show that: the sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers.

Maybe consider the Pell equation to solve this problem. But how to do this? Thanks
 A: Assume that $\frac{p}{q}$ is a convergent of the continued fraction of $\sqrt{2003}$. We have
$$ \left|\sqrt{2003}-\frac{p}{q}\right|\leq \frac{1}{q^2} $$
hence $n=pq$ (or $n=pq\pm 1$) is a good candidate for $\left\lfloor n\sqrt{2003}\right\rfloor$ to be a square, since
$$ \left| pq\sqrt{2003}-p^2\right|\leq \frac{p}{q}\approx\sqrt{2003}. $$
You just have to show that among the best rational approximations given by the continued fraction of $\sqrt{2003}$, there are some "very best" approximations for which
$$ \left|\sqrt{2003}-\frac{p}{q}\right|\leq \frac{C}{q^2} $$
holds with a very small constant $C$. That can be done by explicitly writing the continued fraction of $\sqrt{2003}$
$$\sqrt{2003}=[44;\overline{1,3,12,1,1,6,2,1,2,1,3,6,7,1,\color{red}{43},1,7,6,3,1,2,1,2,6,1,1,12,3,1,\color{red}{88}}]$$
and studying what happens at each step, especially when we meet a large element of such continued fraction.
A: Yes, using the Pell equation is a great idea! In this case, the best Pell equation to use is $x^2-2003y^2 = -2$ (there are no solutions to the $-1$ version of this Pell equation). Given such a solution, note that
$$
xy\sqrt{2003} - x^2 = \frac{2x}{y\sqrt{2003} + x} = \frac{2x}{\sqrt{x^2+2} + x},
$$
and therefore
$$
0 < xy\sqrt{2003} - x^2 < 1,
$$
showing that $\lfloor xy\sqrt{2003} \rfloor = x^2$.
The first solution to the Pell equation $x^2-2003y^2 = -2$ is $(x,y) = (65912269,1472739)$, and an infinite family of solutions can be generated in the usual way using the fundamental solution $(x_1,y_1) = (4344427204728362,97071569134791)$ to $x^2-2003y^2 = 1$. (And indeed, these two solutions come exactly from the convergents indicated by the red numbers in Jack D'Aurizio's answer!)
A: CW again.  Comment on the pretty idea in the other answers: Suppose we have
$$  x^2 - d y^2 = -k, $$
with small integer $k > 0.$
$$  ( x - y \sqrt d) (x + y \sqrt d) = -k, $$
$$ x - y \sqrt d = \frac{-k}{x + y \sqrt d}, $$
$$ x = y \sqrt d - \frac{k}{x + y \sqrt d}, $$
$$ x^2 = xy \sqrt d - \frac{kx}{x + y \sqrt d}, $$
$$  xy \sqrt d = x^2 + \frac{kx}{x + y \sqrt d}. $$
If $kx < x + y \sqrt d,$ then $\lfloor xy \sqrt d \rfloor = x^2.$
When is $kx < x + y \sqrt d ?$ We know $ x - y \sqrt d < 0,$ so that
$y \sqrt d > x.$ That is, 
$$ x + y \sqrt d > 2x.  $$
Therefore, if $k = 1$ or $k = 2,$ we do get $\lfloor xy \sqrt d \rfloor = x^2.$
However, as I found when doing $x^2 - 7 y^2 = -3,$ if $k \geq 3,$ this does not work. Let me prove that, need some more time...
This will do. If $x^2 - d y^2 = -k$ and the very mild $x > \sqrt \frac{k}{3},$ but  $y \sqrt d \geq 2x,$ then $d y^2 \geq 4 x^2.$ Then $x^2 - d y^2 \leq - 3 x^2.$ However, we made the mild bound $x > \sqrt \frac{k}{3},$ so that $3 x^2 > k.$ It follows that $x^2 - d y^2 < k.$ This contradicts the assumption that $y \sqrt d \geq 2x.$ So, except for a small finite number of cases with  $x \leq \sqrt \frac{k}{3},$ we have
$$  y \sqrt d < 2x,  $$
$$ x + y \sqrt d < 3x,  $$
$$ \frac{1}{x + y \sqrt d} > \frac{1}{3x}.  $$ From
$$  xy \sqrt d = x^2 + \frac{kx}{x + y \sqrt d} $$ this says
$$  xy \sqrt d > x^2 + \frac{kx}{3x}, $$
$$  xy \sqrt d > x^2 + \frac{k}{3}. $$
That is, if $k \geq 3,$ unless $x \leq \sqrt \frac{k}{3},$ 
$$ \lfloor xy \sqrt d \rfloor \neq x^2 . $$
