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

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 $x \leq \sqrt \frac{k}{3},$ 
$$ \lfloor xy \sqrt d \rfloor \neq x^2 . $$
