<s>[INST] 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
 [/INST] 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 $x \leq \sqrt \frac{k}{3},$ 
$$ \lfloor xy \sqrt d \rfloor \neq x^2 . $$
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