there are infnitely many postive integer $n$ such $ \lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$ show that: there are  infnitely many postive integer $n$ such   $$ \lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$$
I think use pell equation to solve it. But I can't.

I get that
there are an infinite number of $n$
such that
$\lfloor n\sqrt{d} \rfloor
=k^2-1
$,
not $k^2+1$.
However,
for $d$ such that
there are solutions to
$x^2-dy^2 = -3$,
such as $d=7$,
then there are $n$
such that
$\lfloor n \sqrt{d} \rfloor
= k^2+1
$.
This generalizes to $k^2 \pm j$
depending on the existence
of solutions to
$x^2-dy^2 = \pm m$
for different $m$.
Here we go.
As the OP stated,
the Pell equation comes into it.
We start with the fact that
there are an infinite
number of integer solutions to
$x^2-dy^2 = 1$,
where $d$ is square free.
For each of these,
$1
=x^2-dy^2
=(x-y\sqrt{d})(x+y\sqrt{d})
$
so
$(x-y\sqrt{d})
=\dfrac1{x+y\sqrt{d}}
$
or,
squaring,
$x^2-2xy\sqrt{d}+dy^2
=\dfrac1{(x+y\sqrt{d})^2}
$
or
$2xy\sqrt{d}
=x^2+dy^2-\dfrac1{(x+y\sqrt{d})^2}
=x^2+(x^2-1)-\dfrac1{(x+y\sqrt{d})^2}
=2x^2-1-\dfrac1{(x+y\sqrt{d})^2}
$
or
$xy\sqrt{d}
=x^2-\frac12(1+\dfrac1{(x+y\sqrt{d})^2})
$.
Since
$0 < \dfrac1{(x+y\sqrt{d})^2})
< \frac12$,
$\frac12
< \frac12(1+\dfrac1{(x+y\sqrt{d})^2})
< 1
$
so
$\lfloor xy\sqrt{d} \rfloor
=\lfloor x^2-\frac12(1+\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2-\lfloor\frac12(1+\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2-1
$.
This is not what is asked.
However,
if there is one solution to
$x^2-dy^2 = -1$,
then there are
an infinite number of solutions.
Modifying this calculation
we get
$x^2-2xy\sqrt{d}+dy^2
=\dfrac1{(x+y\sqrt{d})^2}
$
or
$2xy\sqrt{d}
=x^2+dy^2-\dfrac1{(x+y\sqrt{d})^2}
=x^2+(x^2+1)-\dfrac1{(x+y\sqrt{d})^2}
=2x^2+1-\dfrac1{(x+y\sqrt{d})^2}
$
or
$xy\sqrt{d}
=x^2+\frac12(1-\dfrac1{(x+y\sqrt{d})^2})
$
$\lfloor xy\sqrt{d} \rfloor
=\lfloor x^2+\frac12(1-\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2+\lfloor\frac12(1-\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2
$.
However,
there are no solutions to
$x^2-7y^2 = -1$,
so this does not hold.
However,
suppose there are
an infinite number of solutions to
$x^2-dy^2 = -m$.
Modifying this calculation
we get
$-m
=x^2-dy^2
=(x-y\sqrt{d})(x+y\sqrt{d})
$
or
$x-y\sqrt{d}
=\dfrac{-m}{x+y\sqrt{d}}
$.
Squaring,
$x^2-2xy\sqrt{d}+dy^2
=\dfrac{m^2}{(x+y\sqrt{d})^2}
$
or
$2xy\sqrt{d}
=x^2+dy^2-\dfrac{m^2}{(x+y\sqrt{d})^2}
=x^2+(x^2+m)-\dfrac{m^2}{(x+y\sqrt{d})^2}
=2x^2+m-\dfrac{m^2}{(x+y\sqrt{d})^2}
$
or
$xy\sqrt{d}
=x^2+\frac12(m-\dfrac{m^2}{(x+y\sqrt{d})^2})
$
so
$\lfloor xy\sqrt{d} \rfloor
=\lfloor x^2+\frac12(m-\dfrac{m^2}{(x+y\sqrt{d})^2}) \rfloor
= x^2+\lfloor\frac12(m-\dfrac{m^2}{(x+y\sqrt{d})^2}) \rfloor
$.
If $m$ is odd,
$m = 2j+1$,
then
$\lfloor xy\sqrt{d} \rfloor
= x^2+\lfloor\frac12(2j+1-\dfrac{m^2}{(x+y\sqrt{d})^2}) \rfloor
=x^2+j
$
once
$x+y\sqrt{d}
> m
$.
Since there are solutions to
$x^2-7y^2 = -3$
(e.g.,
$5^2-7\cdot 2^2 = -3$)
there are an infinite number of solutions,
so OP's statement is true.
