there are infnitely many postive integer $n$ such $ \lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$

Question

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.

Answer 1

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.