Question

Let $\left[\sqrt{n^2+1}\right]=\left[\sqrt{n^2+\lambda }\right]$, where $n,\lambda \in N$. Then $\lambda$ can have $y\times n$ different values. Find $y$
(Note : $[.]$ represents the greatest integer function.)

Answer 1

We have,

$n^2+1=(n+1{)}^2-2n<(n+1{)}^2;n\in N$
ie,

$\sqrt{n^2+1}<n+1$ or $n<\sqrt{n^2+1}<n+1$

$\Rightarrow \left[\sqrt{n^2+1}\right]=n$

$\therefore \left[\sqrt{n^2+\lambda }\right]=n$

$\Rightarrow n<\sqrt{n^2+\lambda }<(n+1)$
or $n^2<(n^2+\lambda )<(n+1{)}^2$
$\Rightarrow 0<\lambda <2n+1$

Thus, $\lambda$ can take $2n$ different values.

Hence $y=2$