If [√(n^2 +λ)=[√(n^2 +1)]+2, where λ, n∈N then λ can assume

We have,

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

$∴[\sqrt{n^2 +λ}=[\sqrt{n^2 +1}]+2$

$⇒[\sqrt{n^2 +λ}]=n+2$

$⇒n+2≤\sqrt{n^2 +λ} <n+3$

$⇒(n+ 2)^2 ≤n^2 +λ<(n+3)^2$

$⇒4n+4≤λ<6n+9$

$⇒λ=4n+4, 4n+5,...., 6n+ 8$.

Hence, λ can assume $(2n+5)$ distinct values.