Solving an equation involving the integer part

Question

Let $[x]$ denote the integer part of $x\in R$. Let consider the following equation:

$$ [\sqrt {x+1} \,]=[\sqrt{x}],\; x\in N. \;\; (Eq). $$ Using the definition of the integer part, one can show that elements $x\in N$ such that: $x\equiv 1 $ mod $4$ or $x\equiv 2 $ mod $ 4 $ are all solutions of (Eq).

My question is as follows: among integers $x$ such that $x\equiv 0 $ mod $4$, which ones are solutions of (Eq)? Any help is welcome. Thanks

Answer 1

If $x+1$ is a square, then $[\sqrt{x+1}]=[\sqrt{x}]+1\neq[\sqrt{x}]$

Conversely, $[\sqrt{x+1}]\neq[\sqrt{x}]$ and $x\in \mathbb{N}$ implies $x+1$ is a square.

So the solutions of the equation are $\mathbb{N}-\{n^2-1, n\in \mathbb{N}\}=\{1,2,4,5,6,7,9,10...\}$

Answer 2

The sequence of $\lfloor\sqrt n\rfloor$ has a unit jump at every perfect square, and so has the sequence of $\lfloor\sqrt{n+1}\rfloor$, shifted by one.

$$0,1,1,1,2,2,2,2,2,3,3,3,\cdots$$ vs

$$1,1,1,2,2,2,2,2,3,3,3,3\cdots$$