How should I approach solving this floor function?

Question

Prove that for all $n \in \Bbb Z, \lfloor\sqrt {(n)}+ \sqrt {(n+1)} \rfloor = \lfloor \sqrt{4n+2}\rfloor$.

There must be some algebraic substitution?

Answer 1

Hint: for non-negative $n$, note that $$ (\sqrt n + \sqrt{n+1})^2 = 2n + 1 + 2\sqrt{n^2 + n} $$ and that $n^2 \leq n^2 + n < (n+1)^2$