Prove that: ⌊√(n)+√(n+1)⌋=⌊√(4n+2)⌋, for all positive integer n.

Prove that: $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor= \left\lfloor{\sqrt{4n+2}}\right\rfloor$, for all $n\in N$.

My solution:

Step 1:

Note that:

$4n^2+4n\lt 4n^2+4n+1\lt 4n^2+8n+4$

$4n(n+1)\lt (2n+2)^2$

$2\sqrt{n(n+1)}\lt 2n+2$

$2n+1+2\sqrt{n(n+1)}\lt 2n+1+2n+2$

$n+n+1+2\sqrt{n(n+1)}\lt 4n+3$

$(\sqrt{n}+\sqrt{n+1})^2\lt 4n+3$

Step 2:

$n\lt n(n+1)$

$2n\lt 2n(n+1)$

$n+n+1+2n\lt n+n+1+2n(n+1)$

$4n+1\lt (\sqrt{n}+\sqrt{n+1})^2$

Step 3:

Combining both steps from 1 and 2 we have:

$4n+1\lt (\sqrt{n}+\sqrt{n+1})^2\lt 4n+3$

Step 4:

This is the most important progress that could let us finish the problem beautifully.

Note that $4n+1$ and $4n+3$ are neither square at the same time, thus we have $\left\lfloor{4n+1}\right\rfloor=\left\lfloor{4n+2}\right\rfloor=\left\lfloor{4n+3}\right\rfloor$ and therefore, we have proved that:

$\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor= \left\lfloor{\sqrt{4n+2}}\right\rfloor$, for all $n\in N$.