Prove $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{n}+\sqrt{n+2}]$ for all integers $n\ge 1$ This is the full question

Prove that of the two equations $$ [\sqrt{n}+\sqrt{n+1}] = [\sqrt{n}+\sqrt{n+2}] \\ [\sqrt[3]{n}+\sqrt[3]{n+1}] = [\sqrt[3]{n}+\sqrt[3]{n+2}]$$ the first one holds for every positive integer $n$, but the second one does not. Note: $[x]$ denotes the greatest integer function/floor function of $x$.

For the second equation, using python I was able to find two values of $n$ between $1$ and $100$ for which the equation is false. They were $n=15$ and $n=42$. Is there a way to actually calculate these values directly from the equation. Also, I don't have any ideas for proving the first equation. Any hints? 
 A: Assume that there is an integer $m$ such that
$$\sqrt{n}+\sqrt{n+1}<m<\sqrt{n}+\sqrt{n+2}.$$
Squaring quickly proves that 
$$\lceil(\sqrt{n}+\sqrt{n+1})^2\rceil=4n+2$$
and
$$
\lceil(\sqrt{n}+\sqrt{n+2})^2\rceil=4n+4.
$$
These imply that $m^2$ must be equal to either $4n+2$ or $4n+3$, both of which are impossible by considering quadratic residues modulo $4$.

It seems to me that for all $n\ge2$ we have 
$$\lfloor(\root3\of{n}+\root3\of{n+1})^3\rfloor=8n+3$$ and
$$\lfloor(\root3\of{n}+\root3\of{n+2})^3\rfloor=8n+7.$$ Whenever there is a perfect cube in between (allowing equality at the upper limit), there will be a failure with the cube roots. When $n=15$ we have $8n+3=123<5^3<127=8n+7$. When $n=42$ we have
$8n+3=339<7^3=8n+7$. 
The cubes of even numbers are divisible by eight, and won't produce more examples. On the other hand for odd numbers $m$ we have $m^3\equiv m\pmod8$. Therefore the cubes of integers $m\equiv5,7\pmod8$ do give more examples. Indeed, with $m=13$ we have $m^3=8\cdot274+5$, and we get the next counterexample when $n=274$. The next counterexample comes at $n=421$ as
$8\cdot421+3<15^3=8\cdot421+7.$
