$\Bigl\lfloor{\sqrt{n}+1/2\Bigl\rfloor}=\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$ for $n \in \mathbb{N}$

Question

I must prove that for every $n \in \mathbb{N}$

$\Bigl\lfloor{\sqrt{n}+1/2\Bigl\rfloor}=\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$

Can you give me any idea where to start?

$\Bigl\lfloor{x\Bigl\rfloor}$ represents the floor function

I have already received help regarding the problem, in order to be more detailed, I expose the following.

It is reasoned by absurdity, that is to say, it is assumed that $\Bigl\lfloor{\sqrt{n}+1/2\Bigl\rfloor}\neq\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$ for $n \in \mathbb{N}$, then

$\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor>\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$ or $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor<\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$

if $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor<\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$ then $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor<\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}<\sqrt{n-3/4}+1/2<\sqrt{n} + \frac{1}{2}$ which cannot happen since $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor$ is the largest integer that satisfies $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor<\sqrt{n} + \frac{1}{2}$ then compulsorily $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor>\Bigl\lfloor{\sqrt{n-3/4}+1/2\Bigl\rfloor}$ then what @Scott Sobolewski mentions happens

Answer 1

Certainly, $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor$ cannot be less than $\left \lfloor{\sqrt{n - \frac{3}{4}} + \frac{1}{2}}\right \rfloor$. For the other case, assume

$\left \lfloor{\sqrt{n - \frac{3}{4}} + \frac{1}{2}}\right \rfloor$ < $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor$

Inspired by Brian Scott's answer, let $m=\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor$. We now have

$\sqrt{n - \frac{3}{4}} + \frac{1}{2} < m \leq \sqrt{n} + \frac{1}{2}$

Squaring all terms and simplifying,

$n < m^{2}-m+1 \leq n + \frac{3}{4}$

With $n$ being an integer along with $m^{2}-m+1$, this equation has no solution. This is enough to force $\left \lfloor{\sqrt{n} + \frac{1}{2}}\right \rfloor$ and $\left \lfloor{\sqrt{n - \frac{3}{4}} + \frac{1}{2}}\right \rfloor$ into equality.