$\lfloor 1/2+\sqrt{n+1/2}\rfloor =\lfloor 1/2+\sqrt{n+1/2020}\rfloor$

Question

For every $n$ positive odd number show that $\lfloor 1/2+\sqrt{n+1/2}\rfloor=\lfloor 1/2+\sqrt{n+1/2020}\rfloor$

Here is my second solution: Let's first consider the expression $\sqrt{n+1/2}$ and $\sqrt{n+1/2020}$ for positive odd integers $n$.

For any positive odd integer $n$, $n$ can be written as $n = 2k + 1$ for some non-negative integer $k$. Substituting this value into the expressions, we get:

$\sqrt{n + 1/2} = \sqrt{(2k + 1) + 1/2} = \sqrt{2k + 3/2}$ $\sqrt{n + 1/2020} = \sqrt{(2k + 1) + 1/2020} = \sqrt{2k + 2021/2020}$

Now, let's consider the fractional parts of both expressions

For $\sqrt{2k + 3/2}$, the fractional part is between 0 and 1, and it does not exceed 1/2 since $\sqrt{2k + 3/2} - k > 1/2$.

For $\sqrt{2k + 2021/2020}$, the fractional part is also between 0 and 1, and it does not exceed 1/2 since $\sqrt{2k + 2021/2020} - k < 1/2$.

Since the fractional parts of both expressions are less than or equal to 1/2, when we add 1/2 to them and take the floor function, they will produce the same integer value. Therefore, we have shown that $\lfloor 1/2+\sqrt{n+1/2}\rfloor=\lfloor 1/2+\sqrt{n+1/2020}\rfloor$ for every positive odd integer $n$.

İs there any mistake? I will be glad if you report my mistakes.

Answer 1

Let us consider a case where equality is not true for any (not necessarly odd) integer value of $n$. Then, there exists an integer $m$ such that $1/2 + \sqrt{n+1/2} \geq m$ and $1/2 + \sqrt{n+1/2020} < m$. This transforms into $n \geq m^2-m-1/4$ and $n<m^2-m+126/505$. The only integer value between both bounds is $m^2-m$ so we must have $n=m^2-m$. With a little bit more work, one can show that any number $n$ of this form does not satisfy the equality. Now, $m$ and $m^2$ have same parity so the exceptions $m^2-m$ are all even.

This proves that equality is satisfied by all odd integers (and all even integers not of the form $n=m^2-m$).

Answer 2

What about my second attempt? Is it true?

Let $\lfloor{ \frac{1}{2}+\sqrt{n+\frac{1}{2}}}\rfloor = k$ $$k+1 >\frac{1}{2}+\sqrt{n+\frac{1}{2}} >k$$ $k^2+k-1 \geq n \geq k^2-k+1$

We have to show $$\lfloor{ \frac{1}{2}+\sqrt{n+\frac{1}{2020}}}\rfloor = k$$ $\frac{1}{2}+\sqrt{n+\frac{1}{2020}} < k+1$

It is obvious $\sqrt{n+\frac{1}{2020}}<\sqrt{n+\frac{1}{2}} < k+\frac{1}{2}$

We assume that $$\frac{1}{2}+\sqrt{n+\frac{1}{2020}}<k$$ $k^2-k+1+\frac{1}{2020}<n+\frac{1}{2020}<(k-\frac{1}{2})^2$ must be conradiction. So $$\lfloor{ \frac{1}{2}+\sqrt{n+\frac{1}{2020}}}\rfloor = k$$