Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$

Question

$$|{\sqrt{n} - \sqrt{2011}}| < 1$$

What is the number of positive integer $n$ values, which satisfy the above inequality.

My effort:

$ ({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^2<4 \times 2011n \\n^2 -4024n +2010^2 < 0 $

But it seems this won't lead me for desired answer.

Answer 1

$\sqrt{n}\in(\sqrt{2011}-1,\sqrt{2011}+1)\iff n\in ( (\sqrt{2011}-1)^2,(\sqrt{2011}+1)^2)=(2012-2\sqrt{2011},2012+2\sqrt{2011})$

There are $2\lfloor2\sqrt{2011} \rfloor + 1$ integers in this range.

We approximate $\sqrt{2011}$ as $44.8$ and we find the answer is $2\cdot 89 + 1=179$.

Answer 2

$|\sqrt n - \sqrt{2011}| <1 \iff -1 < \sqrt n - \sqrt {2011 }< 1 \iff \sqrt{2011} - 1 <\sqrt n < \sqrt {2011} + 1 \iff (\sqrt{2011} - 1)^2 < n < (\sqrt{2011}+1)^2.$ You should be able to finish it from here.

Answer 3

Brute force says that $(\sqrt{2011}-1)^2\leq n\leq (\sqrt{2011}+1)^2$. So you can use a calculator to find the lower and upper bound integer values: $1923\leq n\leq 2101$, do there are $2101-1923+1$ integer values $n$.

Algebraically, we see that $(\sqrt{2011}-1)^2 = 2012-2\sqrt{2011}$ and $(\sqrt{2011}+1)^2=2012+2\sqrt{2011}$. So a non-calculator value is $1+2\lfloor2\sqrt{2011}\rfloor$.

These both give the value $179$.