2020/016) find n such that $\sqrt{n} + \sqrt{n+2005}$ is a natural number

Let $\sqrt{n+2005} + \sqrt{n} =m \cdots(1)$

We know $(n+2005) - (n) = 2005$
Or $(\sqrt{n+2005})^2 - (\sqrt{n})^2 = 2005$
Or $(\sqrt{n+2005} +  \sqrt{n}) (\sqrt{n+2005} - \sqrt{n}) = 2005\cdots(2)$

Dividing (2) by (1) we get
$(\sqrt{n+2005} - \sqrt{n}) = \frac{2005}{m}\cdots(3)$

Subtracting  (3) from (1) we get $2\sqrt{n}= m - \frac{2005}{m}$

Clearly we have $m^2>=2005$ so we choose m factor of 2005 that is 401, 2005

Taking $m= 2005$ we get $n = (\frac{1}{2}(2005-1)^2 = (1002)^2 = 1004004$

Taking $m= 401$ we get $n = (\frac{1}{2}(401-5)^2 = (198)^2 = 39204$

So we have solution set $(1004004,39204)$