Integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n \in \mathbb{Z}$ Find integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n$ is an integer.
So far, I have just listed some squares which are larger than $2005.$ The first few are $2025,2116, 2209,2304,$ etc. I can just plug in values of $n$ which match up the square minus $2005.$ However, this seems way too bashy. Is there any good way to do this problem? Thanks in advance.
 A: If $a$ and $b$ are non-negative integers, then $\sqrt{a}+\sqrt{b}$ is an integer if and only if $a$ and $b$ are perfect squares.
So if $\sqrt{n}+\sqrt{n+2005}$ is a perfect square, $n = k^2$ and $n+2005 = m^2$ for some non-negative integers $k,m$.
Substituting $n = k^2$ into $n+2005 = m^2$ yields $k^2+2005 = m^2$, i.e. $2005 = m^2-k^2 = (m+k)(m-k)$.
Hence, $m+k$ and $m-k$ must be complimentary factors of $2005$. Since the prime factorization of $2005$ is $2005 = 5 \cdot 401$ and $m+k \ge m-k$, there are only two possibilities.
Case 1: $m+k = 2005$ and $m-k = 1$. Solve this to get $(m,k) = (1003,1002)$, i.e. $n = 1002^2$.
Case 2: $m+k = 401$ and $m-k = 5$. Solve this to get $(m,k) = (203,198)$, i.e. $n = 198^2$.
A: HINT
I would start by noticing that
\begin{align*}
\sqrt{n} + \sqrt{n+2005} = \frac{2005}{\sqrt{n + 2005} - \sqrt{n}}
\end{align*}
Now you can solve the equation $\sqrt{n + 2005} - \sqrt{n} = k$, where $k$ divides $2005$.
Can you take it from here?
A: $$\begin{align}
\sqrt n+\sqrt{n+2005}=k
&\implies \sqrt{n+2005}=k-\sqrt n\\
&\implies n+2005=k^2-2k\sqrt n+n\\
&\implies \sqrt n=\left(k^2-2005\over2k\right)={1\over2}\left(k-{2005\over k}\right)\\
&\implies k\mid2005\\
&\implies k\in\{1,5,401,2005\}
\end{align}$$
But $k\ge\sqrt{n+2005}\gt\sqrt{25}$ (since $n$ must be nonnegative in order for $\sqrt n$ to be real), so we are left with $401$ and $2005$ as the only possible integer values for $\sqrt n+\sqrt{n+2005}$ with $n\in\mathbb{Z}$.
Remark: This derivation essentially contains a proof of the fact that the sum of two square roots (of integers) is an integer if and only if the two square roots are already integers.
