Find number of integral solutions of $\sqrt{n}+\sqrt{n+7259}$ 
How many integers $n$ are there such that $\sqrt{n}+\sqrt{n+7259}$ is an integer?  

No idea on this one.
 A: PERHAPS this is one approach:
Let $n=t^2$ that takes care of the first square root. Then in the second square root we get $t^2+7259$. Set this square root term equal to, say $v^2$ so that we end up with $v^2-t^2=7259$ or $(v-t)(v+t)$=$7259$. Find all factors of $7259$ (which is finite) and figure out possible values of $v$ and $t$.
If my approach is completely wrong, I will take it off.
A: Imranfat's answer is good, however you might be looking for a more rigorous argument to show that both $n$ and $7259+n$ have to be squares. Here it is:
If $\sqrt n+\sqrt{7259+n}$ is integer, then $$\sqrt n-\sqrt{7259+n}=\frac{-7259}{\sqrt n+\sqrt{7259+n}}$$ is rational.
Adding and substracting these two rationals, we get that both
$$(\sqrt n+\sqrt{7259+n})+(\sqrt n-\sqrt{7259+n})=2\sqrt n$$
and $$(\sqrt n+\sqrt{7259+n})-(\sqrt n-\sqrt{7259+n})=2\sqrt{7259+n}$$
are rationals.
This means both $n$ and $7259+n$ are perfect squares. From here, imranfat's solution is the best way to continue.
A: (This assumes $n \in \mathbb{Z}$.)
First, note that $k^2 + (2k + 1) = (k+1)^2$; that is, the difference between two consecutive sqaures is an odd number.  This implies that the difference between any two perfect squares is a sum of consecutive odd numbers.
We start by finding the largest $n$ that satisfies $\sqrt{n} + \sqrt{n+7259} \in \mathbb{Z}$. Let $n = k^2$. Now let $2k+1 = 7259$. This means $k=3629$, and so $n= 3629^2=13169641$. This must be the largest solution, as the difference between any larger consecutive squares would exceed $7259$.
Now assume that the difference between the two squares is the sum of consecutive odd numbers, 
$$\begin{aligned}
\ (2k+1) + (2k + 3) + \ldots + (2k + [2z -1]) &= 2kz + (1 + 3 + \ldots +[2z - 1]) \\
\ &= 2kz + z^2 \\
\ &= (2k + z)z \\
\end{aligned}$$
And since $(2k+z)z = 7259$, we need only check values of $z$ such that $z$ divides $7259$  i.e., $z \in \{7, 17, 61\}$. (Any product of these numbers will yield $z > \sqrt{7259}$ and therefore $k < 1$.) If $z = 7$, then $2k+7 = 17\cdot 61$, and so $k = 515 \implies n = 265225$. Letting $z = 17$ and $61$ yields $n = 42025$ and $841$, respectively.
Therefore, the four solutions to the problem are $n \in \{841, 42025, 265225, 13169641\}$.
