$p_{n}$ be the probability that $C+D$ is perfect square. Compute $\lim\limits_{n \to \infty}\left(\sqrt{n} p_{n}\right)$ 
Assume $C$ and $D$ are randomly chosen from $\{1,2, \cdots, n\}$. Let $p_{n}$ be the probability that $C+D$ is perfect square. Compute $$\lim\limits _{n \to \infty}\left(\sqrt{n} p_{n}\right)$$

My Approach: My assumptions were that $C,D$ can be equal and the order of $C,D$ are not considered [I don't know I assumed correct, but in the question it was not clear.]
The number of ways a number $k$ can be partitioned into $2$ parts is equal to $\Big\lfloor\frac{k}{2}\Big\rfloor$
Now with the numbers $\{1,2, \cdots, n\}$ all the numbers from $2$ to $2 n$ can be made by choosing $2$ numbers and adding them up. So number of perfect squares in this region are $\big\lfloor\sqrt{2 n}\big\rfloor-1$ ($-1$ because we cannot make $1$). So the probability of choosing $C, D$ such that $C+D$ is a perfect square is
$$p_n=
\frac{\displaystyle{\sum_{k=2}^{\big\lfloor\sqrt{2 n}\big\rfloor} \Bigg\lfloor\frac{k^{2}}{2}\Bigg\rfloor}}{n^{2}}
$$
Now I cant progress further.
Tell me if I am correct. Also are my assumptions correct ?
Edit: The answer of the limit is $\frac{4(\sqrt{2}-1)}{3}$

We have $1^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6} \sim n^3/3.$ Now drop the floors and include $k=1$ in the sum to get $\sqrt{n} p_n \sim \sqrt{n} \cdot (\sqrt{2n})^3/(3n^2) = 2\sqrt{2}/3.$ You can go back and clean this up.
