We have a natural number $n\geq4$ and the set A= {$a_{1},a_{2},...,a_{n}$} with natural numbers. We know $a_{k}+k\sqrt{a_{k+1}} \in A $ for every $k \in \{{1,2,...,n-2}\}$. We also know that $a_{1}<a_{2}<...<a_{n}$. Knowing that $\sum_{k=1}^{n}a_{k}$ is a perfect square, find $A$.
So we have $a_{k}+k\sqrt{a_{k+1}}=a_{q}$. We can easily see that $q>k$ so $a_{q}>a_{k}$ for a fixed k. Also we find out that $k\sqrt{a_{k+1}}$ is a natural number so $a_{k+1}$ is a perfect square for every $k \in \{1,2,3,...,n-2 \}$. From here we can say $a_{2},a_{3},...,a_{n-1}$ are perfect squares. Another interesting thing is the fact that $a_{n-2}+(n-2)\sqrt{a_{n-1}} \in \{a_{n-1},a_{n}\}$. Let's say $a_{k}=p_{k}^{2}$, for $k \in \{2,3,...,n-1\}$, and $p_{k}$ a natural number. Obviously if $i>j$, then $p_{i}>p_{j}$. Let's say $\sum_{k=1}^{n}a_{k}=P^{2}$, where $P$ is a natural number. We have $a_{1}+a_{n}+\sum_{k=2}^{n-1}p_{k}^{2}=P$. This is where I got stuck. Can someone give me an idea/hint?
Due to the $a_i$ sequence being strictly increasing, as you already noted, we have
$$a_{k}+k\sqrt{a_{k+1}}=a_{q} \tag{1}\label{eq1A}$$
with $q \gt k$. Note that, as $k$ increases, $q - k$ is a non-decreasing function. Thus, since $q \le n$ at $k = n-2$, then $q-k$ is always either $1$ or $2$. For $k = 1$, consider if
$$a_{1}+\sqrt{a_{2}} = a_{3} \tag{2}\label{eq2A}$$
Since $a_{2}$ and $a_{3}$ are both perfect squares, and $a_{3} \gt a_{2}$, we then have
$$\sqrt{a_{3}} = \sqrt{a_{2}} + j_{2}, \;\; j_2 \ge 1 \tag{3}\label{eq3A}$$
Substituting this into \eqref{eq2A} gives
$$\begin{equation}\begin{aligned} a_{1}+\sqrt{a_{2}} & = (\sqrt{a_{2}} + j_2)^2 \\ a_{1}+\sqrt{a_{2}} & = a_{2} + 2j_2\sqrt{a_{2}} + j_2^2 \\ a_{1} & = a_{2} + (2j_2 - 1)\sqrt{a_{2}} + j_2^2 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, since $(2j_2 - 1)\sqrt{a_{2}} \gt 0$ and $j_2^2 \gt 0$, then \eqref{eq4A} indicates that $a_{1} \gt a_{2}$, contradicting the requirement $a_{2} \gt a_{1}$. Thus, we must instead have that
$$a_1 + \sqrt{a_2} = a_2 \tag{5}\label{eq5A}$$
Next, consider if
$$\begin{equation}\begin{aligned} a_2 + 2\sqrt{a_3} & = a_3 \\ a_2 + 1 & = a_3 - 2\sqrt{a_3} + 1 \\ a_2 + 1 & = (\sqrt{a_3} - 1)^2 \end{aligned}\end{equation}\tag{6}\label{eq6A}$$
However, $0$ is the only perfect square that, when $1$ is added to it, gives another perfect square. Since $a_2 \gt 0$, this is not possible. Thus, we must instead have
$$a_2 + 2\sqrt{a_3} = a_4 \tag{7}\label{eq7A}$$
If $n \gt 4$, then $a_4$ must be a perfect square so, similar to \eqref{eq3A}, we get
$$\sqrt{a_4} = \sqrt{a_3} + j_3, \;\; j_3 \ge 1 \tag{8}\label{eq8A}$$
Using this in \eqref{eq7A} results in
$$\begin{equation}\begin{aligned} a_2 + 2\sqrt{a_3} & = (\sqrt{a_3} + j_3)^2 \\ a_2 + 2\sqrt{a_3} & = a_3 + 2j_3\sqrt{a_3} + j_3^2 \\ a_2 & = a_3 + 2(j_3-1)\sqrt{a_3} + j_3^2 \end{aligned}\end{equation}\tag{9}\label{eq9A}$$
Since $2(j_3-1)\sqrt{a_3} \ge 0$ and $j_3^2 \gt 0$, this gives $a_2 \gt a_3$, which is not allowed. Thus, $a_4$ can't be a perfect square, so $n = 4$. Next, we have
$$a_4 = a_3 + j_4, \;\; j_4 \ge 1 \tag{10}\label{eq10A}$$
Using this in \eqref{eq7A} gives that
$$\begin{equation}\begin{aligned} a_2 & = (a_3 + j_4) - 2\sqrt{a_3} \\ & = (a_3 - 2\sqrt{a_3} + 1) + j_4 - 1 \\ & = (\sqrt{a_3} - 1)^2 + j_4 - 1 \end{aligned}\end{equation}\tag{11}\label{eq11A}$$
Since $\sqrt{a_3} - 1 \ge \sqrt{a_2} \;\;\to\;\; (\sqrt{a_3} - 1)^2 \ge a_2$ and $j_4 - 1 \ge 0$, we must have
$$\sqrt{a_3} = \sqrt{a_2} + 1, \;\; j_4 = 1 \tag{12}\label{eq12A}$$
With $\sum_{k=1}^{n}a_k=m^2$ for some integer $m$, and letting $\sqrt{a_2} = r$ for simpler algebra, then using \eqref{eq5A}, \eqref{eq10A} and \eqref{eq12A} gives
$$\begin{equation}\begin{aligned} m^2 & = a_1 + a_2 + a_3 + a_4 \\ & = (a_2 - \sqrt{a_2}) + a_2 + (\sqrt{a_2} + 1)^2 + ((\sqrt{a_2} + 1)^2 + 1) \\ & = (r^2 - r) + r^2 + (r + 1)^2 + ((r + 1)^2 + 1) \\ & = 4r^2 + 3r + 3 \end{aligned}\end{equation}\tag{13}\label{eq13A}$$
Since for $r \gt 0$ we have $(2r)^2$ being too small and $(2r+2)^2 = 4r^2 + 8r + 4$ being too large, the only possibility is $m = 2r+1 \;\;\to\;\; m^2 = 4r^2 + 4r + 1$, so
$$4r^2 + 4r + 1 = 4r^2 + 3r + 3 \;\;\to\;\; r = 2 \tag{14}\label{eq14A}$$
Thus, $a_1 = 2^2 - 2 = 2$, $a_2 = 2^2 = 4$, $a_3 = (2 + 1)^2 = 9$ and $a_4 = 9 + 1 = 10$, giving that $A = \{2, 4, 9, 10\}$.