Not the solution but Induction may work here.

$\sqrt{a_{n+1}}=\sqrt{a(a_n+1)}+\sqrt{(a+1)a_n}$

amparvardi wrote: Let $a$ be a positive integer and let $\{a_n\}$ be defined by $a_0 = 0$ and \[a_{n+1 }= (a_n + 1)a + (a + 1)a_n + 2 \sqrt{a(a + 1)a_n(a_n + 1)} \qquad (n = 1, 2 ,\dots ).\] Show that for each positive integer $n$, $a_n$ is a positive integer. Lemma : If $\cosh(u)$ is a positive odd integer, then $\cosh(nu)$ is a positive odd integer $\forall n\in\mathbb N$.

Let then the sequence $b_n=2a_n+1$ It's easy to check that $b_n=\cosh(nu)$ for some $u$ such that $\cosh(u)=2a+1$

Using lemma, we get that $b_n$ is a positive odd integer, and so $a_n=\frac{b_n-1}2$ is a positive integer (obviously $b_n>1$) Q.E.D.

See here http://www.artofproblemsolving.com/community/c6h1310224p7020406 also.

Solved with Michelle Liang, along with emotional support from Aiden Wen, Channing Yang, and Ming Yang. Quote: Let $k$ be a positive integer. The sequence $(a_n)_{n \geq 0}$ is defined by $a_0=0$ and \[a_{n+1}=(a_n+1)k+(k+1)a_n+2\sqrt{k(k+1)a_n(a_n+1)}, \qquad n=0,1, \ldots.\]Prove that all $a_n$ are positive integers. Recall the identity $(a+b)^2=a^2+b^2+2ab.$ Then it is easy to see \begin{align*} a_{n+1}&=(a_n+1)k+(k+1)a_n+2\sqrt{k(k+1)a_n(a_n+1)} \\ &= (\sqrt{(a_n+1)k}+\sqrt{(k+1)a_n})^2. \end{align*}Thus, $\sqrt{a_{n+1}}=\sqrt{(a_n+1)k}+\sqrt{(k+1)a_n}.$ Consider $(a_n+1)k+(k+1)a_n+1=2a_nk+a_n+k+1.$ By Simon's Favorite Factoring Trick, $a_nk+a_n+k+1=(a_n+1)(k+1).$ Thus it follows $(a_n+1)k+(k+1)a_n+1=(a_n+1)(k+1)+a_nk.$ Therefore, applying the $(a+b)^2$ identity once again we obtain \begin{align*} a_{n+1}+1 &= (a_n+1)(k+1)+a_nk+2\sqrt{k(k+1)a_n(a_n+1)} \\ &= (\sqrt{(a_n+1)(k+1)}+\sqrt{a_nk})^2. \end{align*}Thus, $\sqrt{a_{n+1}+1} = \sqrt{(a_n+1)(k+1)} + \sqrt{a_nk}.$ Now consider $\sqrt{a_{n+1}}+\sqrt{a_{n+1}+1}.$ Note \begin{align*} \sqrt{a_{n+1}}+\sqrt{a_{n+1}+1} &= \sqrt{(a_n+1)k}+\sqrt{(k+1)a_n} + \sqrt{(a_n+1)(k+1)}+ \sqrt{a_nk} \\ &= \sqrt{a_n+1}\sqrt{k} + \sqrt{a_n+1}\sqrt{k+1} + \sqrt{k+1}\sqrt{a_n} + \sqrt{a_n}\sqrt{k} \\ &= \sqrt{a_n+1}(\sqrt{k} + \sqrt{k+1}) + \sqrt{a_n}(\sqrt{k}+\sqrt{k+1}) \\ &= (\sqrt{a_n+1}+\sqrt{a_n})(\sqrt{k}+\sqrt{k+1}). \end{align*}Notice the left factor of our product is precisely our original $\sqrt{a_{n+1}}+\sqrt{a_{n+1}+1}$ except the indices are shifted down by $1.$ Therefore through induction it follows that $\sqrt{a_{n+1}}+\sqrt{a_{n+1}+1} = (\sqrt{k}+\sqrt{k+1})^{n+1},$ so $\sqrt{a_n} + \sqrt{a_n+1} = (\sqrt{k} + \sqrt{k+1})^n.$ Now consider $\sqrt{a_{n+1}+1} - \sqrt{a_{n+1}}.$ Similarly, computing gives \begin{align*} \sqrt{a_{n+1}+1} - \sqrt{a_{n+1}} &= (\sqrt{(a_n+1)(k+1)} + \sqrt{a_nk}) - (\sqrt{(a_n+1)k} + \sqrt{(k+1)a_n}) \\ &= \sqrt{a_n+1}\sqrt{k+1}-\sqrt{a_n+1}\sqrt{k} + \sqrt{a_n}\sqrt{k}-\sqrt{k+1}\sqrt{a_n} \\ &= \sqrt{a_n+1}(\sqrt{k+1}-\sqrt{k}) + \sqrt{a_n}(\sqrt{k}-\sqrt{k+1}) \\ &= \sqrt{a_n+1}(\sqrt{k+1}-\sqrt{k}) - \sqrt{a_n}(\sqrt{k+1}-\sqrt{k}) \\ &= (\sqrt{a_n+1} - \sqrt{a_n})(\sqrt{k+1}-\sqrt{k}). \end{align*}Once again we can see the left factor of our product is simply our original $\sqrt{a_{n+1}+1}- \sqrt{a_{n+1}}$ but the indices are shifted down by $1.$ Through induction again we see $\sqrt{a_{n+1}+1} - \sqrt{a_{n+1}} = (\sqrt{k+1} - \sqrt{k})^{n+1},$ implying $\sqrt{a_n+1} - \sqrt{a_n} = (\sqrt{k+1}-\sqrt{k})^{n}.$ Now we have two equations where we can find a closed form for $\sqrt{a_n}.$ In particular, $\sqrt{a_n} + \sqrt{a_n+1} = (\sqrt{k} + \sqrt{k+1})^n$ and $\sqrt{a_n+1} - \sqrt{a_n} = (\sqrt{k+1}-\sqrt{k})^{n}.$ Subtracting the two equations, we obtain $2\sqrt{a_n}=(\sqrt{k}+\sqrt{k+1})^n - ( \sqrt{k+1} - \sqrt{k})^n.$ Dividing both sides by $2,$ we have $\sqrt{a_n} = \tfrac{1}{2}((\sqrt{k} + \sqrt{k+1})^n - (\sqrt{k+1} - \sqrt{k})^n).$ Square both sides of the equation and multiply both sides by $4$ to obtain: \begin{align*} 4a_n &=(\sqrt{k+1}+\sqrt{k})^{2n}-2(\sqrt{k+1}+\sqrt{k})^n(\sqrt{k+1}-\sqrt{k})^n+(\sqrt{k+1}-\sqrt{k})^{2n} \\ &=(\sqrt{k+1}+\sqrt{k})^{2n} - 2((\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k}))^n + (\sqrt{k+1}-\sqrt{k})^{2n} \\ &= (\sqrt{k+1}+\sqrt{k})^{2n} - 2((\sqrt{k+1})^2-\sqrt{k}^2)^n + (\sqrt{k+1}-\sqrt{k})^{2n} \\ &= (\sqrt{k+1}+\sqrt{k})^{2n} - 2(1)^n + (\sqrt{k+1}-\sqrt{k})^{2n} \\ &= (\sqrt{k+1}+\sqrt{k})^{2n} + (\sqrt{k+1}-\sqrt{k})^{2n} - 2. \end{align*}We now make use of the Binomial Theorem. Notice that terms such that $i,$ the ``dummy" variable, is odd will be cancelled out due to the $\sqrt{k}^i$ and $(-\sqrt{k})^i$ terms. Therefore, applying, we obtain \begin{align*} 4a_n &= (\sqrt{k+1}+\sqrt{k})^{2n} + (\sqrt{k+1}-\sqrt{k})^{2n} - 2\\ &=\sum_{i=0}^{2n} \binom{2n}{i} \sqrt{k+1}^{2n-i}\sqrt{k}^i + \sum_{i=0}^{2n} \binom{2n}{i} \sqrt{k+1}^{2n-i} (-\sqrt{k})^i -2 \\ &=2\left(\sqrt{k+1}^{2n} + \tbinom{2n}{2} \sqrt{k+1}^{2n-2}\sqrt{k}^2+\cdots + \sqrt{k}^{2n}\right)-2 \\ &=2\left((k+1)^n + \tbinom{2n}{2} (k+1)^{n-1}k+\cdots+ k^n\right)-2. \end{align*}From the second to third equation was when we cancelled out terms. Now note $k$ and $k+1$ are obviously opposite parity, meaning either $k$ is even, $k+1$ is odd, or $k$ is odd, $k+1$ is even. Then, it also follows $(k+1)^n$ and $k^n$ are opposite parity, and the rest of the terms in the expansion are even. This implies the expression $(k+1)^n+\tbinom{2n}{2}(k+1)^{n-1}k+\cdots+k^n$ is odd. It readily follows $4a_n=2((k+1)^n+\tbinom{2n}{2}(k+1)^{n-1}k+\cdots+k^n)-2 \equiv 0 \pmod{4},$ and therefore since $k$ is a positive integer and our sequence is defined for $n=0,1, \ldots,$ we know all $a_n$ are positive integers as well, as desired.

The key is to notice that $a_{n+1} + 1$ is also a perfect square. In particular, set $x_n = \sqrt{a_n}$ and $y_n = \sqrt{a_n+1}$, such that \begin{align*} x_{n+1} &= \sqrt a y_n + \sqrt{a+1} x_n \\ y_{n+1} &= \sqrt a x_n + \sqrt{a+1} y_n. \end{align*}Then,$$x_{n+1} + y_{n+1} = (\sqrt a + \sqrt{a+1})^n.$$It is easy to show (inductively, or by theory of Pell equations) that$$(\sqrt a + \sqrt{a+1})^n = \sqrt k + \sqrt{k+1}$$for some positive integer $k$, so this implies $a_n = k$ always satisfies the recursion, as needed.