Problem: Determine whether or not there exists an infinite sequence
$a_1$, $a_2$, \dots\ of positive integers satisfying
\[ a_{n+2} = a_{n+1} + \sqrt{a_{n+1}+a_{n}} \]
for every positive integer $n$.

Solution: In fact, we will show the following stronger result:
the largest $N$ for which one can find $(a_1, \dots, a_N)$ satisfying
(for all $1 \le n \le N-2$) is actually $N = 5$.
This largest $N$ is obtained
for example by $(a_1,a_2,a_3,a_4,a_5) = (477,7,29,35,43)$.
\begin{remark*}
  Basically, the idea is to choose $a_3$ first; then as long as the number
  \[ s \coloneq \sqrt{a_3 + a_4} = \sqrt{2a_3 + \sqrt{a_2 + a_3}} \]
  is as integer, the rest of the sequence can be chosen to have integer values.
  Unfortunately, $a_1$ may turn out to be negative in this situation.
  But if one experiments with numbers, it should be possible to ensure $a_2 < a_3$,
  and after this no further obstructions arise.

  For example, if one makes the arbitrary starting choice $s = 10$,
  then choosing $a_3 = 46$ gives the nice choice $a_2 = 18$, thus $a_1 = 766$.
  (We picked this number so that $2a_3 + \sqrt{a_3}$ was just under $100$.)
  Meanwhile, moving forward, $a_4 = 46 + \sqrt{46+18} = 54$
  and $a_5 = 54 + \sqrt{46+54} = 64$.
  Hence $(a_1,a_2,a_3,a_4,a_5) = (766, 18, 46, 54, 64)$ is another example.
\end{remark*}

Let
\[ x_n \coloneq a_{n+1}-a_n = \sqrt{a_{n}+a_{n-1}}. \]
We will rewrite everything in terms of the $(x_n)$.
Since the $(a_n)$ are strictly increasing for $n \ge 2$
so are the $(x_n)$ when $n \ge 3$.
For $n \ge 2$ observe that
\[ x_{n+1}^2 - x_n^2 = a_{n+1} - a_{n-1}
  = \left( a_{n+1}-a_{n} \right) + \left( a_{n}-a_{n-1} \right) = x_n + x_{n-1} \]
thus
\[ x_{n+1}-x_n = \frac{x_n+x_{n-1}}{x_{n+1}+x_n}. \]
But suppose $n \ge 4$.
Then since the $x_n$ are supposed to be strictly increasing, the right-hand side is $< 1$.
Yet $x_{n+1} - x_n \ge 1$ as well, which is a contradiction.

Thus it is impossible to have six terms in the sequence.